Mixing
Visually stunning math concepts which are easy to explain
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Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.
Do you know of any other concepts like these?
soft-question education big-list visualization
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Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.
Do you know of any other concepts like these?
soft-question education big-list visualization
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see also, fractals
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– Guy
Mar 31 '14 at 11:55
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It looks like mathpop or demand for math entertainment)
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– rook
Apr 2 '14 at 12:59
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@ColeJohnson the 'transcendentality" is the beauty of it!
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– Guy
Apr 4 '14 at 17:07
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There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words
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– Martin Brandenburg
Jun 20 '14 at 22:13
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@TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site.
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– user147263
Aug 29 '14 at 18:37
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show 7 more comments
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Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.
Do you know of any other concepts like these?
soft-question education big-list visualization
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Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.
Do you know of any other concepts like these?
soft-question education big-list visualization
soft-question education big-list visualization
edited Mar 8 at 13:51
community wiki
13 revs, 8 users 33%
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see also, fractals
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– Guy
Mar 31 '14 at 11:55
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It looks like mathpop or demand for math entertainment)
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– rook
Apr 2 '14 at 12:59
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@ColeJohnson the 'transcendentality" is the beauty of it!
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– Guy
Apr 4 '14 at 17:07
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There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words
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– Martin Brandenburg
Jun 20 '14 at 22:13
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@TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site.
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– user147263
Aug 29 '14 at 18:37
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show 7 more comments
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see also, fractals
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– Guy
Mar 31 '14 at 11:55
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It looks like mathpop or demand for math entertainment)
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– rook
Apr 2 '14 at 12:59
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@ColeJohnson the 'transcendentality" is the beauty of it!
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– Guy
Apr 4 '14 at 17:07
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There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words
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– Martin Brandenburg
Jun 20 '14 at 22:13
7
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@TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site.
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– user147263
Aug 29 '14 at 18:37
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34
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see also, fractals
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– Guy
Mar 31 '14 at 11:55
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see also, fractals
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– Guy
Mar 31 '14 at 11:55
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13
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It looks like mathpop or demand for math entertainment)
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– rook
Apr 2 '14 at 12:59
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It looks like mathpop or demand for math entertainment)
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– rook
Apr 2 '14 at 12:59
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10
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@ColeJohnson the 'transcendentality" is the beauty of it!
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– Guy
Apr 4 '14 at 17:07
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@ColeJohnson the 'transcendentality" is the beauty of it!
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– Guy
Apr 4 '14 at 17:07
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6
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There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words
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– Martin Brandenburg
Jun 20 '14 at 22:13
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There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words
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– Martin Brandenburg
Jun 20 '14 at 22:13
7
7
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@TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site.
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– user147263
Aug 29 '14 at 18:37
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@TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site.
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– user147263
Aug 29 '14 at 18:37
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66 Answers
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I think if you look at this animation and think about it long enough, you'll understand:
- Why circles and right-angle triangles and angles are all related
- Why sine is opposite over hypotenuse and so on
- Why cosine is simply sine but offset by $pi/2$ radians
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@joeA: it's not as smooth as regenerating at a higher framerate, but gfycat.com allows one to view gifs at different speeds: gfycat.com/TintedWatchfulAxisdeer#?speed=0.25
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– Max
Apr 2 '14 at 20:51
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If high school math just used a fraction of the resources here, we'd have way more mathematicians.
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– user148298
Apr 3 '14 at 17:00
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Is the source of this animation available? (It looks like it's in $mathrmTeX$.)
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– bb010g
Apr 3 '14 at 23:31
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This is the normal way of introducing sine and cosine at high schools. At least in our country.
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– Vladimir F
Apr 7 '14 at 15:36
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@joeA: I've made a programmatic version of this on Khan Academy; feel free to change the speed to your liking: khanacademy.org/cs/circle/4597064320155648
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– BHSPitMonkey
Apr 16 '14 at 10:48
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My favorite: tell someone that $$sum_{n=1}^{infty}frac{1}{2^n}=1$$ and they probably won't believe you. However, show them the below:
and suddenly what had been obscure is now obvious.
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My first intuitive visualization of this sum was a circle. I wasn't $100%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :)
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– Guy
Mar 31 '14 at 17:38
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I still don't believe you.
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– Ojonugwa Ochalifu
Apr 2 '14 at 8:51
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Another way to think of this is that 1/2 + 1/4 + 1/8 = 0.111...binary = 0.999...decimal = 1.
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– Justin L.
Apr 7 '14 at 3:02
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You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1.
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– Maxim Umansky
Apr 7 '14 at 4:05
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@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $frac{1}{2}$ and $frac{1}{4}$ in their "standard shape", so what remains must obviously be $frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $frac{1}{2}cdotfrac{1}{4}=frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger).
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– nobody
Apr 9 '14 at 20:09
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This visualisation of the Fourier Transform was very enlightening for me:
The author, LucasVB, has a whole gallery of similar visualisations at his Wikipedia gallery and his tumblr blog.
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@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave.
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– Joel
Apr 7 '14 at 2:15
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I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets).
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– Mikhail
Apr 7 '14 at 3:19
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this is pretty cool too: bl.ocks.org/jinroh/7524988
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– ilia choly
Apr 7 '14 at 20:54
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I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works...
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– Cole Johnson
Apr 11 '14 at 1:21
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I don't find this graphical example helpful in explaining the Fourier Transform. If anything, it is going to add to existing confusion people have with the Fourier Series. The sin and cos expression shown is too general to be meaningful, and the graphic shown for the "transform" is really just a visualization of the magnitudes of the Fourier Series coefficients you would have if the partial sum function were presumed to be infinitely periodic. When I calculate the actual Fourier Transform of the segment of function shown, I get something quite different: imgur.com/eQuerJa
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– Adam Smith
Aug 31 '14 at 18:53
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Here is a classic: the sum of the first $n$ positive odd numbers $= n^2$.
We also see that the sum of the first $n$ positive even numbers $= n(n+1)$ (excluding $0$), by adding a column to the left.
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If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$
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– becko
Apr 3 '14 at 16:29
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I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this.
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– Nick T
Apr 3 '14 at 17:37
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I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit"
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– Guy
Apr 6 '14 at 9:40
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I have noticed an interesting method to calculate a square root based on this fact: disqus.com/home/discussion/geeksforgeeks/…
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– sergiol
Sep 7 '15 at 23:46
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A well-known visual to explain $(a+b)^2 = a^2+2ab+b^2$:
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Always liked this one
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– Almo
Apr 2 '14 at 14:11
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why area of a^2 is greater tham area of b^2?
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– jaczjill
Apr 3 '14 at 7:34
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@jaczjill Because a and b are not the same number.
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– Conor Pender
Apr 3 '14 at 11:24
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@jaczjill the areas of a and b are arbitrary. They could be any size and it would make no difference. It's just an example.
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– theyetiman
Apr 4 '14 at 8:29
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Have you seen the Montessori binomial cube?
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– MJD
Apr 4 '14 at 14:30
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The sum of the exterior angles of any convex polygon will always add up to $360^circ$.
This can be viewed as a zooming out process, as illustrate by the animation below:
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I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take).
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– daviewales
Apr 5 '14 at 12:27
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I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance.
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– Steven Gubkin
Apr 5 '14 at 14:10
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This would be better as an animation (which achieves the zooming effect).
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– Duncan Jones
Apr 7 '14 at 10:31
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Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.)
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– Yiyuan Lee
Apr 7 '14 at 13:36
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this is really,really clever
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– Konstantinos Gaitanas
Aug 3 '17 at 9:07
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While attending an Abstract Algebra course I was given the assignment to write out the multiplication table modulo n. I forgot to do the homework until just before class, but it was so easy to write the program I was able to print the result between classes.
The circular patterns in the tables fascinated me, and compelled me to replace the numbers with colors. The result is a beautiful illustration showing the emergence of primes and symmetry of multiplication.
The colors were chosen to start blue at 1 (cold) and fade to red at n (hot). White is used for zero (frozen), because it communicates the most information about prime factorization.
The interactive version can be found here:
https://web.archive.org/web/20140830110358/http://arapaho.nsuok.edu/~deckar01/Zvis.html
Multiplication of the integers modulo 15:
Multiplication of the integers modulo 512:
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This is awesome. Looks very much like a fractal. Is there a name for this fractal?
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– OmnipotentEntity
Apr 7 '14 at 15:52
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What in specific is it lacking? Correct me if I'm wrong, but if we take the white dots to be the area of the fractal and the red and blue to be the border then it's an object of fractional dimension (either 0d points or 1d lines filling a 2d plane) and self-symmetric.
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– OmnipotentEntity
Apr 7 '14 at 17:42
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@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime.
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– John Dvorak
Apr 7 '14 at 18:48
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I just read about qualitative self-similarity which I think describes the relationship between frames in this example, but it isn't continuous in any dimension.
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– Jared Deckard
Apr 7 '14 at 18:51
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Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200
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– Vikram Saraph
Apr 9 '14 at 23:05
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When I understood Fourier series visually-
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OH. Wow. I just took my signals course and even after I did well on the exam I still didn't actually UNDERSTAND fourier series. Thanks, wow.
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– VicVu
Apr 2 '14 at 22:25
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This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice!
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– Steven Stadnicki
Apr 4 '14 at 6:09
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I don't understand what this is showing: is it showing hte approximation of a square wave by a Fourier series?
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– ShreevatsaR
Apr 5 '14 at 5:18
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@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon.
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– Potato
Apr 5 '14 at 6:01
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@ShreevatsaR: I evaluate the limiting amount of overshoot in the Gibbs Phenomenon in this answer (about $8.9%$ of the size of the discontinuity).
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– robjohn♦
Apr 6 '14 at 12:03
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Simple answer for "what is a radian":
Logarithmic spiral and scale:
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What program produces these images?
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– rschwieb
Aug 11 '14 at 19:49
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@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq
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– Alice Ryhl
Aug 13 '14 at 14:40
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I can't believe how long I used radians in college without realizing what a radian really is.
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– greg7gkb
Jan 9 '15 at 0:28
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The exponential curve has curvature-arc length relation like cycloidal? What does the animation show?
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– Narasimham
Apr 9 '16 at 2:31
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With what software can you make those graphics? They are good.
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– Darío A. Gutiérrez
Mar 31 '17 at 11:18
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This is a neat little proof that the area of a circle is $pi r^2$, which I was first taught aged about 12 and it has stuck with me ever since. The circle is subdivided into equal pieces, then rearranged. As the number of pieces gets larger, the resulting shape gets closer and closer to a rectangle. It is obvious that the short side of this rectangle has length $r$, and a little thought will show that the two long sides each have a length half the circumference, or $pi r$, giving an area for the rectangle of $pi r^2$.
This can also be done physically by taking a paper circle and actually cutting it up and rearranging the pieces. This exercise also offers some introduction to (infinite) sequences.
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This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to $pi r$.
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– Nicholas Wilson
Apr 8 '14 at 16:50
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@NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help?
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– ghoppe
Apr 8 '14 at 19:29
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@ghoppe Nice try! But it's all wiggly. We're lucky that $sin{theta} sim theta$ as $theta rightarrow 0$, so the actual perimeter of the bottom (which is known to be $pi$) does converge to the width of the rectangle. But - that nice property of $sin$ is exactly what we're trying to prove!
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– Nicholas Wilson
Apr 9 '14 at 8:20
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@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof.
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– Nicholas Wilson
Apr 9 '14 at 17:56
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@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent.
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– Ben Rowland
Dec 16 '14 at 22:16
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Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this.
It can be explained as follows. We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ Since distance satisfies condition 5, for any right angle triangle at all, not just ones whose legs are parallel to the axes, the square of the length of its hypotenuse is equal to the sum of the squares of the lengths of its legs.
This image shows that using that definition of distance, for any right angle triangle whose legs are parallel to the axes and have lengths $x in mathbb{R}^+$ and $y in mathbb{R}^+$, the area of a square with the hypotenuse as one of its edges is $(x - y)^2 + 2xy = x^2 + y^2 = (d(x, y))^2$. Combining that result with the fact that distance satisfies condition 5, we can show that for any right angle triangle, even with legs non parallel to the axes, the area of a square with its hypotenuse as its edge has an area equal to the sum of the squares of the lengths of its legs.
Sources:
- The validity of the proofs of the Pythagorean Theorem and the concept of area
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
- https://thept.weebly.com/the-theorem.html
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This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple.
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– Sparr
Apr 7 '14 at 16:06
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Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote.
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– B T
Aug 27 '14 at 21:12
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@BT, do you consider the fact that $1^3 + 2^3 neq 18^3$ a demonstration of Fermat's Last Theorem?
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– Daniel McLaury
Aug 28 '14 at 1:17
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@DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool.
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– Samuel Yusim
Aug 28 '14 at 5:30
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@GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice.
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– Samuel Yusim
Sep 5 '14 at 0:06
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A visual explanation of a Taylor series:
$f(0)+frac {f'(0)}{1!} x+ frac{f''(0)}{2!} x^2+frac{f^{(3)}(0)}{3!}x^3+ cdots$
or
$f(a)+frac {f'(a)}{1!} (x-a)+ frac{f''(a)}{2!} (x-a)^2+frac{f^{(3)}(a)}{3!}(x-a)^3+ cdots$
When you think about it, it's quite beautiful that as you add each term it wraps around the curve.
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It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have acosinefunction built-in, but will fly through an equation such asx => 1 - (x^2)/(2!) + (x^4)/(4!) - ...and from this equation, you could build a function that accepts aprecisionparameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this)
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– sleblanc
Apr 3 '14 at 21:55
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@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally
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– joshlk
Apr 4 '14 at 8:15
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@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas.
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– Cole Johnson
Apr 4 '14 at 16:05
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You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have theFSINinstruction... If anything, I should have said [most] instead. ;-)
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– sleblanc
Apr 12 '14 at 14:31
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@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition.
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– Calmarius
Aug 27 '14 at 16:57
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When I look up "area of a rhombus" on Google images, I find plenty of disappointing images like this one:
which show the formula, but fail to show why the formula works. That's why I really appreciate this image instead:
which, with a little bit of careful thought, illustrates why the product of the diagonals equals twice the area of the rhombus.
EDIT: Some have mentioned in comments that that second diagram is more complicated than it needs to be. Something like this would work as well:
My main objective is to offer students something that encourages them to think about why a formula works, not just what numbers to plug into an equation to get an answer.
As a side note, the following story is not exactly "visually stunning," but it put an indelible imprint on my mind, and affected the way I teach today. A very gifted Jr. High math teacher was teaching us about volume. I suppose just every about school system has a place in the curriculum where students are required learn how to calculate the volume of a pyramid. Sadly, most teachers probably accomplish this by simply writing the formula on the board, and assigning a few plug-and-chug homework problems.
No wonder that, when I ask my college students if they can tell me the formula for the volume of a pyramid, fewer than 5% can.
Instead, building upon lessons from earlier that week, our math teacher began the lesson by saying:
We've learned how to calculate the volume of a prism: we simply multiply the area of the base times the height. That's easy. But what if we don't have a prism? What if we have a pyramid?
At this point, she rummaged through her box of math props, and pulled out a clear plastic cube, and a clear plastic pyramid. She continued by putting the pyramid atop the cube, and then dropping the pyramid, point-side down inside the cube:
She continued:
These have the same base, and they are the same height. How many of these pyramids do you suppose would fit in this cube? Two? Two-and-a-half? Three?
Then she picked one student from the front row, and instructed him to walk them down the hallway:
Go down to the water fountain, and fill this pyramid up with water, and tell us how many it takes to fill up the cube.
The class sat in silence for about a full minute or so, until he walked back in the room. She asked him to give his report.
"Three," he said.
She pressed him, giving him a hard look. "Exactly three?"
"Exactly three," he affirmed.
Then, she looked around the room:
"Who here can tell me the formula I use to get the volume of a pyramid?" she asked.
One girl raised her hand: "One-third the base times the height?"
I've never forgotten that formula, because, instead of having it told to us, we were asked to derive it. Not only have I remembered the formula, I can even tell you the name of boy who went to the water fountain, and the girl who told us all the formula (David and Jill).
Given the upvoted comment, If high school math just used a fraction of the resources here, we'd have way more mathematicians, I hope you don't mind me sharing this story here. Powerful visuals can happen even in the imagination. I never got to see that cube filling up with water, but everything else in the story I vividly remember.
Incidentally, this same teacher introduced us to the concept of pi by asking us to find something circular in our house (“like a plate or a coffee can”), measuring the circumference and the diameter, and dividing the one number by the other. I can still see her studying the data on the chalkboard the next day – all 20 or so numbers just a smidgeon over 3 – marveling how, even though we all probably measured differently-sized circles, the answers were coming out remarkably similar, “as if maybe that ratio is some kind of constant or something...”
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^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways.
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– daviewales
Apr 5 '14 at 12:34
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@daviewales It's been started - the mathematics educators SE.
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– user122283
Apr 5 '14 at 14:45
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Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$frac{1}{2}frac{d_1}{2}d_2=frac{1}{2}d_1frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.)
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– Will Orrick
Apr 6 '14 at 2:36
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The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite.
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– LaC
Apr 7 '14 at 3:47
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@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High.
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– J.R.
Feb 22 '15 at 9:44
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As I was in school, a supply teacher brought a scale to lesson:
He gave us several weights that were labeled and about 4 weights without labels (let's call them $A, B, C, D$). Then he told us we should find out the weight of the unlabeled weights. $A$ was very easy as there was a weight $E$ with weight($A$) = weight($E$).
I think at least two of them had the same weight and we could only get them into balance with a combination of the labeled weights. The last one was harder. We had to put a labeled weight on the side of the last one to get the weight.
Then he told us how this can be solved on paper without having the weights. So he introduced us to the concept of equations. That was a truly amazing day. Such an important concept explained with a neat way.
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I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set.
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– Guillermo Garza
Apr 7 '14 at 0:34
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@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 cdot m_1 = a_2 cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way.
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– Martin Thoma
Apr 7 '14 at 5:58
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I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind!
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– Guillermo Garza
Apr 8 '14 at 0:50
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@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics.
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– Martin Thoma
Apr 8 '14 at 1:59
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This is similar to Aky's answer, but includes a second drawing (and no math.)
To me the second drawing is key to understanding why the $mathrm c^2$ area is equal to the sum of $mathrm a^2+mathrm b^2$.
Edit: comments requested an animation, so a simple gif is attached...
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I've always felt this one worked better in .gif format.
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– Patrick M
Apr 3 '14 at 20:47
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This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles.
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– Nick Retallack
Apr 7 '14 at 0:31
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I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size.
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– Sparr
Apr 7 '14 at 15:58
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@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it.
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– Tobia
Apr 7 '14 at 16:33
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Anyway...THIS was my favourite, so, good that you already did it!
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– MattAllegro
May 12 '14 at 21:49
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How about a line integral of a scalar field by http://1ucasvb.tumblr.com:
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Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;)
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– Luke
Aug 27 '14 at 17:44
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Nice visual intuition, but why is it true that $|mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step)
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– enthdegree
Aug 11 '16 at 5:18
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Isn't this how students usually learn line integrals?
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– user3932000
Mar 9 '17 at 1:50
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@enthdegree the integral is with respect to arc length ($ds$= arc length). $$int_C f(x,y)ds=int_{a}^{b}f(x(t),y(t))sqrt{x'(t)^2+y'(t)^2}dt=int_{a}^{b}f(r(t))|r'(t)|dt$$
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– clathratus
Oct 6 '18 at 1:54
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It's just "u-substitution."
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– Trevor Kafka
Jan 28 at 4:12
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Similarly to eykanal's answer, although demonstrating some interesting facts about medians and geometry as well. It demonstrates that $displaystylesum_{n = 1}^{infty}frac{1}{2^n} = 1$:
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I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut.
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– abnry
Apr 5 '14 at 13:55
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Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $frac{1}{2}$ according to you has actually an area of $frac{1}{2}timesfrac{sqrt{2}}{2}timesfrac{sqrt{2}}{2}=frac{1}{4}$, so what this figure is proving is that $sum_{n=2}^{infty}frac{1}{2^n}=frac{1}{2}$.
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– S4M
Apr 7 '14 at 7:12
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@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition?
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– Andrew Coonce
Apr 7 '14 at 16:51
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@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $frac{1}{4}$ triangle bisects the triangle of area $frac{1}{2}$ that is left after the $frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1.
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– user121055
Apr 8 '14 at 21:17
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@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $frac{sqrt{2}}{2}$.
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– S4M
Apr 9 '14 at 12:27
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This animation shows that a circle's perimeter equals to $2r*pi$. As ShreevatsaR pointed out, this is obvious because $pi$ is by definition the ratio of a circle's circumference to its diameter
In this image we can see how the ratio is calculated. The wheel's diameter is 1. After the perimeter is rolled down we can see that its length equals to $pi$ amount of wheels.
Source
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But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = frac{4}{3} pi r^3$ and a surface area of $SA_3(r) = 4 pi r^2$
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– Cole Johnson
Apr 4 '14 at 16:09
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@ColeJohnson I don't know, just wanted to show this awesome animation:D
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– totymedli
Apr 4 '14 at 21:11
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I just wish someone could make an animation of that. I understand the math behind it, but animations are nice.
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– Cole Johnson
Apr 4 '14 at 22:28
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That the circle's perimeter is $2pi r$ is the definition of $pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $pi$ is about $3.1$.
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– ShreevatsaR
Apr 5 '14 at 5:24
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@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there.
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– Akiva Weinberger
Feb 23 '15 at 1:45
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Steven Wittens presents quite a few math concepts in his talk Making things with math. His slides can be found from his own website.
For example, Bézier curves visually:
He has also created MathBox.js which powers his amazing visualisations in the slides.
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These are beautiful!
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– Cole Johnson
May 15 '14 at 23:16
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Also, if you read the text out loud you get to say "lerp lerp lerp lerp", which has its own aesthetic appeal.
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– msouth
Sep 23 '14 at 21:43
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beautiful, but I don't understand XD
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– Kokizzu
Apr 20 '15 at 11:20
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See also codegolf.stackexchange.com/questions/21178/…
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– SztupY
Jun 21 '16 at 9:24
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Fractal art. Here's an example: "Mandelbrot Island".
The real island of Sark in the (English) Channel Islands looks astonishingly like Mandelbrot island:
Now that I think about it, fractals in general are quite beautiful. Here's a close-up of the Mandelbrot set:
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Are complex numbers easy to explain?
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– adam.r
Mar 31 '14 at 17:15
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Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think.
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– user134824
Mar 31 '14 at 17:16
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Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large).
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– Kaz
Apr 1 '14 at 4:57
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@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list.
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– Szabolcs
Apr 1 '14 at 16:12
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It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example.
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– user134824
Apr 3 '14 at 3:12
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The magnetic pendulum:
An iron pendulum is suspended above a flat surface, with three magnets on it. The magnets are colored red, yellow and blue.
We hold the pendulum above a random point of the surface and let it go, holding our finger on the starting point. After some swinging this way and that, under the attractions of the magnets and gravity, it will come to rest over one of the magnets. We color the starting point (under our finger) with the color of the magnet.
Repeating this for every point on the surface, we get the image shown above.
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How is the shading of the image given?
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– David Zhang
Apr 21 '14 at 14:27
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@DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm.
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– Peter
Apr 21 '14 at 15:27
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Deterministic chaos?
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– Peter Mortensen
Mar 8 '15 at 11:51
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@PeterMortensen Absolutely. In fact, you can see how (in certain areas) a minute change in initial conditions can cause a drastic change in outcome.
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– Peter
Mar 9 '15 at 12:36
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Simple,visual proof of the Pythagorean theorem. Originally from Pythagorean Theorem Proof Without Words 6).
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Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $;b;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1
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– DonAntonio
Mar 31 '14 at 12:31
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Another great pictural proof is 4 triangles in a square: mathalino.com/sites/default/files/images/01-pythagora.jpg
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– Sergey Grinev
Mar 31 '14 at 15:10
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I don't get the original equality. Are you relying on the similarity of two triangles? How is that obvious from the diagram?
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– adam.r
Mar 31 '14 at 17:10
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It's not immediately obvious to me why $frac{c+a}{b} = frac{b}{c-a}$ I'm sure it's very simple and I'll kick myself for asking, sorry.
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– PeteUK
Mar 31 '14 at 17:12
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Well, I looked at that picture and thought: "Err ... how does this work?" I figured it out, but it was far from obvious for me.
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– celtschk
Apr 3 '14 at 18:33
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Topology needs to be represented here, specifically knot theory. The following picture is from the Wikipedia page about Seifert Surfaces and was contributed by Accelerometer. Every link (or knot) is the boundary of a smooth orientable surface in 3D-space. This fact is attributed to Herbert Seifert, since he was the first to give an algorithm for constructing them. The surface we are looking at is bounded by Borromean rings.
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I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question.
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– dfc
Apr 7 '14 at 5:25
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@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles.
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– Slade
Apr 7 '14 at 20:10
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Here's a GIF that I made that demonstrates Phi (golden number)
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That's Beautiful!
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– debap
Sep 4 '16 at 12:36
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Take a look at this great example of Fourier series visualisations written in JavaScript.
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Really amazing....
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– Pacerier
Apr 11 '14 at 16:49
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What is this doing lurking at the bottom of the page? It's brilliant!
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– P i
Feb 25 '15 at 19:39
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The link is brilliant!
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– David Simmons
Jun 14 '15 at 9:50
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If you had captured an animated gif, this would be the top answer!
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– Tobia
Jul 10 '15 at 0:14
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Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall, for different choices of parameter:
The best thing about them is, they satisfy periodic boundary conditions, and so you can pick one of them and set it as a desktop background by tiling it, resulting in a far more spectacular image than just the single unit cells posted above!
The images seem to be a vast interconnected network of lines once you tile them, but in fact the entire picture is actually just a single circle, which has been aliased into a tiling cell thousands of times.
Here is a video of the first couple thosand patterns:
http://www.youtube.com/watch?v=1UVbUWuyNmk
Here is the Mathematica code used to generate and save the images. There are two parameters that are adjustable: mag is the magnification and must be an integer, with 1 generating 600 by 600 images, 2 generating 1200 by 1200 images, etc. i is a parameter which can be any real number between 0 and ~1000, with values between 0 and 500 being typical (most of the preceding images used i values between 200 and 300). By varying i, thousands of unique diagrams can be created. Small values of i create simple patterns (low degree of aliasing), and large values generate complex patterns (high degree of aliasing).
$HistoryLength = 0;
p = {x, y, L};
nnn = 8;
q = 2.0 Table[{Cos[2 [Pi] j/nnn], Sin[2 [Pi] j/nnn], 0}, {j, nnn}];
k = ConstantArray[I, nnn];
n[x_] := Sqrt[x.x];
conjugate[expr_] := expr /. Complex[x_, y_] -> x - I y;
a = Table[k[[i]]/n[p - q[[i]]], {i, nnn}];
[Gamma] = Table[Exp[-I [Omega] n[p - q[[i]]]/c], {i, nnn}];
expr = [Gamma].a /. {L -> 0.1, c -> 1, [Omega] -> 100};
ff = Compile[{{x, _Real}, {y, _Real}}, Evaluate[expr],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
i = 250;
mag = 1;
d = 6 i mag;
[Delta] = 0.02 i;
nn = Floor[Length[Range[-d, d, [Delta]]]/2];
A = Compile[{{x, _Integer}, {y, _Integer}}, Exp[I (x + y)],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}] @@
Transpose[
Outer[List, Range[Length[Range[-d, d, [Delta]]]],
Range[Length[Range[-d, d, [Delta]]]]], {2, 3, 1}];
SaveImage =
Export[CharacterRange["a", "z"][[RandomInteger[{1, 26}, 20]]] <>
".PNG", #] &;
{#, SaveImage@#} &@
Image[RotateRight[
Abs[Fourier[
1 A mag i/
nnn ff @@
Transpose[
Outer[List, Range[-d, d, [Delta]],
Range[-d, d, [Delta]]], {2, 3, 1}]]], {nn, nn}],
Magnification -> 1]
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Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me.
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– gregschlom
Apr 7 '14 at 9:11
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"Visually stunning math concepts which are easy to explain"
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– Rahul
Apr 8 '14 at 1:06
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@Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed.
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– DumpsterDoofus
Apr 8 '14 at 1:17
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This reminds me of the Radon transform and the Hough transform.
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– koo
Apr 8 '14 at 23:59
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Could you post a link to the code here?
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– Catherine Hwang
Apr 9 '14 at 18:46
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A very satisfying visualization of the area of a circle.
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I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious.
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– Lqueryvg
Sep 1 '14 at 10:52
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Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :)
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– msouth
Sep 26 '14 at 0:27
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The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD
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– The Great Duck
Apr 12 '16 at 18:20
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Math is always fun to learn. Here are some of the images that explain some things beautifully visually
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In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes.
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– David K
Aug 28 '14 at 21:25
1
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@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^circ$.
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– Mario Carneiro
Jan 9 '15 at 1:42
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@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all.
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– David K
Jan 9 '15 at 2:56
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When I read the number142857it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!"
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– Qwerty
Jan 11 '15 at 4:20
2
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@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 cdot 12$ and see how it corresponds to the diagram.)
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– Martin Ender
Mar 16 '16 at 15:57
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show 3 more comments
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Francis Galton's Bean machine is interesting as it demonstrates Central Limit Theorem :
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1
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It's one of the best answers.
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– Seifolahi
Mar 16 '16 at 3:42
12
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it probably demonstrates it...
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– enthdegree
Aug 11 '16 at 5:41
add a comment |
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I do not know if this meets your criteria of "visually stunning", but nonetheless -
I like this proof of Pythagoras' Theorem (image taken from www.wisfaq.nl):
The key to understanding this is to realize that the inner quadrilateral must be a square - the sides are equal in length (obviously) and each of its angles is $90^{circ}$ because the two angles on either side sum to $90^{circ}$, and the sum of the three angles is $180^{circ}$. The area of this square is $c^2$.
The outer square's area is $(a + b)^2$, which is $c^2$ plus $2 a b$ which is the total area of the four triangles, each of area $frac{1}{2} a b$.
$(a + b)^2 = c^2 + 2 a b$
$a^2 + b^2 + 2 a b = c^2 + 2 a b$
$a^2 + b^2 = c^2$, which is Pythagoras' theorem.
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1
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+1 finally i can prove Pythagoras theorem without looking into internet or books :)
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– lowtech
Jul 31 '14 at 20:58
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yes it is the most common proof, the easiest to remember
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– reuns
Apr 22 '16 at 21:13
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I think if you look at this animation and think about it long enough, you'll understand:
- Why circles and right-angle triangles and angles are all related
- Why sine is opposite over hypotenuse and so on
- Why cosine is simply sine but offset by $pi/2$ radians
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13
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@joeA: it's not as smooth as regenerating at a higher framerate, but gfycat.com allows one to view gifs at different speeds: gfycat.com/TintedWatchfulAxisdeer#?speed=0.25
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– Max
Apr 2 '14 at 20:51
197
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If high school math just used a fraction of the resources here, we'd have way more mathematicians.
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– user148298
Apr 3 '14 at 17:00
22
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Is the source of this animation available? (It looks like it's in $mathrmTeX$.)
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– bb010g
Apr 3 '14 at 23:31
12
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This is the normal way of introducing sine and cosine at high schools. At least in our country.
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– Vladimir F
Apr 7 '14 at 15:36
12
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@joeA: I've made a programmatic version of this on Khan Academy; feel free to change the speed to your liking: khanacademy.org/cs/circle/4597064320155648
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– BHSPitMonkey
Apr 16 '14 at 10:48
|
show 14 more comments
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I think if you look at this animation and think about it long enough, you'll understand:
- Why circles and right-angle triangles and angles are all related
- Why sine is opposite over hypotenuse and so on
- Why cosine is simply sine but offset by $pi/2$ radians
$endgroup$
13
$begingroup$
@joeA: it's not as smooth as regenerating at a higher framerate, but gfycat.com allows one to view gifs at different speeds: gfycat.com/TintedWatchfulAxisdeer#?speed=0.25
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– Max
Apr 2 '14 at 20:51
197
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If high school math just used a fraction of the resources here, we'd have way more mathematicians.
$endgroup$
– user148298
Apr 3 '14 at 17:00
22
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Is the source of this animation available? (It looks like it's in $mathrmTeX$.)
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– bb010g
Apr 3 '14 at 23:31
12
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This is the normal way of introducing sine and cosine at high schools. At least in our country.
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– Vladimir F
Apr 7 '14 at 15:36
12
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@joeA: I've made a programmatic version of this on Khan Academy; feel free to change the speed to your liking: khanacademy.org/cs/circle/4597064320155648
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– BHSPitMonkey
Apr 16 '14 at 10:48
|
show 14 more comments
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I think if you look at this animation and think about it long enough, you'll understand:
- Why circles and right-angle triangles and angles are all related
- Why sine is opposite over hypotenuse and so on
- Why cosine is simply sine but offset by $pi/2$ radians
$endgroup$
I think if you look at this animation and think about it long enough, you'll understand:
- Why circles and right-angle triangles and angles are all related
- Why sine is opposite over hypotenuse and so on
- Why cosine is simply sine but offset by $pi/2$ radians
edited May 26 '14 at 5:47
community wiki
2 revs, 2 users 95%
congusbongus
13
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@joeA: it's not as smooth as regenerating at a higher framerate, but gfycat.com allows one to view gifs at different speeds: gfycat.com/TintedWatchfulAxisdeer#?speed=0.25
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– Max
Apr 2 '14 at 20:51
197
$begingroup$
If high school math just used a fraction of the resources here, we'd have way more mathematicians.
$endgroup$
– user148298
Apr 3 '14 at 17:00
22
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Is the source of this animation available? (It looks like it's in $mathrmTeX$.)
$endgroup$
– bb010g
Apr 3 '14 at 23:31
12
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This is the normal way of introducing sine and cosine at high schools. At least in our country.
$endgroup$
– Vladimir F
Apr 7 '14 at 15:36
12
$begingroup$
@joeA: I've made a programmatic version of this on Khan Academy; feel free to change the speed to your liking: khanacademy.org/cs/circle/4597064320155648
$endgroup$
– BHSPitMonkey
Apr 16 '14 at 10:48
|
show 14 more comments
13
$begingroup$
@joeA: it's not as smooth as regenerating at a higher framerate, but gfycat.com allows one to view gifs at different speeds: gfycat.com/TintedWatchfulAxisdeer#?speed=0.25
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– Max
Apr 2 '14 at 20:51
197
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If high school math just used a fraction of the resources here, we'd have way more mathematicians.
$endgroup$
– user148298
Apr 3 '14 at 17:00
22
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Is the source of this animation available? (It looks like it's in $mathrmTeX$.)
$endgroup$
– bb010g
Apr 3 '14 at 23:31
12
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This is the normal way of introducing sine and cosine at high schools. At least in our country.
$endgroup$
– Vladimir F
Apr 7 '14 at 15:36
12
$begingroup$
@joeA: I've made a programmatic version of this on Khan Academy; feel free to change the speed to your liking: khanacademy.org/cs/circle/4597064320155648
$endgroup$
– BHSPitMonkey
Apr 16 '14 at 10:48
13
13
$begingroup$
@joeA: it's not as smooth as regenerating at a higher framerate, but gfycat.com allows one to view gifs at different speeds: gfycat.com/TintedWatchfulAxisdeer#?speed=0.25
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– Max
Apr 2 '14 at 20:51
$begingroup$
@joeA: it's not as smooth as regenerating at a higher framerate, but gfycat.com allows one to view gifs at different speeds: gfycat.com/TintedWatchfulAxisdeer#?speed=0.25
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– Max
Apr 2 '14 at 20:51
197
197
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If high school math just used a fraction of the resources here, we'd have way more mathematicians.
$endgroup$
– user148298
Apr 3 '14 at 17:00
$begingroup$
If high school math just used a fraction of the resources here, we'd have way more mathematicians.
$endgroup$
– user148298
Apr 3 '14 at 17:00
22
22
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Is the source of this animation available? (It looks like it's in $mathrmTeX$.)
$endgroup$
– bb010g
Apr 3 '14 at 23:31
$begingroup$
Is the source of this animation available? (It looks like it's in $mathrmTeX$.)
$endgroup$
– bb010g
Apr 3 '14 at 23:31
12
12
$begingroup$
This is the normal way of introducing sine and cosine at high schools. At least in our country.
$endgroup$
– Vladimir F
Apr 7 '14 at 15:36
$begingroup$
This is the normal way of introducing sine and cosine at high schools. At least in our country.
$endgroup$
– Vladimir F
Apr 7 '14 at 15:36
12
12
$begingroup$
@joeA: I've made a programmatic version of this on Khan Academy; feel free to change the speed to your liking: khanacademy.org/cs/circle/4597064320155648
$endgroup$
– BHSPitMonkey
Apr 16 '14 at 10:48
$begingroup$
@joeA: I've made a programmatic version of this on Khan Academy; feel free to change the speed to your liking: khanacademy.org/cs/circle/4597064320155648
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– BHSPitMonkey
Apr 16 '14 at 10:48
|
show 14 more comments
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My favorite: tell someone that $$sum_{n=1}^{infty}frac{1}{2^n}=1$$ and they probably won't believe you. However, show them the below:
and suddenly what had been obscure is now obvious.
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10
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My first intuitive visualization of this sum was a circle. I wasn't $100%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :)
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– Guy
Mar 31 '14 at 17:38
237
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I still don't believe you.
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– Ojonugwa Ochalifu
Apr 2 '14 at 8:51
113
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Another way to think of this is that 1/2 + 1/4 + 1/8 = 0.111...binary = 0.999...decimal = 1.
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– Justin L.
Apr 7 '14 at 3:02
13
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You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1.
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– Maxim Umansky
Apr 7 '14 at 4:05
41
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@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $frac{1}{2}$ and $frac{1}{4}$ in their "standard shape", so what remains must obviously be $frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $frac{1}{2}cdotfrac{1}{4}=frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger).
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– nobody
Apr 9 '14 at 20:09
|
show 10 more comments
$begingroup$
My favorite: tell someone that $$sum_{n=1}^{infty}frac{1}{2^n}=1$$ and they probably won't believe you. However, show them the below:
and suddenly what had been obscure is now obvious.
$endgroup$
10
$begingroup$
My first intuitive visualization of this sum was a circle. I wasn't $100%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :)
$endgroup$
– Guy
Mar 31 '14 at 17:38
237
$begingroup$
I still don't believe you.
$endgroup$
– Ojonugwa Ochalifu
Apr 2 '14 at 8:51
113
$begingroup$
Another way to think of this is that 1/2 + 1/4 + 1/8 = 0.111...binary = 0.999...decimal = 1.
$endgroup$
– Justin L.
Apr 7 '14 at 3:02
13
$begingroup$
You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1.
$endgroup$
– Maxim Umansky
Apr 7 '14 at 4:05
41
$begingroup$
@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $frac{1}{2}$ and $frac{1}{4}$ in their "standard shape", so what remains must obviously be $frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $frac{1}{2}cdotfrac{1}{4}=frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger).
$endgroup$
– nobody
Apr 9 '14 at 20:09
|
show 10 more comments
$begingroup$
My favorite: tell someone that $$sum_{n=1}^{infty}frac{1}{2^n}=1$$ and they probably won't believe you. However, show them the below:
and suddenly what had been obscure is now obvious.
$endgroup$
My favorite: tell someone that $$sum_{n=1}^{infty}frac{1}{2^n}=1$$ and they probably won't believe you. However, show them the below:
and suddenly what had been obscure is now obvious.
edited Apr 7 '14 at 17:27
community wiki
5 revs, 3 users 76%
eykanal
10
$begingroup$
My first intuitive visualization of this sum was a circle. I wasn't $100%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :)
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– Guy
Mar 31 '14 at 17:38
237
$begingroup$
I still don't believe you.
$endgroup$
– Ojonugwa Ochalifu
Apr 2 '14 at 8:51
113
$begingroup$
Another way to think of this is that 1/2 + 1/4 + 1/8 = 0.111...binary = 0.999...decimal = 1.
$endgroup$
– Justin L.
Apr 7 '14 at 3:02
13
$begingroup$
You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1.
$endgroup$
– Maxim Umansky
Apr 7 '14 at 4:05
41
$begingroup$
@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $frac{1}{2}$ and $frac{1}{4}$ in their "standard shape", so what remains must obviously be $frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $frac{1}{2}cdotfrac{1}{4}=frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger).
$endgroup$
– nobody
Apr 9 '14 at 20:09
|
show 10 more comments
10
$begingroup$
My first intuitive visualization of this sum was a circle. I wasn't $100%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :)
$endgroup$
– Guy
Mar 31 '14 at 17:38
237
$begingroup$
I still don't believe you.
$endgroup$
– Ojonugwa Ochalifu
Apr 2 '14 at 8:51
113
$begingroup$
Another way to think of this is that 1/2 + 1/4 + 1/8 = 0.111...binary = 0.999...decimal = 1.
$endgroup$
– Justin L.
Apr 7 '14 at 3:02
13
$begingroup$
You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1.
$endgroup$
– Maxim Umansky
Apr 7 '14 at 4:05
41
$begingroup$
@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $frac{1}{2}$ and $frac{1}{4}$ in their "standard shape", so what remains must obviously be $frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $frac{1}{2}cdotfrac{1}{4}=frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger).
$endgroup$
– nobody
Apr 9 '14 at 20:09
10
10
$begingroup$
My first intuitive visualization of this sum was a circle. I wasn't $100%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :)
$endgroup$
– Guy
Mar 31 '14 at 17:38
$begingroup$
My first intuitive visualization of this sum was a circle. I wasn't $100%$ sure that the answer was $1$(long time back, I had never seen an infinite series before, and 1 was just my first immediate thought) :)
$endgroup$
– Guy
Mar 31 '14 at 17:38
237
237
$begingroup$
I still don't believe you.
$endgroup$
– Ojonugwa Ochalifu
Apr 2 '14 at 8:51
$begingroup$
I still don't believe you.
$endgroup$
– Ojonugwa Ochalifu
Apr 2 '14 at 8:51
113
113
$begingroup$
Another way to think of this is that 1/2 + 1/4 + 1/8 = 0.111...binary = 0.999...decimal = 1.
$endgroup$
– Justin L.
Apr 7 '14 at 3:02
$begingroup$
Another way to think of this is that 1/2 + 1/4 + 1/8 = 0.111...binary = 0.999...decimal = 1.
$endgroup$
– Justin L.
Apr 7 '14 at 3:02
13
13
$begingroup$
You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1.
$endgroup$
– Maxim Umansky
Apr 7 '14 at 4:05
$begingroup$
You don't need to divide the square into such complicated fragments, just use vertical lines (assuming the x coordinate ranges from 0 to 1) at x=1/2, x=1/4, x=1/8 etc. Each time you add 1/2^n to the area, and in the limit you obviously get 1.
$endgroup$
– Maxim Umansky
Apr 7 '14 at 4:05
41
41
$begingroup$
@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $frac{1}{2}$ and $frac{1}{4}$ in their "standard shape", so what remains must obviously be $frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $frac{1}{2}cdotfrac{1}{4}=frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger).
$endgroup$
– nobody
Apr 9 '14 at 20:09
$begingroup$
@MaximUmansky: That way you'd just get lines that get closer together and it'll be not as obvious. Here, you see the fractions $frac{1}{2}$ and $frac{1}{4}$ in their "standard shape", so what remains must obviously be $frac{1}{4}$. Then, put the same shapes inside the remaining square which is of the same proportions as the initial one (and it's easily checked that $frac{1}{2}cdotfrac{1}{4}=frac{1}{8}$); you'll get the next smaller square, hidden deeper in the corner. Repeat, and the square will shrink to a tiny dot (not a whole line, which may intuitively seem larger).
$endgroup$
– nobody
Apr 9 '14 at 20:09
|
show 10 more comments
$begingroup$
This visualisation of the Fourier Transform was very enlightening for me:
The author, LucasVB, has a whole gallery of similar visualisations at his Wikipedia gallery and his tumblr blog.
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11
$begingroup$
@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave.
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– Joel
Apr 7 '14 at 2:15
6
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I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets).
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– Mikhail
Apr 7 '14 at 3:19
8
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this is pretty cool too: bl.ocks.org/jinroh/7524988
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– ilia choly
Apr 7 '14 at 20:54
5
$begingroup$
I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works...
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– Cole Johnson
Apr 11 '14 at 1:21
11
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I don't find this graphical example helpful in explaining the Fourier Transform. If anything, it is going to add to existing confusion people have with the Fourier Series. The sin and cos expression shown is too general to be meaningful, and the graphic shown for the "transform" is really just a visualization of the magnitudes of the Fourier Series coefficients you would have if the partial sum function were presumed to be infinitely periodic. When I calculate the actual Fourier Transform of the segment of function shown, I get something quite different: imgur.com/eQuerJa
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– Adam Smith
Aug 31 '14 at 18:53
|
show 4 more comments
$begingroup$
This visualisation of the Fourier Transform was very enlightening for me:
The author, LucasVB, has a whole gallery of similar visualisations at his Wikipedia gallery and his tumblr blog.
$endgroup$
11
$begingroup$
@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave.
$endgroup$
– Joel
Apr 7 '14 at 2:15
6
$begingroup$
I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets).
$endgroup$
– Mikhail
Apr 7 '14 at 3:19
8
$begingroup$
this is pretty cool too: bl.ocks.org/jinroh/7524988
$endgroup$
– ilia choly
Apr 7 '14 at 20:54
5
$begingroup$
I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works...
$endgroup$
– Cole Johnson
Apr 11 '14 at 1:21
11
$begingroup$
I don't find this graphical example helpful in explaining the Fourier Transform. If anything, it is going to add to existing confusion people have with the Fourier Series. The sin and cos expression shown is too general to be meaningful, and the graphic shown for the "transform" is really just a visualization of the magnitudes of the Fourier Series coefficients you would have if the partial sum function were presumed to be infinitely periodic. When I calculate the actual Fourier Transform of the segment of function shown, I get something quite different: imgur.com/eQuerJa
$endgroup$
– Adam Smith
Aug 31 '14 at 18:53
|
show 4 more comments
$begingroup$
This visualisation of the Fourier Transform was very enlightening for me:
The author, LucasVB, has a whole gallery of similar visualisations at his Wikipedia gallery and his tumblr blog.
$endgroup$
This visualisation of the Fourier Transform was very enlightening for me:
The author, LucasVB, has a whole gallery of similar visualisations at his Wikipedia gallery and his tumblr blog.
edited Apr 7 '14 at 11:12
community wiki
2 revs, 2 users 80%
njp
11
$begingroup$
@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave.
$endgroup$
– Joel
Apr 7 '14 at 2:15
6
$begingroup$
I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets).
$endgroup$
– Mikhail
Apr 7 '14 at 3:19
8
$begingroup$
this is pretty cool too: bl.ocks.org/jinroh/7524988
$endgroup$
– ilia choly
Apr 7 '14 at 20:54
5
$begingroup$
I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works...
$endgroup$
– Cole Johnson
Apr 11 '14 at 1:21
11
$begingroup$
I don't find this graphical example helpful in explaining the Fourier Transform. If anything, it is going to add to existing confusion people have with the Fourier Series. The sin and cos expression shown is too general to be meaningful, and the graphic shown for the "transform" is really just a visualization of the magnitudes of the Fourier Series coefficients you would have if the partial sum function were presumed to be infinitely periodic. When I calculate the actual Fourier Transform of the segment of function shown, I get something quite different: imgur.com/eQuerJa
$endgroup$
– Adam Smith
Aug 31 '14 at 18:53
|
show 4 more comments
11
$begingroup$
@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave.
$endgroup$
– Joel
Apr 7 '14 at 2:15
6
$begingroup$
I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets).
$endgroup$
– Mikhail
Apr 7 '14 at 3:19
8
$begingroup$
this is pretty cool too: bl.ocks.org/jinroh/7524988
$endgroup$
– ilia choly
Apr 7 '14 at 20:54
5
$begingroup$
I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works...
$endgroup$
– Cole Johnson
Apr 11 '14 at 1:21
11
$begingroup$
I don't find this graphical example helpful in explaining the Fourier Transform. If anything, it is going to add to existing confusion people have with the Fourier Series. The sin and cos expression shown is too general to be meaningful, and the graphic shown for the "transform" is really just a visualization of the magnitudes of the Fourier Series coefficients you would have if the partial sum function were presumed to be infinitely periodic. When I calculate the actual Fourier Transform of the segment of function shown, I get something quite different: imgur.com/eQuerJa
$endgroup$
– Adam Smith
Aug 31 '14 at 18:53
11
11
$begingroup$
@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave.
$endgroup$
– Joel
Apr 7 '14 at 2:15
$begingroup$
@gekkostate You're a bit off. f is the sum of multiple simple waves, all with different frequencies and phase angles. The fourier transform takes a complex wave from a given time period, and gives you the phase angles and frequencies of all of the component waves. f^ is the amplitude of each component wave.
$endgroup$
– Joel
Apr 7 '14 at 2:15
6
6
$begingroup$
I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets).
$endgroup$
– Mikhail
Apr 7 '14 at 3:19
$begingroup$
I'm down-voting this because this animation doesn't include phase information. The phase is the difference between a pulse and cw. If you don't want it, you should use another basis (wavelets).
$endgroup$
– Mikhail
Apr 7 '14 at 3:19
8
8
$begingroup$
this is pretty cool too: bl.ocks.org/jinroh/7524988
$endgroup$
– ilia choly
Apr 7 '14 at 20:54
$begingroup$
this is pretty cool too: bl.ocks.org/jinroh/7524988
$endgroup$
– ilia choly
Apr 7 '14 at 20:54
5
5
$begingroup$
I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works...
$endgroup$
– Cole Johnson
Apr 11 '14 at 1:21
$begingroup$
I saw this animation a long time ago, and it helped me finally understand what the Fourier transformation does, but I still am lost as to how it works...
$endgroup$
– Cole Johnson
Apr 11 '14 at 1:21
11
11
$begingroup$
I don't find this graphical example helpful in explaining the Fourier Transform. If anything, it is going to add to existing confusion people have with the Fourier Series. The sin and cos expression shown is too general to be meaningful, and the graphic shown for the "transform" is really just a visualization of the magnitudes of the Fourier Series coefficients you would have if the partial sum function were presumed to be infinitely periodic. When I calculate the actual Fourier Transform of the segment of function shown, I get something quite different: imgur.com/eQuerJa
$endgroup$
– Adam Smith
Aug 31 '14 at 18:53
$begingroup$
I don't find this graphical example helpful in explaining the Fourier Transform. If anything, it is going to add to existing confusion people have with the Fourier Series. The sin and cos expression shown is too general to be meaningful, and the graphic shown for the "transform" is really just a visualization of the magnitudes of the Fourier Series coefficients you would have if the partial sum function were presumed to be infinitely periodic. When I calculate the actual Fourier Transform of the segment of function shown, I get something quite different: imgur.com/eQuerJa
$endgroup$
– Adam Smith
Aug 31 '14 at 18:53
|
show 4 more comments
$begingroup$
Here is a classic: the sum of the first $n$ positive odd numbers $= n^2$.
We also see that the sum of the first $n$ positive even numbers $= n(n+1)$ (excluding $0$), by adding a column to the left.
$endgroup$
19
$begingroup$
If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$
$endgroup$
– becko
Apr 3 '14 at 16:29
27
$begingroup$
I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this.
$endgroup$
– Nick T
Apr 3 '14 at 17:37
15
$begingroup$
I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit"
$endgroup$
– Guy
Apr 6 '14 at 9:40
$begingroup$
I have noticed an interesting method to calculate a square root based on this fact: disqus.com/home/discussion/geeksforgeeks/…
$endgroup$
– sergiol
Sep 7 '15 at 23:46
add a comment |
$begingroup$
Here is a classic: the sum of the first $n$ positive odd numbers $= n^2$.
We also see that the sum of the first $n$ positive even numbers $= n(n+1)$ (excluding $0$), by adding a column to the left.
$endgroup$
19
$begingroup$
If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$
$endgroup$
– becko
Apr 3 '14 at 16:29
27
$begingroup$
I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this.
$endgroup$
– Nick T
Apr 3 '14 at 17:37
15
$begingroup$
I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit"
$endgroup$
– Guy
Apr 6 '14 at 9:40
$begingroup$
I have noticed an interesting method to calculate a square root based on this fact: disqus.com/home/discussion/geeksforgeeks/…
$endgroup$
– sergiol
Sep 7 '15 at 23:46
add a comment |
$begingroup$
Here is a classic: the sum of the first $n$ positive odd numbers $= n^2$.
We also see that the sum of the first $n$ positive even numbers $= n(n+1)$ (excluding $0$), by adding a column to the left.
$endgroup$
Here is a classic: the sum of the first $n$ positive odd numbers $= n^2$.
We also see that the sum of the first $n$ positive even numbers $= n(n+1)$ (excluding $0$), by adding a column to the left.
edited Apr 4 '14 at 13:23
community wiki
7 revs
Yiyuan Lee
19
$begingroup$
If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$
$endgroup$
– becko
Apr 3 '14 at 16:29
27
$begingroup$
I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this.
$endgroup$
– Nick T
Apr 3 '14 at 17:37
15
$begingroup$
I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit"
$endgroup$
– Guy
Apr 6 '14 at 9:40
$begingroup$
I have noticed an interesting method to calculate a square root based on this fact: disqus.com/home/discussion/geeksforgeeks/…
$endgroup$
– sergiol
Sep 7 '15 at 23:46
add a comment |
19
$begingroup$
If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$
$endgroup$
– becko
Apr 3 '14 at 16:29
27
$begingroup$
I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this.
$endgroup$
– Nick T
Apr 3 '14 at 17:37
15
$begingroup$
I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit"
$endgroup$
– Guy
Apr 6 '14 at 9:40
$begingroup$
I have noticed an interesting method to calculate a square root based on this fact: disqus.com/home/discussion/geeksforgeeks/…
$endgroup$
– sergiol
Sep 7 '15 at 23:46
19
19
$begingroup$
If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$
$endgroup$
– becko
Apr 3 '14 at 16:29
$begingroup$
If you remove the leftmost column, you get a proof that the sum of the first $n$ non-negative even numbers (counting 0) is $n(n-1)$
$endgroup$
– becko
Apr 3 '14 at 16:29
27
27
$begingroup$
I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this.
$endgroup$
– Nick T
Apr 3 '14 at 17:37
$begingroup$
I read something about how Galileo noted that the distance a falling body covered over a unit time went as the series of odd numbers and was confused until I realized this.
$endgroup$
– Nick T
Apr 3 '14 at 17:37
15
15
$begingroup$
I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit"
$endgroup$
– Guy
Apr 6 '14 at 9:40
$begingroup$
I remember about 5 years back, it was when we were first learning programming. Just simple kids stuff. Loops. We were given a series to print. $1,4,9,16,dots$ I was a little overworked so I didn't notice the obvious square pattern. I did notice $1,(1+3),(1+3+5),....$ when we were giving in our work, I saw the guy before me had just done $i*i$. I totally went "hoooly shit"
$endgroup$
– Guy
Apr 6 '14 at 9:40
$begingroup$
I have noticed an interesting method to calculate a square root based on this fact: disqus.com/home/discussion/geeksforgeeks/…
$endgroup$
– sergiol
Sep 7 '15 at 23:46
$begingroup$
I have noticed an interesting method to calculate a square root based on this fact: disqus.com/home/discussion/geeksforgeeks/…
$endgroup$
– sergiol
Sep 7 '15 at 23:46
add a comment |
$begingroup$
A well-known visual to explain $(a+b)^2 = a^2+2ab+b^2$:
$endgroup$
1
$begingroup$
Always liked this one
$endgroup$
– Almo
Apr 2 '14 at 14:11
9
$begingroup$
why area of a^2 is greater tham area of b^2?
$endgroup$
– jaczjill
Apr 3 '14 at 7:34
66
$begingroup$
@jaczjill Because a and b are not the same number.
$endgroup$
– Conor Pender
Apr 3 '14 at 11:24
26
$begingroup$
@jaczjill the areas of a and b are arbitrary. They could be any size and it would make no difference. It's just an example.
$endgroup$
– theyetiman
Apr 4 '14 at 8:29
5
$begingroup$
Have you seen the Montessori binomial cube?
$endgroup$
– MJD
Apr 4 '14 at 14:30
|
show 2 more comments
$begingroup$
A well-known visual to explain $(a+b)^2 = a^2+2ab+b^2$:
$endgroup$
1
$begingroup$
Always liked this one
$endgroup$
– Almo
Apr 2 '14 at 14:11
9
$begingroup$
why area of a^2 is greater tham area of b^2?
$endgroup$
– jaczjill
Apr 3 '14 at 7:34
66
$begingroup$
@jaczjill Because a and b are not the same number.
$endgroup$
– Conor Pender
Apr 3 '14 at 11:24
26
$begingroup$
@jaczjill the areas of a and b are arbitrary. They could be any size and it would make no difference. It's just an example.
$endgroup$
– theyetiman
Apr 4 '14 at 8:29
5
$begingroup$
Have you seen the Montessori binomial cube?
$endgroup$
– MJD
Apr 4 '14 at 14:30
|
show 2 more comments
$begingroup$
A well-known visual to explain $(a+b)^2 = a^2+2ab+b^2$:
$endgroup$
A well-known visual to explain $(a+b)^2 = a^2+2ab+b^2$:
answered Apr 1 '14 at 18:48
community wiki
Johannes
1
$begingroup$
Always liked this one
$endgroup$
– Almo
Apr 2 '14 at 14:11
9
$begingroup$
why area of a^2 is greater tham area of b^2?
$endgroup$
– jaczjill
Apr 3 '14 at 7:34
66
$begingroup$
@jaczjill Because a and b are not the same number.
$endgroup$
– Conor Pender
Apr 3 '14 at 11:24
26
$begingroup$
@jaczjill the areas of a and b are arbitrary. They could be any size and it would make no difference. It's just an example.
$endgroup$
– theyetiman
Apr 4 '14 at 8:29
5
$begingroup$
Have you seen the Montessori binomial cube?
$endgroup$
– MJD
Apr 4 '14 at 14:30
|
show 2 more comments
1
$begingroup$
Always liked this one
$endgroup$
– Almo
Apr 2 '14 at 14:11
9
$begingroup$
why area of a^2 is greater tham area of b^2?
$endgroup$
– jaczjill
Apr 3 '14 at 7:34
66
$begingroup$
@jaczjill Because a and b are not the same number.
$endgroup$
– Conor Pender
Apr 3 '14 at 11:24
26
$begingroup$
@jaczjill the areas of a and b are arbitrary. They could be any size and it would make no difference. It's just an example.
$endgroup$
– theyetiman
Apr 4 '14 at 8:29
5
$begingroup$
Have you seen the Montessori binomial cube?
$endgroup$
– MJD
Apr 4 '14 at 14:30
1
1
$begingroup$
Always liked this one
$endgroup$
– Almo
Apr 2 '14 at 14:11
$begingroup$
Always liked this one
$endgroup$
– Almo
Apr 2 '14 at 14:11
9
9
$begingroup$
why area of a^2 is greater tham area of b^2?
$endgroup$
– jaczjill
Apr 3 '14 at 7:34
$begingroup$
why area of a^2 is greater tham area of b^2?
$endgroup$
– jaczjill
Apr 3 '14 at 7:34
66
66
$begingroup$
@jaczjill Because a and b are not the same number.
$endgroup$
– Conor Pender
Apr 3 '14 at 11:24
$begingroup$
@jaczjill Because a and b are not the same number.
$endgroup$
– Conor Pender
Apr 3 '14 at 11:24
26
26
$begingroup$
@jaczjill the areas of a and b are arbitrary. They could be any size and it would make no difference. It's just an example.
$endgroup$
– theyetiman
Apr 4 '14 at 8:29
$begingroup$
@jaczjill the areas of a and b are arbitrary. They could be any size and it would make no difference. It's just an example.
$endgroup$
– theyetiman
Apr 4 '14 at 8:29
5
5
$begingroup$
Have you seen the Montessori binomial cube?
$endgroup$
– MJD
Apr 4 '14 at 14:30
$begingroup$
Have you seen the Montessori binomial cube?
$endgroup$
– MJD
Apr 4 '14 at 14:30
|
show 2 more comments
$begingroup$
The sum of the exterior angles of any convex polygon will always add up to $360^circ$.
This can be viewed as a zooming out process, as illustrate by the animation below:
$endgroup$
49
$begingroup$
I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take).
$endgroup$
– daviewales
Apr 5 '14 at 12:27
16
$begingroup$
I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance.
$endgroup$
– Steven Gubkin
Apr 5 '14 at 14:10
3
$begingroup$
This would be better as an animation (which achieves the zooming effect).
$endgroup$
– Duncan Jones
Apr 7 '14 at 10:31
1
$begingroup$
Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.)
$endgroup$
– Yiyuan Lee
Apr 7 '14 at 13:36
3
$begingroup$
this is really,really clever
$endgroup$
– Konstantinos Gaitanas
Aug 3 '17 at 9:07
|
show 12 more comments
$begingroup$
The sum of the exterior angles of any convex polygon will always add up to $360^circ$.
This can be viewed as a zooming out process, as illustrate by the animation below:
$endgroup$
49
$begingroup$
I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take).
$endgroup$
– daviewales
Apr 5 '14 at 12:27
16
$begingroup$
I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance.
$endgroup$
– Steven Gubkin
Apr 5 '14 at 14:10
3
$begingroup$
This would be better as an animation (which achieves the zooming effect).
$endgroup$
– Duncan Jones
Apr 7 '14 at 10:31
1
$begingroup$
Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.)
$endgroup$
– Yiyuan Lee
Apr 7 '14 at 13:36
3
$begingroup$
this is really,really clever
$endgroup$
– Konstantinos Gaitanas
Aug 3 '17 at 9:07
|
show 12 more comments
$begingroup$
The sum of the exterior angles of any convex polygon will always add up to $360^circ$.
This can be viewed as a zooming out process, as illustrate by the animation below:
$endgroup$
The sum of the exterior angles of any convex polygon will always add up to $360^circ$.
This can be viewed as a zooming out process, as illustrate by the animation below:
edited Apr 13 '14 at 3:48
community wiki
4 revs
Yiyuan Lee
49
$begingroup$
I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take).
$endgroup$
– daviewales
Apr 5 '14 at 12:27
16
$begingroup$
I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance.
$endgroup$
– Steven Gubkin
Apr 5 '14 at 14:10
3
$begingroup$
This would be better as an animation (which achieves the zooming effect).
$endgroup$
– Duncan Jones
Apr 7 '14 at 10:31
1
$begingroup$
Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.)
$endgroup$
– Yiyuan Lee
Apr 7 '14 at 13:36
3
$begingroup$
this is really,really clever
$endgroup$
– Konstantinos Gaitanas
Aug 3 '17 at 9:07
|
show 12 more comments
49
$begingroup$
I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take).
$endgroup$
– daviewales
Apr 5 '14 at 12:27
16
$begingroup$
I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance.
$endgroup$
– Steven Gubkin
Apr 5 '14 at 14:10
3
$begingroup$
This would be better as an animation (which achieves the zooming effect).
$endgroup$
– Duncan Jones
Apr 7 '14 at 10:31
1
$begingroup$
Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.)
$endgroup$
– Yiyuan Lee
Apr 7 '14 at 13:36
3
$begingroup$
this is really,really clever
$endgroup$
– Konstantinos Gaitanas
Aug 3 '17 at 9:07
49
49
$begingroup$
I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take).
$endgroup$
– daviewales
Apr 5 '14 at 12:27
$begingroup$
I always found the easiest way to think about this was that if you move around a convex shape such that you eventually end up at the start, you must have rotated through exactly 360º. If you walk around any shape, and count anticlockwise rotations as negative, but clockwise rotations as positive, I imagine the angles will still add to 360º (or -360º depending on which direction you take).
$endgroup$
– daviewales
Apr 5 '14 at 12:27
16
16
$begingroup$
I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance.
$endgroup$
– Steven Gubkin
Apr 5 '14 at 14:10
$begingroup$
I like to view this as a limiting process. Imagine that the picture on the right is the picture on the left zoomed out a great distance.
$endgroup$
– Steven Gubkin
Apr 5 '14 at 14:10
3
3
$begingroup$
This would be better as an animation (which achieves the zooming effect).
$endgroup$
– Duncan Jones
Apr 7 '14 at 10:31
$begingroup$
This would be better as an animation (which achieves the zooming effect).
$endgroup$
– Duncan Jones
Apr 7 '14 at 10:31
1
1
$begingroup$
Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.)
$endgroup$
– Yiyuan Lee
Apr 7 '14 at 13:36
$begingroup$
Edited. Do have a look. (It may look horrible, since this is the first time I made an animation.)
$endgroup$
– Yiyuan Lee
Apr 7 '14 at 13:36
3
3
$begingroup$
this is really,really clever
$endgroup$
– Konstantinos Gaitanas
Aug 3 '17 at 9:07
$begingroup$
this is really,really clever
$endgroup$
– Konstantinos Gaitanas
Aug 3 '17 at 9:07
|
show 12 more comments
$begingroup$
While attending an Abstract Algebra course I was given the assignment to write out the multiplication table modulo n. I forgot to do the homework until just before class, but it was so easy to write the program I was able to print the result between classes.
The circular patterns in the tables fascinated me, and compelled me to replace the numbers with colors. The result is a beautiful illustration showing the emergence of primes and symmetry of multiplication.
The colors were chosen to start blue at 1 (cold) and fade to red at n (hot). White is used for zero (frozen), because it communicates the most information about prime factorization.
The interactive version can be found here:
https://web.archive.org/web/20140830110358/http://arapaho.nsuok.edu/~deckar01/Zvis.html
Multiplication of the integers modulo 15:
Multiplication of the integers modulo 512:
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8
$begingroup$
This is awesome. Looks very much like a fractal. Is there a name for this fractal?
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 15:52
$begingroup$
What in specific is it lacking? Correct me if I'm wrong, but if we take the white dots to be the area of the fractal and the red and blue to be the border then it's an object of fractional dimension (either 0d points or 1d lines filling a 2d plane) and self-symmetric.
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 17:42
26
$begingroup$
@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime.
$endgroup$
– John Dvorak
Apr 7 '14 at 18:48
$begingroup$
I just read about qualitative self-similarity which I think describes the relationship between frames in this example, but it isn't continuous in any dimension.
$endgroup$
– Jared Deckard
Apr 7 '14 at 18:51
5
$begingroup$
Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200
$endgroup$
– Vikram Saraph
Apr 9 '14 at 23:05
|
show 14 more comments
$begingroup$
While attending an Abstract Algebra course I was given the assignment to write out the multiplication table modulo n. I forgot to do the homework until just before class, but it was so easy to write the program I was able to print the result between classes.
The circular patterns in the tables fascinated me, and compelled me to replace the numbers with colors. The result is a beautiful illustration showing the emergence of primes and symmetry of multiplication.
The colors were chosen to start blue at 1 (cold) and fade to red at n (hot). White is used for zero (frozen), because it communicates the most information about prime factorization.
The interactive version can be found here:
https://web.archive.org/web/20140830110358/http://arapaho.nsuok.edu/~deckar01/Zvis.html
Multiplication of the integers modulo 15:
Multiplication of the integers modulo 512:
$endgroup$
8
$begingroup$
This is awesome. Looks very much like a fractal. Is there a name for this fractal?
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 15:52
$begingroup$
What in specific is it lacking? Correct me if I'm wrong, but if we take the white dots to be the area of the fractal and the red and blue to be the border then it's an object of fractional dimension (either 0d points or 1d lines filling a 2d plane) and self-symmetric.
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 17:42
26
$begingroup$
@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime.
$endgroup$
– John Dvorak
Apr 7 '14 at 18:48
$begingroup$
I just read about qualitative self-similarity which I think describes the relationship between frames in this example, but it isn't continuous in any dimension.
$endgroup$
– Jared Deckard
Apr 7 '14 at 18:51
5
$begingroup$
Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200
$endgroup$
– Vikram Saraph
Apr 9 '14 at 23:05
|
show 14 more comments
$begingroup$
While attending an Abstract Algebra course I was given the assignment to write out the multiplication table modulo n. I forgot to do the homework until just before class, but it was so easy to write the program I was able to print the result between classes.
The circular patterns in the tables fascinated me, and compelled me to replace the numbers with colors. The result is a beautiful illustration showing the emergence of primes and symmetry of multiplication.
The colors were chosen to start blue at 1 (cold) and fade to red at n (hot). White is used for zero (frozen), because it communicates the most information about prime factorization.
The interactive version can be found here:
https://web.archive.org/web/20140830110358/http://arapaho.nsuok.edu/~deckar01/Zvis.html
Multiplication of the integers modulo 15:
Multiplication of the integers modulo 512:
$endgroup$
While attending an Abstract Algebra course I was given the assignment to write out the multiplication table modulo n. I forgot to do the homework until just before class, but it was so easy to write the program I was able to print the result between classes.
The circular patterns in the tables fascinated me, and compelled me to replace the numbers with colors. The result is a beautiful illustration showing the emergence of primes and symmetry of multiplication.
The colors were chosen to start blue at 1 (cold) and fade to red at n (hot). White is used for zero (frozen), because it communicates the most information about prime factorization.
The interactive version can be found here:
https://web.archive.org/web/20140830110358/http://arapaho.nsuok.edu/~deckar01/Zvis.html
Multiplication of the integers modulo 15:
Multiplication of the integers modulo 512:
edited Apr 23 '16 at 22:15
community wiki
3 revs
Jared Deckard
8
$begingroup$
This is awesome. Looks very much like a fractal. Is there a name for this fractal?
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 15:52
$begingroup$
What in specific is it lacking? Correct me if I'm wrong, but if we take the white dots to be the area of the fractal and the red and blue to be the border then it's an object of fractional dimension (either 0d points or 1d lines filling a 2d plane) and self-symmetric.
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 17:42
26
$begingroup$
@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime.
$endgroup$
– John Dvorak
Apr 7 '14 at 18:48
$begingroup$
I just read about qualitative self-similarity which I think describes the relationship between frames in this example, but it isn't continuous in any dimension.
$endgroup$
– Jared Deckard
Apr 7 '14 at 18:51
5
$begingroup$
Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200
$endgroup$
– Vikram Saraph
Apr 9 '14 at 23:05
|
show 14 more comments
8
$begingroup$
This is awesome. Looks very much like a fractal. Is there a name for this fractal?
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 15:52
$begingroup$
What in specific is it lacking? Correct me if I'm wrong, but if we take the white dots to be the area of the fractal and the red and blue to be the border then it's an object of fractional dimension (either 0d points or 1d lines filling a 2d plane) and self-symmetric.
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 17:42
26
$begingroup$
@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime.
$endgroup$
– John Dvorak
Apr 7 '14 at 18:48
$begingroup$
I just read about qualitative self-similarity which I think describes the relationship between frames in this example, but it isn't continuous in any dimension.
$endgroup$
– Jared Deckard
Apr 7 '14 at 18:51
5
$begingroup$
Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200
$endgroup$
– Vikram Saraph
Apr 9 '14 at 23:05
8
8
$begingroup$
This is awesome. Looks very much like a fractal. Is there a name for this fractal?
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 15:52
$begingroup$
This is awesome. Looks very much like a fractal. Is there a name for this fractal?
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 15:52
$begingroup$
What in specific is it lacking? Correct me if I'm wrong, but if we take the white dots to be the area of the fractal and the red and blue to be the border then it's an object of fractional dimension (either 0d points or 1d lines filling a 2d plane) and self-symmetric.
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 17:42
$begingroup$
What in specific is it lacking? Correct me if I'm wrong, but if we take the white dots to be the area of the fractal and the red and blue to be the border then it's an object of fractional dimension (either 0d points or 1d lines filling a 2d plane) and self-symmetric.
$endgroup$
– OmnipotentEntity
Apr 7 '14 at 17:42
26
26
$begingroup$
@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime.
$endgroup$
– John Dvorak
Apr 7 '14 at 18:48
$begingroup$
@OmnipotentEntity it doesn't qualify as having fractional dimension because the underlying space is discrete. If you use a real space instead - the set of x, y such that x * y = 0 mod n, you get a set of hyperboles (linear in dimension). If you take a limit towards infinite squares, then the count of white squares is equal to the length of the edge whenever the length is prime.
$endgroup$
– John Dvorak
Apr 7 '14 at 18:48
$begingroup$
I just read about qualitative self-similarity which I think describes the relationship between frames in this example, but it isn't continuous in any dimension.
$endgroup$
– Jared Deckard
Apr 7 '14 at 18:51
$begingroup$
I just read about qualitative self-similarity which I think describes the relationship between frames in this example, but it isn't continuous in any dimension.
$endgroup$
– Jared Deckard
Apr 7 '14 at 18:51
5
5
$begingroup$
Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200
$endgroup$
– Vikram Saraph
Apr 9 '14 at 23:05
$begingroup$
Although not colored WolframAlpha also does a nice job of illustrating this: wolframalpha.com/input/?i=Z_200
$endgroup$
– Vikram Saraph
Apr 9 '14 at 23:05
|
show 14 more comments
$begingroup$
When I understood Fourier series visually-
$endgroup$
13
$begingroup$
OH. Wow. I just took my signals course and even after I did well on the exam I still didn't actually UNDERSTAND fourier series. Thanks, wow.
$endgroup$
– VicVu
Apr 2 '14 at 22:25
41
$begingroup$
This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice!
$endgroup$
– Steven Stadnicki
Apr 4 '14 at 6:09
8
$begingroup$
I don't understand what this is showing: is it showing hte approximation of a square wave by a Fourier series?
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:18
5
$begingroup$
@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon.
$endgroup$
– Potato
Apr 5 '14 at 6:01
$begingroup$
@ShreevatsaR: I evaluate the limiting amount of overshoot in the Gibbs Phenomenon in this answer (about $8.9%$ of the size of the discontinuity).
$endgroup$
– robjohn♦
Apr 6 '14 at 12:03
|
show 4 more comments
$begingroup$
When I understood Fourier series visually-
$endgroup$
13
$begingroup$
OH. Wow. I just took my signals course and even after I did well on the exam I still didn't actually UNDERSTAND fourier series. Thanks, wow.
$endgroup$
– VicVu
Apr 2 '14 at 22:25
41
$begingroup$
This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice!
$endgroup$
– Steven Stadnicki
Apr 4 '14 at 6:09
8
$begingroup$
I don't understand what this is showing: is it showing hte approximation of a square wave by a Fourier series?
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:18
5
$begingroup$
@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon.
$endgroup$
– Potato
Apr 5 '14 at 6:01
$begingroup$
@ShreevatsaR: I evaluate the limiting amount of overshoot in the Gibbs Phenomenon in this answer (about $8.9%$ of the size of the discontinuity).
$endgroup$
– robjohn♦
Apr 6 '14 at 12:03
|
show 4 more comments
$begingroup$
When I understood Fourier series visually-
$endgroup$
When I understood Fourier series visually-
edited Apr 5 '14 at 5:42
community wiki
2 revs, 2 users 91%
Gopi
13
$begingroup$
OH. Wow. I just took my signals course and even after I did well on the exam I still didn't actually UNDERSTAND fourier series. Thanks, wow.
$endgroup$
– VicVu
Apr 2 '14 at 22:25
41
$begingroup$
This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice!
$endgroup$
– Steven Stadnicki
Apr 4 '14 at 6:09
8
$begingroup$
I don't understand what this is showing: is it showing hte approximation of a square wave by a Fourier series?
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:18
5
$begingroup$
@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon.
$endgroup$
– Potato
Apr 5 '14 at 6:01
$begingroup$
@ShreevatsaR: I evaluate the limiting amount of overshoot in the Gibbs Phenomenon in this answer (about $8.9%$ of the size of the discontinuity).
$endgroup$
– robjohn♦
Apr 6 '14 at 12:03
|
show 4 more comments
13
$begingroup$
OH. Wow. I just took my signals course and even after I did well on the exam I still didn't actually UNDERSTAND fourier series. Thanks, wow.
$endgroup$
– VicVu
Apr 2 '14 at 22:25
41
$begingroup$
This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice!
$endgroup$
– Steven Stadnicki
Apr 4 '14 at 6:09
8
$begingroup$
I don't understand what this is showing: is it showing hte approximation of a square wave by a Fourier series?
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:18
5
$begingroup$
@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon.
$endgroup$
– Potato
Apr 5 '14 at 6:01
$begingroup$
@ShreevatsaR: I evaluate the limiting amount of overshoot in the Gibbs Phenomenon in this answer (about $8.9%$ of the size of the discontinuity).
$endgroup$
– robjohn♦
Apr 6 '14 at 12:03
13
13
$begingroup$
OH. Wow. I just took my signals course and even after I did well on the exam I still didn't actually UNDERSTAND fourier series. Thanks, wow.
$endgroup$
– VicVu
Apr 2 '14 at 22:25
$begingroup$
OH. Wow. I just took my signals course and even after I did well on the exam I still didn't actually UNDERSTAND fourier series. Thanks, wow.
$endgroup$
– VicVu
Apr 2 '14 at 22:25
41
41
$begingroup$
This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice!
$endgroup$
– Steven Stadnicki
Apr 4 '14 at 6:09
$begingroup$
This diagram is a particularly good illustration of the Gibbs Phenomenon, too. Nice!
$endgroup$
– Steven Stadnicki
Apr 4 '14 at 6:09
8
8
$begingroup$
I don't understand what this is showing: is it showing hte approximation of a square wave by a Fourier series?
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:18
$begingroup$
I don't understand what this is showing: is it showing hte approximation of a square wave by a Fourier series?
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:18
5
5
$begingroup$
@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon.
$endgroup$
– Potato
Apr 5 '14 at 6:01
$begingroup$
@ShreevatsaR Yes. It's an illustration of the Gibbs Phenomenon.
$endgroup$
– Potato
Apr 5 '14 at 6:01
$begingroup$
@ShreevatsaR: I evaluate the limiting amount of overshoot in the Gibbs Phenomenon in this answer (about $8.9%$ of the size of the discontinuity).
$endgroup$
– robjohn♦
Apr 6 '14 at 12:03
$begingroup$
@ShreevatsaR: I evaluate the limiting amount of overshoot in the Gibbs Phenomenon in this answer (about $8.9%$ of the size of the discontinuity).
$endgroup$
– robjohn♦
Apr 6 '14 at 12:03
|
show 4 more comments
$begingroup$
Simple answer for "what is a radian":
Logarithmic spiral and scale:
$endgroup$
$begingroup$
What program produces these images?
$endgroup$
– rschwieb
Aug 11 '14 at 19:49
5
$begingroup$
@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq
$endgroup$
– Alice Ryhl
Aug 13 '14 at 14:40
10
$begingroup$
I can't believe how long I used radians in college without realizing what a radian really is.
$endgroup$
– greg7gkb
Jan 9 '15 at 0:28
7
$begingroup$
The exponential curve has curvature-arc length relation like cycloidal? What does the animation show?
$endgroup$
– Narasimham
Apr 9 '16 at 2:31
$begingroup$
With what software can you make those graphics? They are good.
$endgroup$
– Darío A. Gutiérrez
Mar 31 '17 at 11:18
add a comment |
$begingroup$
Simple answer for "what is a radian":
Logarithmic spiral and scale:
$endgroup$
$begingroup$
What program produces these images?
$endgroup$
– rschwieb
Aug 11 '14 at 19:49
5
$begingroup$
@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq
$endgroup$
– Alice Ryhl
Aug 13 '14 at 14:40
10
$begingroup$
I can't believe how long I used radians in college without realizing what a radian really is.
$endgroup$
– greg7gkb
Jan 9 '15 at 0:28
7
$begingroup$
The exponential curve has curvature-arc length relation like cycloidal? What does the animation show?
$endgroup$
– Narasimham
Apr 9 '16 at 2:31
$begingroup$
With what software can you make those graphics? They are good.
$endgroup$
– Darío A. Gutiérrez
Mar 31 '17 at 11:18
add a comment |
$begingroup$
Simple answer for "what is a radian":
Logarithmic spiral and scale:
$endgroup$
Simple answer for "what is a radian":
Logarithmic spiral and scale:
edited May 26 '17 at 12:08
community wiki
3 revs, 3 users 67%
user2804123
$begingroup$
What program produces these images?
$endgroup$
– rschwieb
Aug 11 '14 at 19:49
5
$begingroup$
@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq
$endgroup$
– Alice Ryhl
Aug 13 '14 at 14:40
10
$begingroup$
I can't believe how long I used radians in college without realizing what a radian really is.
$endgroup$
– greg7gkb
Jan 9 '15 at 0:28
7
$begingroup$
The exponential curve has curvature-arc length relation like cycloidal? What does the animation show?
$endgroup$
– Narasimham
Apr 9 '16 at 2:31
$begingroup$
With what software can you make those graphics? They are good.
$endgroup$
– Darío A. Gutiérrez
Mar 31 '17 at 11:18
add a comment |
$begingroup$
What program produces these images?
$endgroup$
– rschwieb
Aug 11 '14 at 19:49
5
$begingroup$
@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq
$endgroup$
– Alice Ryhl
Aug 13 '14 at 14:40
10
$begingroup$
I can't believe how long I used radians in college without realizing what a radian really is.
$endgroup$
– greg7gkb
Jan 9 '15 at 0:28
7
$begingroup$
The exponential curve has curvature-arc length relation like cycloidal? What does the animation show?
$endgroup$
– Narasimham
Apr 9 '16 at 2:31
$begingroup$
With what software can you make those graphics? They are good.
$endgroup$
– Darío A. Gutiérrez
Mar 31 '17 at 11:18
$begingroup$
What program produces these images?
$endgroup$
– rschwieb
Aug 11 '14 at 19:49
$begingroup$
What program produces these images?
$endgroup$
– rschwieb
Aug 11 '14 at 19:49
5
5
$begingroup$
@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq
$endgroup$
– Alice Ryhl
Aug 13 '14 at 14:40
$begingroup$
@rschwieb They're made by 1ucasvb, he's got an faq here: 1ucasvb.tumblr.com/faq
$endgroup$
– Alice Ryhl
Aug 13 '14 at 14:40
10
10
$begingroup$
I can't believe how long I used radians in college without realizing what a radian really is.
$endgroup$
– greg7gkb
Jan 9 '15 at 0:28
$begingroup$
I can't believe how long I used radians in college without realizing what a radian really is.
$endgroup$
– greg7gkb
Jan 9 '15 at 0:28
7
7
$begingroup$
The exponential curve has curvature-arc length relation like cycloidal? What does the animation show?
$endgroup$
– Narasimham
Apr 9 '16 at 2:31
$begingroup$
The exponential curve has curvature-arc length relation like cycloidal? What does the animation show?
$endgroup$
– Narasimham
Apr 9 '16 at 2:31
$begingroup$
With what software can you make those graphics? They are good.
$endgroup$
– Darío A. Gutiérrez
Mar 31 '17 at 11:18
$begingroup$
With what software can you make those graphics? They are good.
$endgroup$
– Darío A. Gutiérrez
Mar 31 '17 at 11:18
add a comment |
$begingroup$
This is a neat little proof that the area of a circle is $pi r^2$, which I was first taught aged about 12 and it has stuck with me ever since. The circle is subdivided into equal pieces, then rearranged. As the number of pieces gets larger, the resulting shape gets closer and closer to a rectangle. It is obvious that the short side of this rectangle has length $r$, and a little thought will show that the two long sides each have a length half the circumference, or $pi r$, giving an area for the rectangle of $pi r^2$.
This can also be done physically by taking a paper circle and actually cutting it up and rearranging the pieces. This exercise also offers some introduction to (infinite) sequences.
$endgroup$
26
$begingroup$
This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to $pi r$.
$endgroup$
– Nicholas Wilson
Apr 8 '14 at 16:50
11
$begingroup$
@NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help?
$endgroup$
– ghoppe
Apr 8 '14 at 19:29
44
$begingroup$
@ghoppe Nice try! But it's all wiggly. We're lucky that $sin{theta} sim theta$ as $theta rightarrow 0$, so the actual perimeter of the bottom (which is known to be $pi$) does converge to the width of the rectangle. But - that nice property of $sin$ is exactly what we're trying to prove!
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 8:20
12
$begingroup$
@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof.
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 17:56
9
$begingroup$
@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent.
$endgroup$
– Ben Rowland
Dec 16 '14 at 22:16
|
show 9 more comments
$begingroup$
This is a neat little proof that the area of a circle is $pi r^2$, which I was first taught aged about 12 and it has stuck with me ever since. The circle is subdivided into equal pieces, then rearranged. As the number of pieces gets larger, the resulting shape gets closer and closer to a rectangle. It is obvious that the short side of this rectangle has length $r$, and a little thought will show that the two long sides each have a length half the circumference, or $pi r$, giving an area for the rectangle of $pi r^2$.
This can also be done physically by taking a paper circle and actually cutting it up and rearranging the pieces. This exercise also offers some introduction to (infinite) sequences.
$endgroup$
26
$begingroup$
This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to $pi r$.
$endgroup$
– Nicholas Wilson
Apr 8 '14 at 16:50
11
$begingroup$
@NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help?
$endgroup$
– ghoppe
Apr 8 '14 at 19:29
44
$begingroup$
@ghoppe Nice try! But it's all wiggly. We're lucky that $sin{theta} sim theta$ as $theta rightarrow 0$, so the actual perimeter of the bottom (which is known to be $pi$) does converge to the width of the rectangle. But - that nice property of $sin$ is exactly what we're trying to prove!
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 8:20
12
$begingroup$
@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof.
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 17:56
9
$begingroup$
@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent.
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– Ben Rowland
Dec 16 '14 at 22:16
|
show 9 more comments
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This is a neat little proof that the area of a circle is $pi r^2$, which I was first taught aged about 12 and it has stuck with me ever since. The circle is subdivided into equal pieces, then rearranged. As the number of pieces gets larger, the resulting shape gets closer and closer to a rectangle. It is obvious that the short side of this rectangle has length $r$, and a little thought will show that the two long sides each have a length half the circumference, or $pi r$, giving an area for the rectangle of $pi r^2$.
This can also be done physically by taking a paper circle and actually cutting it up and rearranging the pieces. This exercise also offers some introduction to (infinite) sequences.
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This is a neat little proof that the area of a circle is $pi r^2$, which I was first taught aged about 12 and it has stuck with me ever since. The circle is subdivided into equal pieces, then rearranged. As the number of pieces gets larger, the resulting shape gets closer and closer to a rectangle. It is obvious that the short side of this rectangle has length $r$, and a little thought will show that the two long sides each have a length half the circumference, or $pi r$, giving an area for the rectangle of $pi r^2$.
This can also be done physically by taking a paper circle and actually cutting it up and rearranging the pieces. This exercise also offers some introduction to (infinite) sequences.
edited Dec 30 '14 at 1:45
community wiki
2 revs, 2 users 93%
Ben Rowland
26
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This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to $pi r$.
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– Nicholas Wilson
Apr 8 '14 at 16:50
11
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@NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help?
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– ghoppe
Apr 8 '14 at 19:29
44
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@ghoppe Nice try! But it's all wiggly. We're lucky that $sin{theta} sim theta$ as $theta rightarrow 0$, so the actual perimeter of the bottom (which is known to be $pi$) does converge to the width of the rectangle. But - that nice property of $sin$ is exactly what we're trying to prove!
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– Nicholas Wilson
Apr 9 '14 at 8:20
12
$begingroup$
@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof.
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– Nicholas Wilson
Apr 9 '14 at 17:56
9
$begingroup$
@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent.
$endgroup$
– Ben Rowland
Dec 16 '14 at 22:16
|
show 9 more comments
26
$begingroup$
This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to $pi r$.
$endgroup$
– Nicholas Wilson
Apr 8 '14 at 16:50
11
$begingroup$
@NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help?
$endgroup$
– ghoppe
Apr 8 '14 at 19:29
44
$begingroup$
@ghoppe Nice try! But it's all wiggly. We're lucky that $sin{theta} sim theta$ as $theta rightarrow 0$, so the actual perimeter of the bottom (which is known to be $pi$) does converge to the width of the rectangle. But - that nice property of $sin$ is exactly what we're trying to prove!
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 8:20
12
$begingroup$
@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof.
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 17:56
9
$begingroup$
@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent.
$endgroup$
– Ben Rowland
Dec 16 '14 at 22:16
26
26
$begingroup$
This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to $pi r$.
$endgroup$
– Nicholas Wilson
Apr 8 '14 at 16:50
$begingroup$
This doesn't work for me :( Sadly, I find it non-obvious that the length of the bottom really will converge helpfully to $pi r$.
$endgroup$
– Nicholas Wilson
Apr 8 '14 at 16:50
11
11
$begingroup$
@NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help?
$endgroup$
– ghoppe
Apr 8 '14 at 19:29
$begingroup$
@NicholasWilson By definition, the circumference of a circle is π times the diameter of the circle. Here, we've cut the circle into slices, half flipped one way, half flipped the other. Therefore half the circumference (πd) is on the bottom. Half of the diameter is the radius, r. Does that help?
$endgroup$
– ghoppe
Apr 8 '14 at 19:29
44
44
$begingroup$
@ghoppe Nice try! But it's all wiggly. We're lucky that $sin{theta} sim theta$ as $theta rightarrow 0$, so the actual perimeter of the bottom (which is known to be $pi$) does converge to the width of the rectangle. But - that nice property of $sin$ is exactly what we're trying to prove!
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 8:20
$begingroup$
@ghoppe Nice try! But it's all wiggly. We're lucky that $sin{theta} sim theta$ as $theta rightarrow 0$, so the actual perimeter of the bottom (which is known to be $pi$) does converge to the width of the rectangle. But - that nice property of $sin$ is exactly what we're trying to prove!
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 8:20
12
12
$begingroup$
@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof.
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 17:56
$begingroup$
@ghoppe What would you visually infer about the perimeter of a Koch snowflake? I've had people assure me "It must be bounded!" Visual inferences are susceptible to error :( You need to invoke some limit arguments to convince me those arcs on the segments do converge. Similarly, inscribing polygons to determine the circumference sounds very dangerous (think what would happen if you tried that with a Koch snowflake) -- but bounding the area above and below by inscribed/exscribed polygons is definitely a sound proof.
$endgroup$
– Nicholas Wilson
Apr 9 '14 at 17:56
9
9
$begingroup$
@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent.
$endgroup$
– Ben Rowland
Dec 16 '14 at 22:16
$begingroup$
@NicholasWilson coming back to this discussion rather late, but perhaps I can convince you. The length of the bottom is always πr, it doesn't change with iteration so all we have to show is that the shape ends up as a rectangle. This is equivalent to stating that the line between two points on a circle tends to the tangent as the two points get closer together, which I believe is at least one definition of the tangent.
$endgroup$
– Ben Rowland
Dec 16 '14 at 22:16
|
show 9 more comments
$begingroup$
Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this.
It can be explained as follows. We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ Since distance satisfies condition 5, for any right angle triangle at all, not just ones whose legs are parallel to the axes, the square of the length of its hypotenuse is equal to the sum of the squares of the lengths of its legs.
This image shows that using that definition of distance, for any right angle triangle whose legs are parallel to the axes and have lengths $x in mathbb{R}^+$ and $y in mathbb{R}^+$, the area of a square with the hypotenuse as one of its edges is $(x - y)^2 + 2xy = x^2 + y^2 = (d(x, y))^2$. Combining that result with the fact that distance satisfies condition 5, we can show that for any right angle triangle, even with legs non parallel to the axes, the area of a square with its hypotenuse as its edge has an area equal to the sum of the squares of the lengths of its legs.
Sources:
- The validity of the proofs of the Pythagorean Theorem and the concept of area
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
- https://thept.weebly.com/the-theorem.html
$endgroup$
86
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This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple.
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– Sparr
Apr 7 '14 at 16:06
113
$begingroup$
Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote.
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– B T
Aug 27 '14 at 21:12
12
$begingroup$
@BT, do you consider the fact that $1^3 + 2^3 neq 18^3$ a demonstration of Fermat's Last Theorem?
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– Daniel McLaury
Aug 28 '14 at 1:17
20
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@DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool.
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– Samuel Yusim
Aug 28 '14 at 5:30
8
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@GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice.
$endgroup$
– Samuel Yusim
Sep 5 '14 at 0:06
|
show 13 more comments
$begingroup$
Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this.
It can be explained as follows. We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ Since distance satisfies condition 5, for any right angle triangle at all, not just ones whose legs are parallel to the axes, the square of the length of its hypotenuse is equal to the sum of the squares of the lengths of its legs.
This image shows that using that definition of distance, for any right angle triangle whose legs are parallel to the axes and have lengths $x in mathbb{R}^+$ and $y in mathbb{R}^+$, the area of a square with the hypotenuse as one of its edges is $(x - y)^2 + 2xy = x^2 + y^2 = (d(x, y))^2$. Combining that result with the fact that distance satisfies condition 5, we can show that for any right angle triangle, even with legs non parallel to the axes, the area of a square with its hypotenuse as its edge has an area equal to the sum of the squares of the lengths of its legs.
Sources:
- The validity of the proofs of the Pythagorean Theorem and the concept of area
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
- https://thept.weebly.com/the-theorem.html
$endgroup$
86
$begingroup$
This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple.
$endgroup$
– Sparr
Apr 7 '14 at 16:06
113
$begingroup$
Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote.
$endgroup$
– B T
Aug 27 '14 at 21:12
12
$begingroup$
@BT, do you consider the fact that $1^3 + 2^3 neq 18^3$ a demonstration of Fermat's Last Theorem?
$endgroup$
– Daniel McLaury
Aug 28 '14 at 1:17
20
$begingroup$
@DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool.
$endgroup$
– Samuel Yusim
Aug 28 '14 at 5:30
8
$begingroup$
@GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice.
$endgroup$
– Samuel Yusim
Sep 5 '14 at 0:06
|
show 13 more comments
$begingroup$
Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this.
It can be explained as follows. We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ Since distance satisfies condition 5, for any right angle triangle at all, not just ones whose legs are parallel to the axes, the square of the length of its hypotenuse is equal to the sum of the squares of the lengths of its legs.
This image shows that using that definition of distance, for any right angle triangle whose legs are parallel to the axes and have lengths $x in mathbb{R}^+$ and $y in mathbb{R}^+$, the area of a square with the hypotenuse as one of its edges is $(x - y)^2 + 2xy = x^2 + y^2 = (d(x, y))^2$. Combining that result with the fact that distance satisfies condition 5, we can show that for any right angle triangle, even with legs non parallel to the axes, the area of a square with its hypotenuse as its edge has an area equal to the sum of the squares of the lengths of its legs.
Sources:
- The validity of the proofs of the Pythagorean Theorem and the concept of area
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
- https://thept.weebly.com/the-theorem.html
$endgroup$
Here is a very insightful waterproof demonstration of the Pythagorean theorem. Also there is a video about this.
It can be explained as follows. We seek a definition of distance from any point in $mathbb{R}^2$ to $mathbb{R}^2$, a function from $(mathbb{R}^2)^2$ to $mathbb{R}$ that satisfies the following properties.
- For any points $(x, y)$ and $(z, w)$, $d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- For any point $(x, y)$, $d((0, 0), (x, y))$ is nonnegative
- For any nonnegative real number $x$, $d((0, 0), (x, 0)) = x$
- For any point $(x, y)$, $d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- For any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
Suppose a function $d$ from $(mathbb{R}^2)^2$ to $mathbb{R}$ satisfies those conditions, then for any point $(x, y)$, $d((0, 0), (x, y))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$ so $d((0, 0), (x, y)) = sqrt{x^2 + y^2}$ so for any points $(x, y)$ and $(z, w)$, $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ Now I will show that $d((x, y), (z, w)) = sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties. It's trivial to show that it satisfies the first 4 conditions. It also satisfies the fifth condition because for any points $(x, y)$ and $(z, w)$, $d((0, 0), (xz - yw, xw + yz)) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$
As a result of this, from now on, I will define the distance from any point $(x, y)$ to any point $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$ and denote it as $d((x, y), (z, w))$. I will also use $d(x, y)$ as shorthand for $d((0, 0), (x, y))$ Since distance satisfies condition 5, for any right angle triangle at all, not just ones whose legs are parallel to the axes, the square of the length of its hypotenuse is equal to the sum of the squares of the lengths of its legs.
This image shows that using that definition of distance, for any right angle triangle whose legs are parallel to the axes and have lengths $x in mathbb{R}^+$ and $y in mathbb{R}^+$, the area of a square with the hypotenuse as one of its edges is $(x - y)^2 + 2xy = x^2 + y^2 = (d(x, y))^2$. Combining that result with the fact that distance satisfies condition 5, we can show that for any right angle triangle, even with legs non parallel to the axes, the area of a square with its hypotenuse as its edge has an area equal to the sum of the squares of the lengths of its legs.
Sources:
- The validity of the proofs of the Pythagorean Theorem and the concept of area
- https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
- https://thept.weebly.com/the-theorem.html
edited Jan 28 at 6:09
community wiki
3 revs, 3 users 70%
Timothy
86
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This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple.
$endgroup$
– Sparr
Apr 7 '14 at 16:06
113
$begingroup$
Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote.
$endgroup$
– B T
Aug 27 '14 at 21:12
12
$begingroup$
@BT, do you consider the fact that $1^3 + 2^3 neq 18^3$ a demonstration of Fermat's Last Theorem?
$endgroup$
– Daniel McLaury
Aug 28 '14 at 1:17
20
$begingroup$
@DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool.
$endgroup$
– Samuel Yusim
Aug 28 '14 at 5:30
8
$begingroup$
@GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice.
$endgroup$
– Samuel Yusim
Sep 5 '14 at 0:06
|
show 13 more comments
86
$begingroup$
This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple.
$endgroup$
– Sparr
Apr 7 '14 at 16:06
113
$begingroup$
Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote.
$endgroup$
– B T
Aug 27 '14 at 21:12
12
$begingroup$
@BT, do you consider the fact that $1^3 + 2^3 neq 18^3$ a demonstration of Fermat's Last Theorem?
$endgroup$
– Daniel McLaury
Aug 28 '14 at 1:17
20
$begingroup$
@DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool.
$endgroup$
– Samuel Yusim
Aug 28 '14 at 5:30
8
$begingroup$
@GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice.
$endgroup$
– Samuel Yusim
Sep 5 '14 at 0:06
86
86
$begingroup$
This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple.
$endgroup$
– Sparr
Apr 7 '14 at 16:06
$begingroup$
This does not actually demonstrate the pythagorean theorem. It is showing that for one single right triangle, the theorem holds. This could be a coincidence. Someone seeing this might think that 3,4,5 is the only pythagorean triple.
$endgroup$
– Sparr
Apr 7 '14 at 16:06
113
113
$begingroup$
Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote.
$endgroup$
– B T
Aug 27 '14 at 21:12
$begingroup$
Sparr, you're incorrect. This does indeed demonstrate the pythagorean theorem. What you mean is that it doesn't prove it. All the demonstrations on this page are simply examples, not the infinity of all possibilities. Think before you speak. And all you upvoters, think before you upvote.
$endgroup$
– B T
Aug 27 '14 at 21:12
12
12
$begingroup$
@BT, do you consider the fact that $1^3 + 2^3 neq 18^3$ a demonstration of Fermat's Last Theorem?
$endgroup$
– Daniel McLaury
Aug 28 '14 at 1:17
$begingroup$
@BT, do you consider the fact that $1^3 + 2^3 neq 18^3$ a demonstration of Fermat's Last Theorem?
$endgroup$
– Daniel McLaury
Aug 28 '14 at 1:17
20
20
$begingroup$
@DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool.
$endgroup$
– Samuel Yusim
Aug 28 '14 at 5:30
$begingroup$
@DanielMcLaury Sure it's a demonstration of FLT, but it's not an interesting one. This gif, however, is interesting because it looks cool.
$endgroup$
– Samuel Yusim
Aug 28 '14 at 5:30
8
8
$begingroup$
@GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice.
$endgroup$
– Samuel Yusim
Sep 5 '14 at 0:06
$begingroup$
@GrumpyParsnip Before you learn why something is true, you might want to do a sanity check to verify that it probably is. This does that, while looking cool. The only thing that I really care about is that it looks cool, though. Mathematics is a very aesthetic thing, so I'm allowed to like something just because it looks nice.
$endgroup$
– Samuel Yusim
Sep 5 '14 at 0:06
|
show 13 more comments
$begingroup$
A visual explanation of a Taylor series:
$f(0)+frac {f'(0)}{1!} x+ frac{f''(0)}{2!} x^2+frac{f^{(3)}(0)}{3!}x^3+ cdots$
or
$f(a)+frac {f'(a)}{1!} (x-a)+ frac{f''(a)}{2!} (x-a)^2+frac{f^{(3)}(a)}{3!}(x-a)^3+ cdots$
When you think about it, it's quite beautiful that as you add each term it wraps around the curve.
$endgroup$
17
$begingroup$
It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have acosinefunction built-in, but will fly through an equation such asx => 1 - (x^2)/(2!) + (x^4)/(4!) - ...and from this equation, you could build a function that accepts aprecisionparameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this)
$endgroup$
– sleblanc
Apr 3 '14 at 21:55
1
$begingroup$
@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally
$endgroup$
– joshlk
Apr 4 '14 at 8:15
3
$begingroup$
@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas.
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:05
$begingroup$
You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have theFSINinstruction... If anything, I should have said [most] instead. ;-)
$endgroup$
– sleblanc
Apr 12 '14 at 14:31
2
$begingroup$
@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition.
$endgroup$
– Calmarius
Aug 27 '14 at 16:57
add a comment |
$begingroup$
A visual explanation of a Taylor series:
$f(0)+frac {f'(0)}{1!} x+ frac{f''(0)}{2!} x^2+frac{f^{(3)}(0)}{3!}x^3+ cdots$
or
$f(a)+frac {f'(a)}{1!} (x-a)+ frac{f''(a)}{2!} (x-a)^2+frac{f^{(3)}(a)}{3!}(x-a)^3+ cdots$
When you think about it, it's quite beautiful that as you add each term it wraps around the curve.
$endgroup$
17
$begingroup$
It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have acosinefunction built-in, but will fly through an equation such asx => 1 - (x^2)/(2!) + (x^4)/(4!) - ...and from this equation, you could build a function that accepts aprecisionparameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this)
$endgroup$
– sleblanc
Apr 3 '14 at 21:55
1
$begingroup$
@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally
$endgroup$
– joshlk
Apr 4 '14 at 8:15
3
$begingroup$
@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas.
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:05
$begingroup$
You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have theFSINinstruction... If anything, I should have said [most] instead. ;-)
$endgroup$
– sleblanc
Apr 12 '14 at 14:31
2
$begingroup$
@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition.
$endgroup$
– Calmarius
Aug 27 '14 at 16:57
add a comment |
$begingroup$
A visual explanation of a Taylor series:
$f(0)+frac {f'(0)}{1!} x+ frac{f''(0)}{2!} x^2+frac{f^{(3)}(0)}{3!}x^3+ cdots$
or
$f(a)+frac {f'(a)}{1!} (x-a)+ frac{f''(a)}{2!} (x-a)^2+frac{f^{(3)}(a)}{3!}(x-a)^3+ cdots$
When you think about it, it's quite beautiful that as you add each term it wraps around the curve.
$endgroup$
A visual explanation of a Taylor series:
$f(0)+frac {f'(0)}{1!} x+ frac{f''(0)}{2!} x^2+frac{f^{(3)}(0)}{3!}x^3+ cdots$
or
$f(a)+frac {f'(a)}{1!} (x-a)+ frac{f''(a)}{2!} (x-a)^2+frac{f^{(3)}(a)}{3!}(x-a)^3+ cdots$
When you think about it, it's quite beautiful that as you add each term it wraps around the curve.
edited Mar 8 '15 at 10:50
community wiki
4 revs, 2 users 63%
Josh
17
$begingroup$
It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have acosinefunction built-in, but will fly through an equation such asx => 1 - (x^2)/(2!) + (x^4)/(4!) - ...and from this equation, you could build a function that accepts aprecisionparameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this)
$endgroup$
– sleblanc
Apr 3 '14 at 21:55
1
$begingroup$
@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally
$endgroup$
– joshlk
Apr 4 '14 at 8:15
3
$begingroup$
@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas.
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:05
$begingroup$
You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have theFSINinstruction... If anything, I should have said [most] instead. ;-)
$endgroup$
– sleblanc
Apr 12 '14 at 14:31
2
$begingroup$
@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition.
$endgroup$
– Calmarius
Aug 27 '14 at 16:57
add a comment |
17
$begingroup$
It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have acosinefunction built-in, but will fly through an equation such asx => 1 - (x^2)/(2!) + (x^4)/(4!) - ...and from this equation, you could build a function that accepts aprecisionparameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this)
$endgroup$
– sleblanc
Apr 3 '14 at 21:55
1
$begingroup$
@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally
$endgroup$
– joshlk
Apr 4 '14 at 8:15
3
$begingroup$
@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas.
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:05
$begingroup$
You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have theFSINinstruction... If anything, I should have said [most] instead. ;-)
$endgroup$
– sleblanc
Apr 12 '14 at 14:31
2
$begingroup$
@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition.
$endgroup$
– Calmarius
Aug 27 '14 at 16:57
17
17
$begingroup$
It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have a
cosine function built-in, but will fly through an equation such as x => 1 - (x^2)/(2!) + (x^4)/(4!) - ... and from this equation, you could build a function that accepts a precision parameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this)$endgroup$
– sleblanc
Apr 3 '14 at 21:55
$begingroup$
It is interesting to note that Taylor series are relatively very precise near their center, which makes them especially useful in computer science when you need a precise but still fast approximation. For example, processors do not have a
cosine function built-in, but will fly through an equation such as x => 1 - (x^2)/(2!) + (x^4)/(4!) - ... and from this equation, you could build a function that accepts a precision parameter so that the call recurses until higher accuracy is achieved. (note: there are many more efficient ways to do this)$endgroup$
– sleblanc
Apr 3 '14 at 21:55
1
1
$begingroup$
@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally
$endgroup$
– joshlk
Apr 4 '14 at 8:15
$begingroup$
@sebleblanc It is just amazing that petty much any function can represented with a Taylor series locally
$endgroup$
– joshlk
Apr 4 '14 at 8:15
3
3
$begingroup$
@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas.
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:05
$begingroup$
@sebleblanc Should have said "[some] processors" instead. x86 processors have had a cosine function for some time now with the FCOS instruction. It doesn't do it with a taylor series as far as I can tell. What I assume it does is use a lookup table followed by some sum/difference formulas.
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:05
$begingroup$
You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have the
FSIN instruction... If anything, I should have said [most] instead. ;-)$endgroup$
– sleblanc
Apr 12 '14 at 14:31
$begingroup$
You are right that X87 co-processors are omni-present, but there are far more embedded CPUs in the world than there are X86/87 CPUs. Just think of how many devices you interact with that are not strictly computers. ///// From my desk, I can see three computer monitors, one TV, two video game consoles, one oscilloscope, one modern refrigerator, two cordless phones, a wireless mouse, a microcontroller dev board, a Raspberry Pi, my cell phone, my credit card. All of these have a processor that does not have the
FSIN instruction... If anything, I should have said [most] instead. ;-)$endgroup$
– sleblanc
Apr 12 '14 at 14:31
2
2
$begingroup$
@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition.
$endgroup$
– Calmarius
Aug 27 '14 at 16:57
$begingroup$
@sebleblanc The x86/87 FPU uses CORDIC which lets you compute trigonometric functions in hardware using only bit shifts and addition.
$endgroup$
– Calmarius
Aug 27 '14 at 16:57
add a comment |
$begingroup$
When I look up "area of a rhombus" on Google images, I find plenty of disappointing images like this one:
which show the formula, but fail to show why the formula works. That's why I really appreciate this image instead:
which, with a little bit of careful thought, illustrates why the product of the diagonals equals twice the area of the rhombus.
EDIT: Some have mentioned in comments that that second diagram is more complicated than it needs to be. Something like this would work as well:
My main objective is to offer students something that encourages them to think about why a formula works, not just what numbers to plug into an equation to get an answer.
As a side note, the following story is not exactly "visually stunning," but it put an indelible imprint on my mind, and affected the way I teach today. A very gifted Jr. High math teacher was teaching us about volume. I suppose just every about school system has a place in the curriculum where students are required learn how to calculate the volume of a pyramid. Sadly, most teachers probably accomplish this by simply writing the formula on the board, and assigning a few plug-and-chug homework problems.
No wonder that, when I ask my college students if they can tell me the formula for the volume of a pyramid, fewer than 5% can.
Instead, building upon lessons from earlier that week, our math teacher began the lesson by saying:
We've learned how to calculate the volume of a prism: we simply multiply the area of the base times the height. That's easy. But what if we don't have a prism? What if we have a pyramid?
At this point, she rummaged through her box of math props, and pulled out a clear plastic cube, and a clear plastic pyramid. She continued by putting the pyramid atop the cube, and then dropping the pyramid, point-side down inside the cube:
She continued:
These have the same base, and they are the same height. How many of these pyramids do you suppose would fit in this cube? Two? Two-and-a-half? Three?
Then she picked one student from the front row, and instructed him to walk them down the hallway:
Go down to the water fountain, and fill this pyramid up with water, and tell us how many it takes to fill up the cube.
The class sat in silence for about a full minute or so, until he walked back in the room. She asked him to give his report.
"Three," he said.
She pressed him, giving him a hard look. "Exactly three?"
"Exactly three," he affirmed.
Then, she looked around the room:
"Who here can tell me the formula I use to get the volume of a pyramid?" she asked.
One girl raised her hand: "One-third the base times the height?"
I've never forgotten that formula, because, instead of having it told to us, we were asked to derive it. Not only have I remembered the formula, I can even tell you the name of boy who went to the water fountain, and the girl who told us all the formula (David and Jill).
Given the upvoted comment, If high school math just used a fraction of the resources here, we'd have way more mathematicians, I hope you don't mind me sharing this story here. Powerful visuals can happen even in the imagination. I never got to see that cube filling up with water, but everything else in the story I vividly remember.
Incidentally, this same teacher introduced us to the concept of pi by asking us to find something circular in our house (“like a plate or a coffee can”), measuring the circumference and the diameter, and dividing the one number by the other. I can still see her studying the data on the chalkboard the next day – all 20 or so numbers just a smidgeon over 3 – marveling how, even though we all probably measured differently-sized circles, the answers were coming out remarkably similar, “as if maybe that ratio is some kind of constant or something...”
$endgroup$
24
$begingroup$
^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways.
$endgroup$
– daviewales
Apr 5 '14 at 12:34
21
$begingroup$
@daviewales It's been started - the mathematics educators SE.
$endgroup$
– user122283
Apr 5 '14 at 14:45
8
$begingroup$
Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$frac{1}{2}frac{d_1}{2}d_2=frac{1}{2}d_1frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.)
$endgroup$
– Will Orrick
Apr 6 '14 at 2:36
20
$begingroup$
The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite.
$endgroup$
– LaC
Apr 7 '14 at 3:47
5
$begingroup$
@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High.
$endgroup$
– J.R.
Feb 22 '15 at 9:44
|
show 21 more comments
$begingroup$
When I look up "area of a rhombus" on Google images, I find plenty of disappointing images like this one:
which show the formula, but fail to show why the formula works. That's why I really appreciate this image instead:
which, with a little bit of careful thought, illustrates why the product of the diagonals equals twice the area of the rhombus.
EDIT: Some have mentioned in comments that that second diagram is more complicated than it needs to be. Something like this would work as well:
My main objective is to offer students something that encourages them to think about why a formula works, not just what numbers to plug into an equation to get an answer.
As a side note, the following story is not exactly "visually stunning," but it put an indelible imprint on my mind, and affected the way I teach today. A very gifted Jr. High math teacher was teaching us about volume. I suppose just every about school system has a place in the curriculum where students are required learn how to calculate the volume of a pyramid. Sadly, most teachers probably accomplish this by simply writing the formula on the board, and assigning a few plug-and-chug homework problems.
No wonder that, when I ask my college students if they can tell me the formula for the volume of a pyramid, fewer than 5% can.
Instead, building upon lessons from earlier that week, our math teacher began the lesson by saying:
We've learned how to calculate the volume of a prism: we simply multiply the area of the base times the height. That's easy. But what if we don't have a prism? What if we have a pyramid?
At this point, she rummaged through her box of math props, and pulled out a clear plastic cube, and a clear plastic pyramid. She continued by putting the pyramid atop the cube, and then dropping the pyramid, point-side down inside the cube:
She continued:
These have the same base, and they are the same height. How many of these pyramids do you suppose would fit in this cube? Two? Two-and-a-half? Three?
Then she picked one student from the front row, and instructed him to walk them down the hallway:
Go down to the water fountain, and fill this pyramid up with water, and tell us how many it takes to fill up the cube.
The class sat in silence for about a full minute or so, until he walked back in the room. She asked him to give his report.
"Three," he said.
She pressed him, giving him a hard look. "Exactly three?"
"Exactly three," he affirmed.
Then, she looked around the room:
"Who here can tell me the formula I use to get the volume of a pyramid?" she asked.
One girl raised her hand: "One-third the base times the height?"
I've never forgotten that formula, because, instead of having it told to us, we were asked to derive it. Not only have I remembered the formula, I can even tell you the name of boy who went to the water fountain, and the girl who told us all the formula (David and Jill).
Given the upvoted comment, If high school math just used a fraction of the resources here, we'd have way more mathematicians, I hope you don't mind me sharing this story here. Powerful visuals can happen even in the imagination. I never got to see that cube filling up with water, but everything else in the story I vividly remember.
Incidentally, this same teacher introduced us to the concept of pi by asking us to find something circular in our house (“like a plate or a coffee can”), measuring the circumference and the diameter, and dividing the one number by the other. I can still see her studying the data on the chalkboard the next day – all 20 or so numbers just a smidgeon over 3 – marveling how, even though we all probably measured differently-sized circles, the answers were coming out remarkably similar, “as if maybe that ratio is some kind of constant or something...”
$endgroup$
24
$begingroup$
^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways.
$endgroup$
– daviewales
Apr 5 '14 at 12:34
21
$begingroup$
@daviewales It's been started - the mathematics educators SE.
$endgroup$
– user122283
Apr 5 '14 at 14:45
8
$begingroup$
Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$frac{1}{2}frac{d_1}{2}d_2=frac{1}{2}d_1frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.)
$endgroup$
– Will Orrick
Apr 6 '14 at 2:36
20
$begingroup$
The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite.
$endgroup$
– LaC
Apr 7 '14 at 3:47
5
$begingroup$
@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High.
$endgroup$
– J.R.
Feb 22 '15 at 9:44
|
show 21 more comments
$begingroup$
When I look up "area of a rhombus" on Google images, I find plenty of disappointing images like this one:
which show the formula, but fail to show why the formula works. That's why I really appreciate this image instead:
which, with a little bit of careful thought, illustrates why the product of the diagonals equals twice the area of the rhombus.
EDIT: Some have mentioned in comments that that second diagram is more complicated than it needs to be. Something like this would work as well:
My main objective is to offer students something that encourages them to think about why a formula works, not just what numbers to plug into an equation to get an answer.
As a side note, the following story is not exactly "visually stunning," but it put an indelible imprint on my mind, and affected the way I teach today. A very gifted Jr. High math teacher was teaching us about volume. I suppose just every about school system has a place in the curriculum where students are required learn how to calculate the volume of a pyramid. Sadly, most teachers probably accomplish this by simply writing the formula on the board, and assigning a few plug-and-chug homework problems.
No wonder that, when I ask my college students if they can tell me the formula for the volume of a pyramid, fewer than 5% can.
Instead, building upon lessons from earlier that week, our math teacher began the lesson by saying:
We've learned how to calculate the volume of a prism: we simply multiply the area of the base times the height. That's easy. But what if we don't have a prism? What if we have a pyramid?
At this point, she rummaged through her box of math props, and pulled out a clear plastic cube, and a clear plastic pyramid. She continued by putting the pyramid atop the cube, and then dropping the pyramid, point-side down inside the cube:
She continued:
These have the same base, and they are the same height. How many of these pyramids do you suppose would fit in this cube? Two? Two-and-a-half? Three?
Then she picked one student from the front row, and instructed him to walk them down the hallway:
Go down to the water fountain, and fill this pyramid up with water, and tell us how many it takes to fill up the cube.
The class sat in silence for about a full minute or so, until he walked back in the room. She asked him to give his report.
"Three," he said.
She pressed him, giving him a hard look. "Exactly three?"
"Exactly three," he affirmed.
Then, she looked around the room:
"Who here can tell me the formula I use to get the volume of a pyramid?" she asked.
One girl raised her hand: "One-third the base times the height?"
I've never forgotten that formula, because, instead of having it told to us, we were asked to derive it. Not only have I remembered the formula, I can even tell you the name of boy who went to the water fountain, and the girl who told us all the formula (David and Jill).
Given the upvoted comment, If high school math just used a fraction of the resources here, we'd have way more mathematicians, I hope you don't mind me sharing this story here. Powerful visuals can happen even in the imagination. I never got to see that cube filling up with water, but everything else in the story I vividly remember.
Incidentally, this same teacher introduced us to the concept of pi by asking us to find something circular in our house (“like a plate or a coffee can”), measuring the circumference and the diameter, and dividing the one number by the other. I can still see her studying the data on the chalkboard the next day – all 20 or so numbers just a smidgeon over 3 – marveling how, even though we all probably measured differently-sized circles, the answers were coming out remarkably similar, “as if maybe that ratio is some kind of constant or something...”
$endgroup$
When I look up "area of a rhombus" on Google images, I find plenty of disappointing images like this one:
which show the formula, but fail to show why the formula works. That's why I really appreciate this image instead:
which, with a little bit of careful thought, illustrates why the product of the diagonals equals twice the area of the rhombus.
EDIT: Some have mentioned in comments that that second diagram is more complicated than it needs to be. Something like this would work as well:
My main objective is to offer students something that encourages them to think about why a formula works, not just what numbers to plug into an equation to get an answer.
As a side note, the following story is not exactly "visually stunning," but it put an indelible imprint on my mind, and affected the way I teach today. A very gifted Jr. High math teacher was teaching us about volume. I suppose just every about school system has a place in the curriculum where students are required learn how to calculate the volume of a pyramid. Sadly, most teachers probably accomplish this by simply writing the formula on the board, and assigning a few plug-and-chug homework problems.
No wonder that, when I ask my college students if they can tell me the formula for the volume of a pyramid, fewer than 5% can.
Instead, building upon lessons from earlier that week, our math teacher began the lesson by saying:
We've learned how to calculate the volume of a prism: we simply multiply the area of the base times the height. That's easy. But what if we don't have a prism? What if we have a pyramid?
At this point, she rummaged through her box of math props, and pulled out a clear plastic cube, and a clear plastic pyramid. She continued by putting the pyramid atop the cube, and then dropping the pyramid, point-side down inside the cube:
She continued:
These have the same base, and they are the same height. How many of these pyramids do you suppose would fit in this cube? Two? Two-and-a-half? Three?
Then she picked one student from the front row, and instructed him to walk them down the hallway:
Go down to the water fountain, and fill this pyramid up with water, and tell us how many it takes to fill up the cube.
The class sat in silence for about a full minute or so, until he walked back in the room. She asked him to give his report.
"Three," he said.
She pressed him, giving him a hard look. "Exactly three?"
"Exactly three," he affirmed.
Then, she looked around the room:
"Who here can tell me the formula I use to get the volume of a pyramid?" she asked.
One girl raised her hand: "One-third the base times the height?"
I've never forgotten that formula, because, instead of having it told to us, we were asked to derive it. Not only have I remembered the formula, I can even tell you the name of boy who went to the water fountain, and the girl who told us all the formula (David and Jill).
Given the upvoted comment, If high school math just used a fraction of the resources here, we'd have way more mathematicians, I hope you don't mind me sharing this story here. Powerful visuals can happen even in the imagination. I never got to see that cube filling up with water, but everything else in the story I vividly remember.
Incidentally, this same teacher introduced us to the concept of pi by asking us to find something circular in our house (“like a plate or a coffee can”), measuring the circumference and the diameter, and dividing the one number by the other. I can still see her studying the data on the chalkboard the next day – all 20 or so numbers just a smidgeon over 3 – marveling how, even though we all probably measured differently-sized circles, the answers were coming out remarkably similar, “as if maybe that ratio is some kind of constant or something...”
edited Aug 27 '14 at 18:25
community wiki
4 revs, 3 users 99%
J.R.
24
$begingroup$
^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways.
$endgroup$
– daviewales
Apr 5 '14 at 12:34
21
$begingroup$
@daviewales It's been started - the mathematics educators SE.
$endgroup$
– user122283
Apr 5 '14 at 14:45
8
$begingroup$
Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$frac{1}{2}frac{d_1}{2}d_2=frac{1}{2}d_1frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.)
$endgroup$
– Will Orrick
Apr 6 '14 at 2:36
20
$begingroup$
The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite.
$endgroup$
– LaC
Apr 7 '14 at 3:47
5
$begingroup$
@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High.
$endgroup$
– J.R.
Feb 22 '15 at 9:44
|
show 21 more comments
24
$begingroup$
^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways.
$endgroup$
– daviewales
Apr 5 '14 at 12:34
21
$begingroup$
@daviewales It's been started - the mathematics educators SE.
$endgroup$
– user122283
Apr 5 '14 at 14:45
8
$begingroup$
Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$frac{1}{2}frac{d_1}{2}d_2=frac{1}{2}d_1frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.)
$endgroup$
– Will Orrick
Apr 6 '14 at 2:36
20
$begingroup$
The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite.
$endgroup$
– LaC
Apr 7 '14 at 3:47
5
$begingroup$
@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High.
$endgroup$
– J.R.
Feb 22 '15 at 9:44
24
24
$begingroup$
^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways.
$endgroup$
– daviewales
Apr 5 '14 at 12:34
$begingroup$
^^1 million upvotes for the stories about maths education. I am writing both of these down for future reference. In fact, I think we should start a website with instructions to help primary teachers teach mathematical concepts in interesting ways.
$endgroup$
– daviewales
Apr 5 '14 at 12:34
21
21
$begingroup$
@daviewales It's been started - the mathematics educators SE.
$endgroup$
– user122283
Apr 5 '14 at 14:45
$begingroup$
@daviewales It's been started - the mathematics educators SE.
$endgroup$
– user122283
Apr 5 '14 at 14:45
8
8
$begingroup$
Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$frac{1}{2}frac{d_1}{2}d_2=frac{1}{2}d_1frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.)
$endgroup$
– Will Orrick
Apr 6 '14 at 2:36
$begingroup$
Great stories. I do, however, think your first rhombus is fine. I actually find it easier to understand the result using that image than the second. Either diagonal splits the rhombus into two triangles, each of area $$frac{1}{2}frac{d_1}{2}d_2=frac{1}{2}d_1frac{d_2}{2}.$$ (Is there some aspect-ratio issue with your image? The white rectangle should be a square, but doesn't look like it on my screen.)
$endgroup$
– Will Orrick
Apr 6 '14 at 2:36
20
20
$begingroup$
The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite.
$endgroup$
– LaC
Apr 7 '14 at 3:47
$begingroup$
The first rhombus picture is fine. Just draw a rectangle around it, and it's plain to see that the four triangles that form the rhombus cover half of the rectangle. In contrast, the second picture is so confusing that you yourself got it wrong: you write that it "illustrates why the product of the diagonals equals half of the area of the rhombus", while the ratio is the opposite.
$endgroup$
– LaC
Apr 7 '14 at 3:47
5
5
$begingroup$
@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High.
$endgroup$
– J.R.
Feb 22 '15 at 9:44
$begingroup$
@Peter - I'll admit that I'm using the word derive a bit loosely for a mathematics forum (it's a soft answer to a soft question), but it's too bad you see this as "dumbing down" and "a waste of time." Most 7th-grade teachers charged with teaching volume would simply write the formula on the board, and it would be well-forgotten by the end of summer. Her technique might have been weak insofar as mathematical rigor goes, but the pedagogy was very strong. I assure you, this woman was not one to "dumb down" anything; I remember re-learning concepts in 11th-grade that she taught us in Jr High.
$endgroup$
– J.R.
Feb 22 '15 at 9:44
|
show 21 more comments
$begingroup$
As I was in school, a supply teacher brought a scale to lesson:
He gave us several weights that were labeled and about 4 weights without labels (let's call them $A, B, C, D$). Then he told us we should find out the weight of the unlabeled weights. $A$ was very easy as there was a weight $E$ with weight($A$) = weight($E$).
I think at least two of them had the same weight and we could only get them into balance with a combination of the labeled weights. The last one was harder. We had to put a labeled weight on the side of the last one to get the weight.
Then he told us how this can be solved on paper without having the weights. So he introduced us to the concept of equations. That was a truly amazing day. Such an important concept explained with a neat way.
$endgroup$
$begingroup$
I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set.
$endgroup$
– Guillermo Garza
Apr 7 '14 at 0:34
13
$begingroup$
@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 cdot m_1 = a_2 cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way.
$endgroup$
– Martin Thoma
Apr 7 '14 at 5:58
2
$begingroup$
I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind!
$endgroup$
– Guillermo Garza
Apr 8 '14 at 0:50
$begingroup$
@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics.
$endgroup$
– Martin Thoma
Apr 8 '14 at 1:59
add a comment |
$begingroup$
As I was in school, a supply teacher brought a scale to lesson:
He gave us several weights that were labeled and about 4 weights without labels (let's call them $A, B, C, D$). Then he told us we should find out the weight of the unlabeled weights. $A$ was very easy as there was a weight $E$ with weight($A$) = weight($E$).
I think at least two of them had the same weight and we could only get them into balance with a combination of the labeled weights. The last one was harder. We had to put a labeled weight on the side of the last one to get the weight.
Then he told us how this can be solved on paper without having the weights. So he introduced us to the concept of equations. That was a truly amazing day. Such an important concept explained with a neat way.
$endgroup$
$begingroup$
I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set.
$endgroup$
– Guillermo Garza
Apr 7 '14 at 0:34
13
$begingroup$
@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 cdot m_1 = a_2 cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way.
$endgroup$
– Martin Thoma
Apr 7 '14 at 5:58
2
$begingroup$
I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind!
$endgroup$
– Guillermo Garza
Apr 8 '14 at 0:50
$begingroup$
@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics.
$endgroup$
– Martin Thoma
Apr 8 '14 at 1:59
add a comment |
$begingroup$
As I was in school, a supply teacher brought a scale to lesson:
He gave us several weights that were labeled and about 4 weights without labels (let's call them $A, B, C, D$). Then he told us we should find out the weight of the unlabeled weights. $A$ was very easy as there was a weight $E$ with weight($A$) = weight($E$).
I think at least two of them had the same weight and we could only get them into balance with a combination of the labeled weights. The last one was harder. We had to put a labeled weight on the side of the last one to get the weight.
Then he told us how this can be solved on paper without having the weights. So he introduced us to the concept of equations. That was a truly amazing day. Such an important concept explained with a neat way.
$endgroup$
As I was in school, a supply teacher brought a scale to lesson:
He gave us several weights that were labeled and about 4 weights without labels (let's call them $A, B, C, D$). Then he told us we should find out the weight of the unlabeled weights. $A$ was very easy as there was a weight $E$ with weight($A$) = weight($E$).
I think at least two of them had the same weight and we could only get them into balance with a combination of the labeled weights. The last one was harder. We had to put a labeled weight on the side of the last one to get the weight.
Then he told us how this can be solved on paper without having the weights. So he introduced us to the concept of equations. That was a truly amazing day. Such an important concept explained with a neat way.
edited Sep 9 '14 at 21:58
community wiki
2 revs, 2 users 95%
moose
$begingroup$
I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set.
$endgroup$
– Guillermo Garza
Apr 7 '14 at 0:34
13
$begingroup$
@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 cdot m_1 = a_2 cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way.
$endgroup$
– Martin Thoma
Apr 7 '14 at 5:58
2
$begingroup$
I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind!
$endgroup$
– Guillermo Garza
Apr 8 '14 at 0:50
$begingroup$
@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics.
$endgroup$
– Martin Thoma
Apr 8 '14 at 1:59
add a comment |
$begingroup$
I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set.
$endgroup$
– Guillermo Garza
Apr 7 '14 at 0:34
13
$begingroup$
@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 cdot m_1 = a_2 cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way.
$endgroup$
– Martin Thoma
Apr 7 '14 at 5:58
2
$begingroup$
I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind!
$endgroup$
– Guillermo Garza
Apr 8 '14 at 0:50
$begingroup$
@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics.
$endgroup$
– Martin Thoma
Apr 8 '14 at 1:59
$begingroup$
I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set.
$endgroup$
– Guillermo Garza
Apr 7 '14 at 0:34
$begingroup$
I do the same thing... except with drawings instead of an actual scale and weights. I need to get this set.
$endgroup$
– Guillermo Garza
Apr 7 '14 at 0:34
13
13
$begingroup$
@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 cdot m_1 = a_2 cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way.
$endgroup$
– Martin Thoma
Apr 7 '14 at 5:58
$begingroup$
@MeemoGarza although drawings are nice, too, I think that an actual device has some advantages: Pupils rather get excited by a real device than by a drawing; it's easier to accept something that you can touch and play with where you instantly get feedback if what you did was correct; as soon as pupils got familiar with the concept of equations, you can explain the physics behind scales ($a_1 cdot m_1 = a_2 cdot m_2$ where $a$ is the length of the part of the scale and $m$ is the mass on that side). That way students can see that math is also about describing the real world in a formal way.
$endgroup$
– Martin Thoma
Apr 7 '14 at 5:58
2
2
$begingroup$
I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind!
$endgroup$
– Guillermo Garza
Apr 8 '14 at 0:50
$begingroup$
I agree completely, but I don't have the apparatus. I'll look into purchasing one before the next semester. Thanks for helping me make up my mind!
$endgroup$
– Guillermo Garza
Apr 8 '14 at 0:50
$begingroup$
@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics.
$endgroup$
– Martin Thoma
Apr 8 '14 at 1:59
$begingroup$
@MeemoGarza Meanwhile you might be interested in this: walter-fendt.de/ph14d/hebel.htm - it's a site created by a physics teacher I had in school. This website contains lots of Java applets that show things in physics.
$endgroup$
– Martin Thoma
Apr 8 '14 at 1:59
add a comment |
$begingroup$
This is similar to Aky's answer, but includes a second drawing (and no math.)
To me the second drawing is key to understanding why the $mathrm c^2$ area is equal to the sum of $mathrm a^2+mathrm b^2$.
Edit: comments requested an animation, so a simple gif is attached...
$endgroup$
11
$begingroup$
I've always felt this one worked better in .gif format.
$endgroup$
– Patrick M
Apr 3 '14 at 20:47
2
$begingroup$
This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles.
$endgroup$
– Nick Retallack
Apr 7 '14 at 0:31
2
$begingroup$
I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size.
$endgroup$
– Sparr
Apr 7 '14 at 15:58
2
$begingroup$
@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it.
$endgroup$
– Tobia
Apr 7 '14 at 16:33
$begingroup$
Anyway...THIS was my favourite, so, good that you already did it!
$endgroup$
– MattAllegro
May 12 '14 at 21:49
|
show 1 more comment
$begingroup$
This is similar to Aky's answer, but includes a second drawing (and no math.)
To me the second drawing is key to understanding why the $mathrm c^2$ area is equal to the sum of $mathrm a^2+mathrm b^2$.
Edit: comments requested an animation, so a simple gif is attached...
$endgroup$
11
$begingroup$
I've always felt this one worked better in .gif format.
$endgroup$
– Patrick M
Apr 3 '14 at 20:47
2
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This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles.
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– Nick Retallack
Apr 7 '14 at 0:31
2
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I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size.
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– Sparr
Apr 7 '14 at 15:58
2
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@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it.
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– Tobia
Apr 7 '14 at 16:33
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Anyway...THIS was my favourite, so, good that you already did it!
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– MattAllegro
May 12 '14 at 21:49
|
show 1 more comment
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This is similar to Aky's answer, but includes a second drawing (and no math.)
To me the second drawing is key to understanding why the $mathrm c^2$ area is equal to the sum of $mathrm a^2+mathrm b^2$.
Edit: comments requested an animation, so a simple gif is attached...
$endgroup$
This is similar to Aky's answer, but includes a second drawing (and no math.)
To me the second drawing is key to understanding why the $mathrm c^2$ area is equal to the sum of $mathrm a^2+mathrm b^2$.
Edit: comments requested an animation, so a simple gif is attached...
edited Jun 9 '16 at 19:50
community wiki
3 revs, 3 users 70%
Tobia
11
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I've always felt this one worked better in .gif format.
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– Patrick M
Apr 3 '14 at 20:47
2
$begingroup$
This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles.
$endgroup$
– Nick Retallack
Apr 7 '14 at 0:31
2
$begingroup$
I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size.
$endgroup$
– Sparr
Apr 7 '14 at 15:58
2
$begingroup$
@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it.
$endgroup$
– Tobia
Apr 7 '14 at 16:33
$begingroup$
Anyway...THIS was my favourite, so, good that you already did it!
$endgroup$
– MattAllegro
May 12 '14 at 21:49
|
show 1 more comment
11
$begingroup$
I've always felt this one worked better in .gif format.
$endgroup$
– Patrick M
Apr 3 '14 at 20:47
2
$begingroup$
This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles.
$endgroup$
– Nick Retallack
Apr 7 '14 at 0:31
2
$begingroup$
I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size.
$endgroup$
– Sparr
Apr 7 '14 at 15:58
2
$begingroup$
@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it.
$endgroup$
– Tobia
Apr 7 '14 at 16:33
$begingroup$
Anyway...THIS was my favourite, so, good that you already did it!
$endgroup$
– MattAllegro
May 12 '14 at 21:49
11
11
$begingroup$
I've always felt this one worked better in .gif format.
$endgroup$
– Patrick M
Apr 3 '14 at 20:47
$begingroup$
I've always felt this one worked better in .gif format.
$endgroup$
– Patrick M
Apr 3 '14 at 20:47
2
2
$begingroup$
This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles.
$endgroup$
– Nick Retallack
Apr 7 '14 at 0:31
$begingroup$
This would be amazing if you animated the change from the first to the second by having the triangles on the right flop over onto the ones on the left to create those rectangles.
$endgroup$
– Nick Retallack
Apr 7 '14 at 0:31
2
2
$begingroup$
I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size.
$endgroup$
– Sparr
Apr 7 '14 at 15:58
$begingroup$
I think this one is lacking because it is not explained in the left picture that the blue and red areas are the same size.
$endgroup$
– Sparr
Apr 7 '14 at 15:58
2
2
$begingroup$
@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it.
$endgroup$
– Tobia
Apr 7 '14 at 16:33
$begingroup$
@Sparr they are not. The salmon area is just 4 times the size of the triangle. The only equivalence is bewteen the blue area on the left (c²) and the blue area on the right (a² + b²) which is just Pythagoras' formula. I will upload a better picture to show it.
$endgroup$
– Tobia
Apr 7 '14 at 16:33
$begingroup$
Anyway...THIS was my favourite, so, good that you already did it!
$endgroup$
– MattAllegro
May 12 '14 at 21:49
$begingroup$
Anyway...THIS was my favourite, so, good that you already did it!
$endgroup$
– MattAllegro
May 12 '14 at 21:49
|
show 1 more comment
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How about a line integral of a scalar field by http://1ucasvb.tumblr.com:
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Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;)
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– Luke
Aug 27 '14 at 17:44
4
$begingroup$
Nice visual intuition, but why is it true that $|mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step)
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– enthdegree
Aug 11 '16 at 5:18
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Isn't this how students usually learn line integrals?
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– user3932000
Mar 9 '17 at 1:50
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@enthdegree the integral is with respect to arc length ($ds$= arc length). $$int_C f(x,y)ds=int_{a}^{b}f(x(t),y(t))sqrt{x'(t)^2+y'(t)^2}dt=int_{a}^{b}f(r(t))|r'(t)|dt$$
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– clathratus
Oct 6 '18 at 1:54
$begingroup$
It's just "u-substitution."
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– Trevor Kafka
Jan 28 at 4:12
add a comment |
$begingroup$
How about a line integral of a scalar field by http://1ucasvb.tumblr.com:
$endgroup$
$begingroup$
Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;)
$endgroup$
– Luke
Aug 27 '14 at 17:44
4
$begingroup$
Nice visual intuition, but why is it true that $|mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step)
$endgroup$
– enthdegree
Aug 11 '16 at 5:18
$begingroup$
Isn't this how students usually learn line integrals?
$endgroup$
– user3932000
Mar 9 '17 at 1:50
$begingroup$
@enthdegree the integral is with respect to arc length ($ds$= arc length). $$int_C f(x,y)ds=int_{a}^{b}f(x(t),y(t))sqrt{x'(t)^2+y'(t)^2}dt=int_{a}^{b}f(r(t))|r'(t)|dt$$
$endgroup$
– clathratus
Oct 6 '18 at 1:54
$begingroup$
It's just "u-substitution."
$endgroup$
– Trevor Kafka
Jan 28 at 4:12
add a comment |
$begingroup$
How about a line integral of a scalar field by http://1ucasvb.tumblr.com:
$endgroup$
How about a line integral of a scalar field by http://1ucasvb.tumblr.com:
answered Apr 7 '14 at 5:55
community wiki
Jure Triglav
$begingroup$
Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;)
$endgroup$
– Luke
Aug 27 '14 at 17:44
4
$begingroup$
Nice visual intuition, but why is it true that $|mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step)
$endgroup$
– enthdegree
Aug 11 '16 at 5:18
$begingroup$
Isn't this how students usually learn line integrals?
$endgroup$
– user3932000
Mar 9 '17 at 1:50
$begingroup$
@enthdegree the integral is with respect to arc length ($ds$= arc length). $$int_C f(x,y)ds=int_{a}^{b}f(x(t),y(t))sqrt{x'(t)^2+y'(t)^2}dt=int_{a}^{b}f(r(t))|r'(t)|dt$$
$endgroup$
– clathratus
Oct 6 '18 at 1:54
$begingroup$
It's just "u-substitution."
$endgroup$
– Trevor Kafka
Jan 28 at 4:12
add a comment |
$begingroup$
Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;)
$endgroup$
– Luke
Aug 27 '14 at 17:44
4
$begingroup$
Nice visual intuition, but why is it true that $|mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step)
$endgroup$
– enthdegree
Aug 11 '16 at 5:18
$begingroup$
Isn't this how students usually learn line integrals?
$endgroup$
– user3932000
Mar 9 '17 at 1:50
$begingroup$
@enthdegree the integral is with respect to arc length ($ds$= arc length). $$int_C f(x,y)ds=int_{a}^{b}f(x(t),y(t))sqrt{x'(t)^2+y'(t)^2}dt=int_{a}^{b}f(r(t))|r'(t)|dt$$
$endgroup$
– clathratus
Oct 6 '18 at 1:54
$begingroup$
It's just "u-substitution."
$endgroup$
– Trevor Kafka
Jan 28 at 4:12
$begingroup$
Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;)
$endgroup$
– Luke
Aug 27 '14 at 17:44
$begingroup$
Awesome — this will be a great introduction to fields & the usage of integrals which I am about to explain to a friend of mine! ;)
$endgroup$
– Luke
Aug 27 '14 at 17:44
4
4
$begingroup$
Nice visual intuition, but why is it true that $|mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step)
$endgroup$
– enthdegree
Aug 11 '16 at 5:18
$begingroup$
Nice visual intuition, but why is it true that $|mathbf{r}'|$ is the right integrator to use? (i.e. the 'flattening/projection to the real line' step)
$endgroup$
– enthdegree
Aug 11 '16 at 5:18
$begingroup$
Isn't this how students usually learn line integrals?
$endgroup$
– user3932000
Mar 9 '17 at 1:50
$begingroup$
Isn't this how students usually learn line integrals?
$endgroup$
– user3932000
Mar 9 '17 at 1:50
$begingroup$
@enthdegree the integral is with respect to arc length ($ds$= arc length). $$int_C f(x,y)ds=int_{a}^{b}f(x(t),y(t))sqrt{x'(t)^2+y'(t)^2}dt=int_{a}^{b}f(r(t))|r'(t)|dt$$
$endgroup$
– clathratus
Oct 6 '18 at 1:54
$begingroup$
@enthdegree the integral is with respect to arc length ($ds$= arc length). $$int_C f(x,y)ds=int_{a}^{b}f(x(t),y(t))sqrt{x'(t)^2+y'(t)^2}dt=int_{a}^{b}f(r(t))|r'(t)|dt$$
$endgroup$
– clathratus
Oct 6 '18 at 1:54
$begingroup$
It's just "u-substitution."
$endgroup$
– Trevor Kafka
Jan 28 at 4:12
$begingroup$
It's just "u-substitution."
$endgroup$
– Trevor Kafka
Jan 28 at 4:12
add a comment |
$begingroup$
Similarly to eykanal's answer, although demonstrating some interesting facts about medians and geometry as well. It demonstrates that $displaystylesum_{n = 1}^{infty}frac{1}{2^n} = 1$:
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10
$begingroup$
I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut.
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– abnry
Apr 5 '14 at 13:55
$begingroup$
Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $frac{1}{2}$ according to you has actually an area of $frac{1}{2}timesfrac{sqrt{2}}{2}timesfrac{sqrt{2}}{2}=frac{1}{4}$, so what this figure is proving is that $sum_{n=2}^{infty}frac{1}{2^n}=frac{1}{2}$.
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– S4M
Apr 7 '14 at 7:12
3
$begingroup$
@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition?
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– Andrew Coonce
Apr 7 '14 at 16:51
$begingroup$
@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $frac{1}{4}$ triangle bisects the triangle of area $frac{1}{2}$ that is left after the $frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1.
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– user121055
Apr 8 '14 at 21:17
1
$begingroup$
@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $frac{sqrt{2}}{2}$.
$endgroup$
– S4M
Apr 9 '14 at 12:27
|
show 1 more comment
$begingroup$
Similarly to eykanal's answer, although demonstrating some interesting facts about medians and geometry as well. It demonstrates that $displaystylesum_{n = 1}^{infty}frac{1}{2^n} = 1$:
$endgroup$
10
$begingroup$
I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut.
$endgroup$
– abnry
Apr 5 '14 at 13:55
$begingroup$
Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $frac{1}{2}$ according to you has actually an area of $frac{1}{2}timesfrac{sqrt{2}}{2}timesfrac{sqrt{2}}{2}=frac{1}{4}$, so what this figure is proving is that $sum_{n=2}^{infty}frac{1}{2^n}=frac{1}{2}$.
$endgroup$
– S4M
Apr 7 '14 at 7:12
3
$begingroup$
@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition?
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– Andrew Coonce
Apr 7 '14 at 16:51
$begingroup$
@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $frac{1}{4}$ triangle bisects the triangle of area $frac{1}{2}$ that is left after the $frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1.
$endgroup$
– user121055
Apr 8 '14 at 21:17
1
$begingroup$
@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $frac{sqrt{2}}{2}$.
$endgroup$
– S4M
Apr 9 '14 at 12:27
|
show 1 more comment
$begingroup$
Similarly to eykanal's answer, although demonstrating some interesting facts about medians and geometry as well. It demonstrates that $displaystylesum_{n = 1}^{infty}frac{1}{2^n} = 1$:
$endgroup$
Similarly to eykanal's answer, although demonstrating some interesting facts about medians and geometry as well. It demonstrates that $displaystylesum_{n = 1}^{infty}frac{1}{2^n} = 1$:
answered Apr 4 '14 at 14:28
community wiki
user121055
10
$begingroup$
I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut.
$endgroup$
– abnry
Apr 5 '14 at 13:55
$begingroup$
Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $frac{1}{2}$ according to you has actually an area of $frac{1}{2}timesfrac{sqrt{2}}{2}timesfrac{sqrt{2}}{2}=frac{1}{4}$, so what this figure is proving is that $sum_{n=2}^{infty}frac{1}{2^n}=frac{1}{2}$.
$endgroup$
– S4M
Apr 7 '14 at 7:12
3
$begingroup$
@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition?
$endgroup$
– Andrew Coonce
Apr 7 '14 at 16:51
$begingroup$
@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $frac{1}{4}$ triangle bisects the triangle of area $frac{1}{2}$ that is left after the $frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1.
$endgroup$
– user121055
Apr 8 '14 at 21:17
1
$begingroup$
@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $frac{sqrt{2}}{2}$.
$endgroup$
– S4M
Apr 9 '14 at 12:27
|
show 1 more comment
10
$begingroup$
I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut.
$endgroup$
– abnry
Apr 5 '14 at 13:55
$begingroup$
Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $frac{1}{2}$ according to you has actually an area of $frac{1}{2}timesfrac{sqrt{2}}{2}timesfrac{sqrt{2}}{2}=frac{1}{4}$, so what this figure is proving is that $sum_{n=2}^{infty}frac{1}{2^n}=frac{1}{2}$.
$endgroup$
– S4M
Apr 7 '14 at 7:12
3
$begingroup$
@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition?
$endgroup$
– Andrew Coonce
Apr 7 '14 at 16:51
$begingroup$
@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $frac{1}{4}$ triangle bisects the triangle of area $frac{1}{2}$ that is left after the $frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1.
$endgroup$
– user121055
Apr 8 '14 at 21:17
1
$begingroup$
@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $frac{sqrt{2}}{2}$.
$endgroup$
– S4M
Apr 9 '14 at 12:27
10
10
$begingroup$
I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut.
$endgroup$
– abnry
Apr 5 '14 at 13:55
$begingroup$
I think this one better. In the square one you have to make a choice between dividing horizontally or vertically. With the triangle there is only one dividing cut.
$endgroup$
– abnry
Apr 5 '14 at 13:55
$begingroup$
Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $frac{1}{2}$ according to you has actually an area of $frac{1}{2}timesfrac{sqrt{2}}{2}timesfrac{sqrt{2}}{2}=frac{1}{4}$, so what this figure is proving is that $sum_{n=2}^{infty}frac{1}{2^n}=frac{1}{2}$.
$endgroup$
– S4M
Apr 7 '14 at 7:12
$begingroup$
Just nitpicking, but you got the areas of the triangles wrong. For example, the one of are $frac{1}{2}$ according to you has actually an area of $frac{1}{2}timesfrac{sqrt{2}}{2}timesfrac{sqrt{2}}{2}=frac{1}{4}$, so what this figure is proving is that $sum_{n=2}^{infty}frac{1}{2^n}=frac{1}{2}$.
$endgroup$
– S4M
Apr 7 '14 at 7:12
3
3
$begingroup$
@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition?
$endgroup$
– Andrew Coonce
Apr 7 '14 at 16:51
$begingroup$
@S4M: The area of the first triangle is defined as $1/2$; how is the calculation wrong when there is no calculation... only definition?
$endgroup$
– Andrew Coonce
Apr 7 '14 at 16:51
$begingroup$
@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $frac{1}{4}$ triangle bisects the triangle of area $frac{1}{2}$ that is left after the $frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1.
$endgroup$
– user121055
Apr 8 '14 at 21:17
$begingroup$
@S4M: Each new triangle bisects the previous one. The full triangle has area 1. The $frac{1}{4}$ triangle bisects the triangle of area $frac{1}{2}$ that is left after the $frac{1}{2}$ triangle and so on. I'm not trying to prove that the right side of the largest bisection line equals the other side (the summation you have), but that the entire collection of triangles approaches 1.
$endgroup$
– user121055
Apr 8 '14 at 21:17
1
1
$begingroup$
@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $frac{sqrt{2}}{2}$.
$endgroup$
– S4M
Apr 9 '14 at 12:27
$begingroup$
@AndrewCoonce I was wrong in my previous comment. I asscumend that the side of the main triangle was $1$. It's all fine if its side is $frac{sqrt{2}}{2}$.
$endgroup$
– S4M
Apr 9 '14 at 12:27
|
show 1 more comment
$begingroup$
This animation shows that a circle's perimeter equals to $2r*pi$. As ShreevatsaR pointed out, this is obvious because $pi$ is by definition the ratio of a circle's circumference to its diameter
In this image we can see how the ratio is calculated. The wheel's diameter is 1. After the perimeter is rolled down we can see that its length equals to $pi$ amount of wheels.
Source
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5
$begingroup$
But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = frac{4}{3} pi r^3$ and a surface area of $SA_3(r) = 4 pi r^2$
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– Cole Johnson
Apr 4 '14 at 16:09
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@ColeJohnson I don't know, just wanted to show this awesome animation:D
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– totymedli
Apr 4 '14 at 21:11
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I just wish someone could make an animation of that. I understand the math behind it, but animations are nice.
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– Cole Johnson
Apr 4 '14 at 22:28
42
$begingroup$
That the circle's perimeter is $2pi r$ is the definition of $pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $pi$ is about $3.1$.
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– ShreevatsaR
Apr 5 '14 at 5:24
$begingroup$
@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there.
$endgroup$
– Akiva Weinberger
Feb 23 '15 at 1:45
add a comment |
$begingroup$
This animation shows that a circle's perimeter equals to $2r*pi$. As ShreevatsaR pointed out, this is obvious because $pi$ is by definition the ratio of a circle's circumference to its diameter
In this image we can see how the ratio is calculated. The wheel's diameter is 1. After the perimeter is rolled down we can see that its length equals to $pi$ amount of wheels.
Source
$endgroup$
5
$begingroup$
But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = frac{4}{3} pi r^3$ and a surface area of $SA_3(r) = 4 pi r^2$
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:09
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@ColeJohnson I don't know, just wanted to show this awesome animation:D
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– totymedli
Apr 4 '14 at 21:11
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I just wish someone could make an animation of that. I understand the math behind it, but animations are nice.
$endgroup$
– Cole Johnson
Apr 4 '14 at 22:28
42
$begingroup$
That the circle's perimeter is $2pi r$ is the definition of $pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $pi$ is about $3.1$.
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:24
$begingroup$
@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there.
$endgroup$
– Akiva Weinberger
Feb 23 '15 at 1:45
add a comment |
$begingroup$
This animation shows that a circle's perimeter equals to $2r*pi$. As ShreevatsaR pointed out, this is obvious because $pi$ is by definition the ratio of a circle's circumference to its diameter
In this image we can see how the ratio is calculated. The wheel's diameter is 1. After the perimeter is rolled down we can see that its length equals to $pi$ amount of wheels.
Source
$endgroup$
This animation shows that a circle's perimeter equals to $2r*pi$. As ShreevatsaR pointed out, this is obvious because $pi$ is by definition the ratio of a circle's circumference to its diameter
In this image we can see how the ratio is calculated. The wheel's diameter is 1. After the perimeter is rolled down we can see that its length equals to $pi$ amount of wheels.
Source
edited Apr 13 '17 at 12:21
community wiki
4 revs
totymedli
5
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But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = frac{4}{3} pi r^3$ and a surface area of $SA_3(r) = 4 pi r^2$
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– Cole Johnson
Apr 4 '14 at 16:09
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@ColeJohnson I don't know, just wanted to show this awesome animation:D
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– totymedli
Apr 4 '14 at 21:11
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I just wish someone could make an animation of that. I understand the math behind it, but animations are nice.
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– Cole Johnson
Apr 4 '14 at 22:28
42
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That the circle's perimeter is $2pi r$ is the definition of $pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $pi$ is about $3.1$.
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– ShreevatsaR
Apr 5 '14 at 5:24
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@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there.
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– Akiva Weinberger
Feb 23 '15 at 1:45
add a comment |
5
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But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = frac{4}{3} pi r^3$ and a surface area of $SA_3(r) = 4 pi r^2$
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– Cole Johnson
Apr 4 '14 at 16:09
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@ColeJohnson I don't know, just wanted to show this awesome animation:D
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– totymedli
Apr 4 '14 at 21:11
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I just wish someone could make an animation of that. I understand the math behind it, but animations are nice.
$endgroup$
– Cole Johnson
Apr 4 '14 at 22:28
42
$begingroup$
That the circle's perimeter is $2pi r$ is the definition of $pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $pi$ is about $3.1$.
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:24
$begingroup$
@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there.
$endgroup$
– Akiva Weinberger
Feb 23 '15 at 1:45
5
5
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But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = frac{4}{3} pi r^3$ and a surface area of $SA_3(r) = 4 pi r^2$
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:09
$begingroup$
But can you show that a sphere with radius, $r$, has a volume of $V_3(r) = frac{4}{3} pi r^3$ and a surface area of $SA_3(r) = 4 pi r^2$
$endgroup$
– Cole Johnson
Apr 4 '14 at 16:09
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@ColeJohnson I don't know, just wanted to show this awesome animation:D
$endgroup$
– totymedli
Apr 4 '14 at 21:11
$begingroup$
@ColeJohnson I don't know, just wanted to show this awesome animation:D
$endgroup$
– totymedli
Apr 4 '14 at 21:11
$begingroup$
I just wish someone could make an animation of that. I understand the math behind it, but animations are nice.
$endgroup$
– Cole Johnson
Apr 4 '14 at 22:28
$begingroup$
I just wish someone could make an animation of that. I understand the math behind it, but animations are nice.
$endgroup$
– Cole Johnson
Apr 4 '14 at 22:28
42
42
$begingroup$
That the circle's perimeter is $2pi r$ is the definition of $pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $pi$ is about $3.1$.
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:24
$begingroup$
That the circle's perimeter is $2pi r$ is the definition of $pi$, so I wouldn't say this is an explanation of the fact; rather it's an illustration of what the definition means, and that the value of $pi$ is about $3.1$.
$endgroup$
– ShreevatsaR
Apr 5 '14 at 5:24
$begingroup$
@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there.
$endgroup$
– Akiva Weinberger
Feb 23 '15 at 1:45
$begingroup$
@ColeJohnson There's a neat way to find the volume of a sphere with Cavalieri's principle and Pythagoras. Just look up Cavalieri's principle on Wikipedia, I think it's on there.
$endgroup$
– Akiva Weinberger
Feb 23 '15 at 1:45
add a comment |
$begingroup$
Steven Wittens presents quite a few math concepts in his talk Making things with math. His slides can be found from his own website.
For example, Bézier curves visually:
He has also created MathBox.js which powers his amazing visualisations in the slides.
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4
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These are beautiful!
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– Cole Johnson
May 15 '14 at 23:16
17
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Also, if you read the text out loud you get to say "lerp lerp lerp lerp", which has its own aesthetic appeal.
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– msouth
Sep 23 '14 at 21:43
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beautiful, but I don't understand XD
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– Kokizzu
Apr 20 '15 at 11:20
1
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See also codegolf.stackexchange.com/questions/21178/…
$endgroup$
– SztupY
Jun 21 '16 at 9:24
add a comment |
$begingroup$
Steven Wittens presents quite a few math concepts in his talk Making things with math. His slides can be found from his own website.
For example, Bézier curves visually:
He has also created MathBox.js which powers his amazing visualisations in the slides.
$endgroup$
4
$begingroup$
These are beautiful!
$endgroup$
– Cole Johnson
May 15 '14 at 23:16
17
$begingroup$
Also, if you read the text out loud you get to say "lerp lerp lerp lerp", which has its own aesthetic appeal.
$endgroup$
– msouth
Sep 23 '14 at 21:43
$begingroup$
beautiful, but I don't understand XD
$endgroup$
– Kokizzu
Apr 20 '15 at 11:20
1
$begingroup$
See also codegolf.stackexchange.com/questions/21178/…
$endgroup$
– SztupY
Jun 21 '16 at 9:24
add a comment |
$begingroup$
Steven Wittens presents quite a few math concepts in his talk Making things with math. His slides can be found from his own website.
For example, Bézier curves visually:
He has also created MathBox.js which powers his amazing visualisations in the slides.
$endgroup$
Steven Wittens presents quite a few math concepts in his talk Making things with math. His slides can be found from his own website.
For example, Bézier curves visually:
He has also created MathBox.js which powers his amazing visualisations in the slides.
edited Mar 8 '15 at 11:46
community wiki
2 revs, 2 users 75%
Kimmo
4
$begingroup$
These are beautiful!
$endgroup$
– Cole Johnson
May 15 '14 at 23:16
17
$begingroup$
Also, if you read the text out loud you get to say "lerp lerp lerp lerp", which has its own aesthetic appeal.
$endgroup$
– msouth
Sep 23 '14 at 21:43
$begingroup$
beautiful, but I don't understand XD
$endgroup$
– Kokizzu
Apr 20 '15 at 11:20
1
$begingroup$
See also codegolf.stackexchange.com/questions/21178/…
$endgroup$
– SztupY
Jun 21 '16 at 9:24
add a comment |
4
$begingroup$
These are beautiful!
$endgroup$
– Cole Johnson
May 15 '14 at 23:16
17
$begingroup$
Also, if you read the text out loud you get to say "lerp lerp lerp lerp", which has its own aesthetic appeal.
$endgroup$
– msouth
Sep 23 '14 at 21:43
$begingroup$
beautiful, but I don't understand XD
$endgroup$
– Kokizzu
Apr 20 '15 at 11:20
1
$begingroup$
See also codegolf.stackexchange.com/questions/21178/…
$endgroup$
– SztupY
Jun 21 '16 at 9:24
4
4
$begingroup$
These are beautiful!
$endgroup$
– Cole Johnson
May 15 '14 at 23:16
$begingroup$
These are beautiful!
$endgroup$
– Cole Johnson
May 15 '14 at 23:16
17
17
$begingroup$
Also, if you read the text out loud you get to say "lerp lerp lerp lerp", which has its own aesthetic appeal.
$endgroup$
– msouth
Sep 23 '14 at 21:43
$begingroup$
Also, if you read the text out loud you get to say "lerp lerp lerp lerp", which has its own aesthetic appeal.
$endgroup$
– msouth
Sep 23 '14 at 21:43
$begingroup$
beautiful, but I don't understand XD
$endgroup$
– Kokizzu
Apr 20 '15 at 11:20
$begingroup$
beautiful, but I don't understand XD
$endgroup$
– Kokizzu
Apr 20 '15 at 11:20
1
1
$begingroup$
See also codegolf.stackexchange.com/questions/21178/…
$endgroup$
– SztupY
Jun 21 '16 at 9:24
$begingroup$
See also codegolf.stackexchange.com/questions/21178/…
$endgroup$
– SztupY
Jun 21 '16 at 9:24
add a comment |
$begingroup$
Fractal art. Here's an example: "Mandelbrot Island".
The real island of Sark in the (English) Channel Islands looks astonishingly like Mandelbrot island:
Now that I think about it, fractals in general are quite beautiful. Here's a close-up of the Mandelbrot set:
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36
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Are complex numbers easy to explain?
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– adam.r
Mar 31 '14 at 17:15
6
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Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think.
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– user134824
Mar 31 '14 at 17:16
9
$begingroup$
Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large).
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– Kaz
Apr 1 '14 at 4:57
4
$begingroup$
@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list.
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– Szabolcs
Apr 1 '14 at 16:12
3
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It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example.
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– user134824
Apr 3 '14 at 3:12
|
show 9 more comments
$begingroup$
Fractal art. Here's an example: "Mandelbrot Island".
The real island of Sark in the (English) Channel Islands looks astonishingly like Mandelbrot island:
Now that I think about it, fractals in general are quite beautiful. Here's a close-up of the Mandelbrot set:
$endgroup$
36
$begingroup$
Are complex numbers easy to explain?
$endgroup$
– adam.r
Mar 31 '14 at 17:15
6
$begingroup$
Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think.
$endgroup$
– user134824
Mar 31 '14 at 17:16
9
$begingroup$
Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large).
$endgroup$
– Kaz
Apr 1 '14 at 4:57
4
$begingroup$
@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list.
$endgroup$
– Szabolcs
Apr 1 '14 at 16:12
3
$begingroup$
It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example.
$endgroup$
– user134824
Apr 3 '14 at 3:12
|
show 9 more comments
$begingroup$
Fractal art. Here's an example: "Mandelbrot Island".
The real island of Sark in the (English) Channel Islands looks astonishingly like Mandelbrot island:
Now that I think about it, fractals in general are quite beautiful. Here's a close-up of the Mandelbrot set:
$endgroup$
Fractal art. Here's an example: "Mandelbrot Island".
The real island of Sark in the (English) Channel Islands looks astonishingly like Mandelbrot island:
Now that I think about it, fractals in general are quite beautiful. Here's a close-up of the Mandelbrot set:
edited Mar 9 '17 at 17:31
community wiki
3 revs
user134824
36
$begingroup$
Are complex numbers easy to explain?
$endgroup$
– adam.r
Mar 31 '14 at 17:15
6
$begingroup$
Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think.
$endgroup$
– user134824
Mar 31 '14 at 17:16
9
$begingroup$
Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large).
$endgroup$
– Kaz
Apr 1 '14 at 4:57
4
$begingroup$
@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list.
$endgroup$
– Szabolcs
Apr 1 '14 at 16:12
3
$begingroup$
It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example.
$endgroup$
– user134824
Apr 3 '14 at 3:12
|
show 9 more comments
36
$begingroup$
Are complex numbers easy to explain?
$endgroup$
– adam.r
Mar 31 '14 at 17:15
6
$begingroup$
Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think.
$endgroup$
– user134824
Mar 31 '14 at 17:16
9
$begingroup$
Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large).
$endgroup$
– Kaz
Apr 1 '14 at 4:57
4
$begingroup$
@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list.
$endgroup$
– Szabolcs
Apr 1 '14 at 16:12
3
$begingroup$
It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example.
$endgroup$
– user134824
Apr 3 '14 at 3:12
36
36
$begingroup$
Are complex numbers easy to explain?
$endgroup$
– adam.r
Mar 31 '14 at 17:15
$begingroup$
Are complex numbers easy to explain?
$endgroup$
– adam.r
Mar 31 '14 at 17:15
6
6
$begingroup$
Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think.
$endgroup$
– user134824
Mar 31 '14 at 17:16
$begingroup$
Probably not. But the idea that an object "looks the same if you zoom in on it" is easy to explain, I think.
$endgroup$
– user134824
Mar 31 '14 at 17:16
9
9
$begingroup$
Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large).
$endgroup$
– Kaz
Apr 1 '14 at 4:57
$begingroup$
Whereas complex numbers might not be easy to explain, it is easy to explain the idea idea that each pixel is calculated by repeating a simple calculation on two numbers (which are initially the coordinates of the pixel), and the color is a measure of how the resulting pair of numbers "escapes" (grows large).
$endgroup$
– Kaz
Apr 1 '14 at 4:57
4
4
$begingroup$
@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list.
$endgroup$
– Szabolcs
Apr 1 '14 at 16:12
$begingroup$
@adam.r I don't recall precisely when I learned about complex numbers in school, but it was around the 9th grade (maybe 8th, maybe 10th). To understand the Mandelbrot set, nothing more is needed than basic arithmetic with complex numbers. That should put it within reach for anyone who graduated high school. It's certainly a good candidate for this list.
$endgroup$
– Szabolcs
Apr 1 '14 at 16:12
3
3
$begingroup$
It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example.
$endgroup$
– user134824
Apr 3 '14 at 3:12
$begingroup$
It is a close up. The Mandelbrot set is not completely self-similar, but there are some self-similarities as you zoom. See this zoom sequence for an example.
$endgroup$
– user134824
Apr 3 '14 at 3:12
|
show 9 more comments
$begingroup$
The magnetic pendulum:
An iron pendulum is suspended above a flat surface, with three magnets on it. The magnets are colored red, yellow and blue.
We hold the pendulum above a random point of the surface and let it go, holding our finger on the starting point. After some swinging this way and that, under the attractions of the magnets and gravity, it will come to rest over one of the magnets. We color the starting point (under our finger) with the color of the magnet.
Repeating this for every point on the surface, we get the image shown above.
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1
$begingroup$
How is the shading of the image given?
$endgroup$
– David Zhang
Apr 21 '14 at 14:27
3
$begingroup$
@DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm.
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– Peter
Apr 21 '14 at 15:27
$begingroup$
Deterministic chaos?
$endgroup$
– Peter Mortensen
Mar 8 '15 at 11:51
4
$begingroup$
@PeterMortensen Absolutely. In fact, you can see how (in certain areas) a minute change in initial conditions can cause a drastic change in outcome.
$endgroup$
– Peter
Mar 9 '15 at 12:36
add a comment |
$begingroup$
The magnetic pendulum:
An iron pendulum is suspended above a flat surface, with three magnets on it. The magnets are colored red, yellow and blue.
We hold the pendulum above a random point of the surface and let it go, holding our finger on the starting point. After some swinging this way and that, under the attractions of the magnets and gravity, it will come to rest over one of the magnets. We color the starting point (under our finger) with the color of the magnet.
Repeating this for every point on the surface, we get the image shown above.
$endgroup$
1
$begingroup$
How is the shading of the image given?
$endgroup$
– David Zhang
Apr 21 '14 at 14:27
3
$begingroup$
@DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm.
$endgroup$
– Peter
Apr 21 '14 at 15:27
$begingroup$
Deterministic chaos?
$endgroup$
– Peter Mortensen
Mar 8 '15 at 11:51
4
$begingroup$
@PeterMortensen Absolutely. In fact, you can see how (in certain areas) a minute change in initial conditions can cause a drastic change in outcome.
$endgroup$
– Peter
Mar 9 '15 at 12:36
add a comment |
$begingroup$
The magnetic pendulum:
An iron pendulum is suspended above a flat surface, with three magnets on it. The magnets are colored red, yellow and blue.
We hold the pendulum above a random point of the surface and let it go, holding our finger on the starting point. After some swinging this way and that, under the attractions of the magnets and gravity, it will come to rest over one of the magnets. We color the starting point (under our finger) with the color of the magnet.
Repeating this for every point on the surface, we get the image shown above.
$endgroup$
The magnetic pendulum:
An iron pendulum is suspended above a flat surface, with three magnets on it. The magnets are colored red, yellow and blue.
We hold the pendulum above a random point of the surface and let it go, holding our finger on the starting point. After some swinging this way and that, under the attractions of the magnets and gravity, it will come to rest over one of the magnets. We color the starting point (under our finger) with the color of the magnet.
Repeating this for every point on the surface, we get the image shown above.
answered Apr 8 '14 at 18:03
community wiki
Peter
1
$begingroup$
How is the shading of the image given?
$endgroup$
– David Zhang
Apr 21 '14 at 14:27
3
$begingroup$
@DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm.
$endgroup$
– Peter
Apr 21 '14 at 15:27
$begingroup$
Deterministic chaos?
$endgroup$
– Peter Mortensen
Mar 8 '15 at 11:51
4
$begingroup$
@PeterMortensen Absolutely. In fact, you can see how (in certain areas) a minute change in initial conditions can cause a drastic change in outcome.
$endgroup$
– Peter
Mar 9 '15 at 12:36
add a comment |
1
$begingroup$
How is the shading of the image given?
$endgroup$
– David Zhang
Apr 21 '14 at 14:27
3
$begingroup$
@DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm.
$endgroup$
– Peter
Apr 21 '14 at 15:27
$begingroup$
Deterministic chaos?
$endgroup$
– Peter Mortensen
Mar 8 '15 at 11:51
4
$begingroup$
@PeterMortensen Absolutely. In fact, you can see how (in certain areas) a minute change in initial conditions can cause a drastic change in outcome.
$endgroup$
– Peter
Mar 9 '15 at 12:36
1
1
$begingroup$
How is the shading of the image given?
$endgroup$
– David Zhang
Apr 21 '14 at 14:27
$begingroup$
How is the shading of the image given?
$endgroup$
– David Zhang
Apr 21 '14 at 14:27
3
3
$begingroup$
@DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm.
$endgroup$
– Peter
Apr 21 '14 at 15:27
$begingroup$
@DavidZhang, the longer the pendulum swings, the darker the point. I don't know the exact function used for this image, but it you click on the link, you'll find the algorithm.
$endgroup$
– Peter
Apr 21 '14 at 15:27
$begingroup$
Deterministic chaos?
$endgroup$
– Peter Mortensen
Mar 8 '15 at 11:51
$begingroup$
Deterministic chaos?
$endgroup$
– Peter Mortensen
Mar 8 '15 at 11:51
4
4
$begingroup$
@PeterMortensen Absolutely. In fact, you can see how (in certain areas) a minute change in initial conditions can cause a drastic change in outcome.
$endgroup$
– Peter
Mar 9 '15 at 12:36
$begingroup$
@PeterMortensen Absolutely. In fact, you can see how (in certain areas) a minute change in initial conditions can cause a drastic change in outcome.
$endgroup$
– Peter
Mar 9 '15 at 12:36
add a comment |
$begingroup$
Simple,visual proof of the Pythagorean theorem. Originally from Pythagorean Theorem Proof Without Words 6).
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13
$begingroup$
Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $;b;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1
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– DonAntonio
Mar 31 '14 at 12:31
5
$begingroup$
Another great pictural proof is 4 triangles in a square: mathalino.com/sites/default/files/images/01-pythagora.jpg
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– Sergey Grinev
Mar 31 '14 at 15:10
6
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I don't get the original equality. Are you relying on the similarity of two triangles? How is that obvious from the diagram?
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– adam.r
Mar 31 '14 at 17:10
56
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It's not immediately obvious to me why $frac{c+a}{b} = frac{b}{c-a}$ I'm sure it's very simple and I'll kick myself for asking, sorry.
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– PeteUK
Mar 31 '14 at 17:12
9
$begingroup$
Well, I looked at that picture and thought: "Err ... how does this work?" I figured it out, but it was far from obvious for me.
$endgroup$
– celtschk
Apr 3 '14 at 18:33
|
show 13 more comments
$begingroup$
Simple,visual proof of the Pythagorean theorem. Originally from Pythagorean Theorem Proof Without Words 6).
$endgroup$
13
$begingroup$
Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $;b;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1
$endgroup$
– DonAntonio
Mar 31 '14 at 12:31
5
$begingroup$
Another great pictural proof is 4 triangles in a square: mathalino.com/sites/default/files/images/01-pythagora.jpg
$endgroup$
– Sergey Grinev
Mar 31 '14 at 15:10
6
$begingroup$
I don't get the original equality. Are you relying on the similarity of two triangles? How is that obvious from the diagram?
$endgroup$
– adam.r
Mar 31 '14 at 17:10
56
$begingroup$
It's not immediately obvious to me why $frac{c+a}{b} = frac{b}{c-a}$ I'm sure it's very simple and I'll kick myself for asking, sorry.
$endgroup$
– PeteUK
Mar 31 '14 at 17:12
9
$begingroup$
Well, I looked at that picture and thought: "Err ... how does this work?" I figured it out, but it was far from obvious for me.
$endgroup$
– celtschk
Apr 3 '14 at 18:33
|
show 13 more comments
$begingroup$
Simple,visual proof of the Pythagorean theorem. Originally from Pythagorean Theorem Proof Without Words 6).
$endgroup$
Simple,visual proof of the Pythagorean theorem. Originally from Pythagorean Theorem Proof Without Words 6).
edited Mar 8 '15 at 10:21
community wiki
3 revs, 3 users 63%
Sabyasachi
13
$begingroup$
Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $;b;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1
$endgroup$
– DonAntonio
Mar 31 '14 at 12:31
5
$begingroup$
Another great pictural proof is 4 triangles in a square: mathalino.com/sites/default/files/images/01-pythagora.jpg
$endgroup$
– Sergey Grinev
Mar 31 '14 at 15:10
6
$begingroup$
I don't get the original equality. Are you relying on the similarity of two triangles? How is that obvious from the diagram?
$endgroup$
– adam.r
Mar 31 '14 at 17:10
56
$begingroup$
It's not immediately obvious to me why $frac{c+a}{b} = frac{b}{c-a}$ I'm sure it's very simple and I'll kick myself for asking, sorry.
$endgroup$
– PeteUK
Mar 31 '14 at 17:12
9
$begingroup$
Well, I looked at that picture and thought: "Err ... how does this work?" I figured it out, but it was far from obvious for me.
$endgroup$
– celtschk
Apr 3 '14 at 18:33
|
show 13 more comments
13
$begingroup$
Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $;b;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1
$endgroup$
– DonAntonio
Mar 31 '14 at 12:31
5
$begingroup$
Another great pictural proof is 4 triangles in a square: mathalino.com/sites/default/files/images/01-pythagora.jpg
$endgroup$
– Sergey Grinev
Mar 31 '14 at 15:10
6
$begingroup$
I don't get the original equality. Are you relying on the similarity of two triangles? How is that obvious from the diagram?
$endgroup$
– adam.r
Mar 31 '14 at 17:10
56
$begingroup$
It's not immediately obvious to me why $frac{c+a}{b} = frac{b}{c-a}$ I'm sure it's very simple and I'll kick myself for asking, sorry.
$endgroup$
– PeteUK
Mar 31 '14 at 17:12
9
$begingroup$
Well, I looked at that picture and thought: "Err ... how does this work?" I figured it out, but it was far from obvious for me.
$endgroup$
– celtschk
Apr 3 '14 at 18:33
13
13
$begingroup$
Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $;b;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1
$endgroup$
– DonAntonio
Mar 31 '14 at 12:31
$begingroup$
Perhaps I'm trying to oversimplify, but this visual proof would be way easier (perhaps even trivial) if the legs of the big straight angle (S.A.) triangle (hypotenuse = the circle's diameter, third vertex on the top of that leg of length $;b;$) were drawn, and then from basic geometry: " In a S.A. triangle, the height to the hypotenuse divides the triangle in two triangles similar to each other and also to the big triangle". Nicely brought. +1
$endgroup$
– DonAntonio
Mar 31 '14 at 12:31
5
5
$begingroup$
Another great pictural proof is 4 triangles in a square: mathalino.com/sites/default/files/images/01-pythagora.jpg
$endgroup$
– Sergey Grinev
Mar 31 '14 at 15:10
$begingroup$
Another great pictural proof is 4 triangles in a square: mathalino.com/sites/default/files/images/01-pythagora.jpg
$endgroup$
– Sergey Grinev
Mar 31 '14 at 15:10
6
6
$begingroup$
I don't get the original equality. Are you relying on the similarity of two triangles? How is that obvious from the diagram?
$endgroup$
– adam.r
Mar 31 '14 at 17:10
$begingroup$
I don't get the original equality. Are you relying on the similarity of two triangles? How is that obvious from the diagram?
$endgroup$
– adam.r
Mar 31 '14 at 17:10
56
56
$begingroup$
It's not immediately obvious to me why $frac{c+a}{b} = frac{b}{c-a}$ I'm sure it's very simple and I'll kick myself for asking, sorry.
$endgroup$
– PeteUK
Mar 31 '14 at 17:12
$begingroup$
It's not immediately obvious to me why $frac{c+a}{b} = frac{b}{c-a}$ I'm sure it's very simple and I'll kick myself for asking, sorry.
$endgroup$
– PeteUK
Mar 31 '14 at 17:12
9
9
$begingroup$
Well, I looked at that picture and thought: "Err ... how does this work?" I figured it out, but it was far from obvious for me.
$endgroup$
– celtschk
Apr 3 '14 at 18:33
$begingroup$
Well, I looked at that picture and thought: "Err ... how does this work?" I figured it out, but it was far from obvious for me.
$endgroup$
– celtschk
Apr 3 '14 at 18:33
|
show 13 more comments
$begingroup$
Topology needs to be represented here, specifically knot theory. The following picture is from the Wikipedia page about Seifert Surfaces and was contributed by Accelerometer. Every link (or knot) is the boundary of a smooth orientable surface in 3D-space. This fact is attributed to Herbert Seifert, since he was the first to give an algorithm for constructing them. The surface we are looking at is bounded by Borromean rings.
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57
$begingroup$
I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question.
$endgroup$
– dfc
Apr 7 '14 at 5:25
4
$begingroup$
@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles.
$endgroup$
– Slade
Apr 7 '14 at 20:10
add a comment |
$begingroup$
Topology needs to be represented here, specifically knot theory. The following picture is from the Wikipedia page about Seifert Surfaces and was contributed by Accelerometer. Every link (or knot) is the boundary of a smooth orientable surface in 3D-space. This fact is attributed to Herbert Seifert, since he was the first to give an algorithm for constructing them. The surface we are looking at is bounded by Borromean rings.
$endgroup$
57
$begingroup$
I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question.
$endgroup$
– dfc
Apr 7 '14 at 5:25
4
$begingroup$
@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles.
$endgroup$
– Slade
Apr 7 '14 at 20:10
add a comment |
$begingroup$
Topology needs to be represented here, specifically knot theory. The following picture is from the Wikipedia page about Seifert Surfaces and was contributed by Accelerometer. Every link (or knot) is the boundary of a smooth orientable surface in 3D-space. This fact is attributed to Herbert Seifert, since he was the first to give an algorithm for constructing them. The surface we are looking at is bounded by Borromean rings.
$endgroup$
Topology needs to be represented here, specifically knot theory. The following picture is from the Wikipedia page about Seifert Surfaces and was contributed by Accelerometer. Every link (or knot) is the boundary of a smooth orientable surface in 3D-space. This fact is attributed to Herbert Seifert, since he was the first to give an algorithm for constructing them. The surface we are looking at is bounded by Borromean rings.
edited Mar 8 '15 at 10:53
community wiki
2 revs, 2 users 71%
N. Owad
57
$begingroup$
I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question.
$endgroup$
– dfc
Apr 7 '14 at 5:25
4
$begingroup$
@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles.
$endgroup$
– Slade
Apr 7 '14 at 20:10
add a comment |
57
$begingroup$
I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question.
$endgroup$
– dfc
Apr 7 '14 at 5:25
4
$begingroup$
@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles.
$endgroup$
– Slade
Apr 7 '14 at 20:10
57
57
$begingroup$
I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question.
$endgroup$
– dfc
Apr 7 '14 at 5:25
$begingroup$
I do not know much about topology so I will take your word for it that this is a beautiful idea/concept. However this picture and your description explain nothing to me. It seems like you missed the "easy to explain" bit in the question.
$endgroup$
– dfc
Apr 7 '14 at 5:25
4
4
$begingroup$
@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles.
$endgroup$
– Slade
Apr 7 '14 at 20:10
$begingroup$
@dfc I don't know, it seems like you can convey most of the meat here using soap bubbles.
$endgroup$
– Slade
Apr 7 '14 at 20:10
add a comment |
$begingroup$
Here's a GIF that I made that demonstrates Phi (golden number)
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6
$begingroup$
That's Beautiful!
$endgroup$
– debap
Sep 4 '16 at 12:36
add a comment |
$begingroup$
Here's a GIF that I made that demonstrates Phi (golden number)
$endgroup$
6
$begingroup$
That's Beautiful!
$endgroup$
– debap
Sep 4 '16 at 12:36
add a comment |
$begingroup$
Here's a GIF that I made that demonstrates Phi (golden number)
$endgroup$
Here's a GIF that I made that demonstrates Phi (golden number)
answered Apr 7 '14 at 21:34
community wiki
Steve Lewis
6
$begingroup$
That's Beautiful!
$endgroup$
– debap
Sep 4 '16 at 12:36
add a comment |
6
$begingroup$
That's Beautiful!
$endgroup$
– debap
Sep 4 '16 at 12:36
6
6
$begingroup$
That's Beautiful!
$endgroup$
– debap
Sep 4 '16 at 12:36
$begingroup$
That's Beautiful!
$endgroup$
– debap
Sep 4 '16 at 12:36
add a comment |
$begingroup$
Take a look at this great example of Fourier series visualisations written in JavaScript.
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$begingroup$
Really amazing....
$endgroup$
– Pacerier
Apr 11 '14 at 16:49
$begingroup$
What is this doing lurking at the bottom of the page? It's brilliant!
$endgroup$
– P i
Feb 25 '15 at 19:39
3
$begingroup$
The link is brilliant!
$endgroup$
– David Simmons
Jun 14 '15 at 9:50
10
$begingroup$
If you had captured an animated gif, this would be the top answer!
$endgroup$
– Tobia
Jul 10 '15 at 0:14
add a comment |
$begingroup$
Take a look at this great example of Fourier series visualisations written in JavaScript.
$endgroup$
$begingroup$
Really amazing....
$endgroup$
– Pacerier
Apr 11 '14 at 16:49
$begingroup$
What is this doing lurking at the bottom of the page? It's brilliant!
$endgroup$
– P i
Feb 25 '15 at 19:39
3
$begingroup$
The link is brilliant!
$endgroup$
– David Simmons
Jun 14 '15 at 9:50
10
$begingroup$
If you had captured an animated gif, this would be the top answer!
$endgroup$
– Tobia
Jul 10 '15 at 0:14
add a comment |
$begingroup$
Take a look at this great example of Fourier series visualisations written in JavaScript.
$endgroup$
Take a look at this great example of Fourier series visualisations written in JavaScript.
edited Mar 8 '15 at 11:32
community wiki
3 revs, 3 users 43%
Gandi
$begingroup$
Really amazing....
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– Pacerier
Apr 11 '14 at 16:49
$begingroup$
What is this doing lurking at the bottom of the page? It's brilliant!
$endgroup$
– P i
Feb 25 '15 at 19:39
3
$begingroup$
The link is brilliant!
$endgroup$
– David Simmons
Jun 14 '15 at 9:50
10
$begingroup$
If you had captured an animated gif, this would be the top answer!
$endgroup$
– Tobia
Jul 10 '15 at 0:14
add a comment |
$begingroup$
Really amazing....
$endgroup$
– Pacerier
Apr 11 '14 at 16:49
$begingroup$
What is this doing lurking at the bottom of the page? It's brilliant!
$endgroup$
– P i
Feb 25 '15 at 19:39
3
$begingroup$
The link is brilliant!
$endgroup$
– David Simmons
Jun 14 '15 at 9:50
10
$begingroup$
If you had captured an animated gif, this would be the top answer!
$endgroup$
– Tobia
Jul 10 '15 at 0:14
$begingroup$
Really amazing....
$endgroup$
– Pacerier
Apr 11 '14 at 16:49
$begingroup$
Really amazing....
$endgroup$
– Pacerier
Apr 11 '14 at 16:49
$begingroup$
What is this doing lurking at the bottom of the page? It's brilliant!
$endgroup$
– P i
Feb 25 '15 at 19:39
$begingroup$
What is this doing lurking at the bottom of the page? It's brilliant!
$endgroup$
– P i
Feb 25 '15 at 19:39
3
3
$begingroup$
The link is brilliant!
$endgroup$
– David Simmons
Jun 14 '15 at 9:50
$begingroup$
The link is brilliant!
$endgroup$
– David Simmons
Jun 14 '15 at 9:50
10
10
$begingroup$
If you had captured an animated gif, this would be the top answer!
$endgroup$
– Tobia
Jul 10 '15 at 0:14
$begingroup$
If you had captured an animated gif, this would be the top answer!
$endgroup$
– Tobia
Jul 10 '15 at 0:14
add a comment |
$begingroup$
Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall, for different choices of parameter:
The best thing about them is, they satisfy periodic boundary conditions, and so you can pick one of them and set it as a desktop background by tiling it, resulting in a far more spectacular image than just the single unit cells posted above!
The images seem to be a vast interconnected network of lines once you tile them, but in fact the entire picture is actually just a single circle, which has been aliased into a tiling cell thousands of times.
Here is a video of the first couple thosand patterns:
http://www.youtube.com/watch?v=1UVbUWuyNmk
Here is the Mathematica code used to generate and save the images. There are two parameters that are adjustable: mag is the magnification and must be an integer, with 1 generating 600 by 600 images, 2 generating 1200 by 1200 images, etc. i is a parameter which can be any real number between 0 and ~1000, with values between 0 and 500 being typical (most of the preceding images used i values between 200 and 300). By varying i, thousands of unique diagrams can be created. Small values of i create simple patterns (low degree of aliasing), and large values generate complex patterns (high degree of aliasing).
$HistoryLength = 0;
p = {x, y, L};
nnn = 8;
q = 2.0 Table[{Cos[2 [Pi] j/nnn], Sin[2 [Pi] j/nnn], 0}, {j, nnn}];
k = ConstantArray[I, nnn];
n[x_] := Sqrt[x.x];
conjugate[expr_] := expr /. Complex[x_, y_] -> x - I y;
a = Table[k[[i]]/n[p - q[[i]]], {i, nnn}];
[Gamma] = Table[Exp[-I [Omega] n[p - q[[i]]]/c], {i, nnn}];
expr = [Gamma].a /. {L -> 0.1, c -> 1, [Omega] -> 100};
ff = Compile[{{x, _Real}, {y, _Real}}, Evaluate[expr],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
i = 250;
mag = 1;
d = 6 i mag;
[Delta] = 0.02 i;
nn = Floor[Length[Range[-d, d, [Delta]]]/2];
A = Compile[{{x, _Integer}, {y, _Integer}}, Exp[I (x + y)],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}] @@
Transpose[
Outer[List, Range[Length[Range[-d, d, [Delta]]]],
Range[Length[Range[-d, d, [Delta]]]]], {2, 3, 1}];
SaveImage =
Export[CharacterRange["a", "z"][[RandomInteger[{1, 26}, 20]]] <>
".PNG", #] &;
{#, SaveImage@#} &@
Image[RotateRight[
Abs[Fourier[
1 A mag i/
nnn ff @@
Transpose[
Outer[List, Range[-d, d, [Delta]],
Range[-d, d, [Delta]]], {2, 3, 1}]]], {nn, nn}],
Magnification -> 1]
$endgroup$
42
$begingroup$
Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me.
$endgroup$
– gregschlom
Apr 7 '14 at 9:11
2
$begingroup$
"Visually stunning math concepts which are easy to explain"
$endgroup$
– Rahul
Apr 8 '14 at 1:06
3
$begingroup$
@Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed.
$endgroup$
– DumpsterDoofus
Apr 8 '14 at 1:17
$begingroup$
This reminds me of the Radon transform and the Hough transform.
$endgroup$
– koo
Apr 8 '14 at 23:59
1
$begingroup$
Could you post a link to the code here?
$endgroup$
– Catherine Hwang
Apr 9 '14 at 18:46
|
show 3 more comments
$begingroup$
Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall, for different choices of parameter:
The best thing about them is, they satisfy periodic boundary conditions, and so you can pick one of them and set it as a desktop background by tiling it, resulting in a far more spectacular image than just the single unit cells posted above!
The images seem to be a vast interconnected network of lines once you tile them, but in fact the entire picture is actually just a single circle, which has been aliased into a tiling cell thousands of times.
Here is a video of the first couple thosand patterns:
http://www.youtube.com/watch?v=1UVbUWuyNmk
Here is the Mathematica code used to generate and save the images. There are two parameters that are adjustable: mag is the magnification and must be an integer, with 1 generating 600 by 600 images, 2 generating 1200 by 1200 images, etc. i is a parameter which can be any real number between 0 and ~1000, with values between 0 and 500 being typical (most of the preceding images used i values between 200 and 300). By varying i, thousands of unique diagrams can be created. Small values of i create simple patterns (low degree of aliasing), and large values generate complex patterns (high degree of aliasing).
$HistoryLength = 0;
p = {x, y, L};
nnn = 8;
q = 2.0 Table[{Cos[2 [Pi] j/nnn], Sin[2 [Pi] j/nnn], 0}, {j, nnn}];
k = ConstantArray[I, nnn];
n[x_] := Sqrt[x.x];
conjugate[expr_] := expr /. Complex[x_, y_] -> x - I y;
a = Table[k[[i]]/n[p - q[[i]]], {i, nnn}];
[Gamma] = Table[Exp[-I [Omega] n[p - q[[i]]]/c], {i, nnn}];
expr = [Gamma].a /. {L -> 0.1, c -> 1, [Omega] -> 100};
ff = Compile[{{x, _Real}, {y, _Real}}, Evaluate[expr],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
i = 250;
mag = 1;
d = 6 i mag;
[Delta] = 0.02 i;
nn = Floor[Length[Range[-d, d, [Delta]]]/2];
A = Compile[{{x, _Integer}, {y, _Integer}}, Exp[I (x + y)],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}] @@
Transpose[
Outer[List, Range[Length[Range[-d, d, [Delta]]]],
Range[Length[Range[-d, d, [Delta]]]]], {2, 3, 1}];
SaveImage =
Export[CharacterRange["a", "z"][[RandomInteger[{1, 26}, 20]]] <>
".PNG", #] &;
{#, SaveImage@#} &@
Image[RotateRight[
Abs[Fourier[
1 A mag i/
nnn ff @@
Transpose[
Outer[List, Range[-d, d, [Delta]],
Range[-d, d, [Delta]]], {2, 3, 1}]]], {nn, nn}],
Magnification -> 1]
$endgroup$
42
$begingroup$
Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me.
$endgroup$
– gregschlom
Apr 7 '14 at 9:11
2
$begingroup$
"Visually stunning math concepts which are easy to explain"
$endgroup$
– Rahul
Apr 8 '14 at 1:06
3
$begingroup$
@Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed.
$endgroup$
– DumpsterDoofus
Apr 8 '14 at 1:17
$begingroup$
This reminds me of the Radon transform and the Hough transform.
$endgroup$
– koo
Apr 8 '14 at 23:59
1
$begingroup$
Could you post a link to the code here?
$endgroup$
– Catherine Hwang
Apr 9 '14 at 18:46
|
show 3 more comments
$begingroup$
Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall, for different choices of parameter:
The best thing about them is, they satisfy periodic boundary conditions, and so you can pick one of them and set it as a desktop background by tiling it, resulting in a far more spectacular image than just the single unit cells posted above!
The images seem to be a vast interconnected network of lines once you tile them, but in fact the entire picture is actually just a single circle, which has been aliased into a tiling cell thousands of times.
Here is a video of the first couple thosand patterns:
http://www.youtube.com/watch?v=1UVbUWuyNmk
Here is the Mathematica code used to generate and save the images. There are two parameters that are adjustable: mag is the magnification and must be an integer, with 1 generating 600 by 600 images, 2 generating 1200 by 1200 images, etc. i is a parameter which can be any real number between 0 and ~1000, with values between 0 and 500 being typical (most of the preceding images used i values between 200 and 300). By varying i, thousands of unique diagrams can be created. Small values of i create simple patterns (low degree of aliasing), and large values generate complex patterns (high degree of aliasing).
$HistoryLength = 0;
p = {x, y, L};
nnn = 8;
q = 2.0 Table[{Cos[2 [Pi] j/nnn], Sin[2 [Pi] j/nnn], 0}, {j, nnn}];
k = ConstantArray[I, nnn];
n[x_] := Sqrt[x.x];
conjugate[expr_] := expr /. Complex[x_, y_] -> x - I y;
a = Table[k[[i]]/n[p - q[[i]]], {i, nnn}];
[Gamma] = Table[Exp[-I [Omega] n[p - q[[i]]]/c], {i, nnn}];
expr = [Gamma].a /. {L -> 0.1, c -> 1, [Omega] -> 100};
ff = Compile[{{x, _Real}, {y, _Real}}, Evaluate[expr],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
i = 250;
mag = 1;
d = 6 i mag;
[Delta] = 0.02 i;
nn = Floor[Length[Range[-d, d, [Delta]]]/2];
A = Compile[{{x, _Integer}, {y, _Integer}}, Exp[I (x + y)],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}] @@
Transpose[
Outer[List, Range[Length[Range[-d, d, [Delta]]]],
Range[Length[Range[-d, d, [Delta]]]]], {2, 3, 1}];
SaveImage =
Export[CharacterRange["a", "z"][[RandomInteger[{1, 26}, 20]]] <>
".PNG", #] &;
{#, SaveImage@#} &@
Image[RotateRight[
Abs[Fourier[
1 A mag i/
nnn ff @@
Transpose[
Outer[List, Range[-d, d, [Delta]],
Range[-d, d, [Delta]]], {2, 3, 1}]]], {nn, nn}],
Magnification -> 1]
$endgroup$
Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall, for different choices of parameter:
The best thing about them is, they satisfy periodic boundary conditions, and so you can pick one of them and set it as a desktop background by tiling it, resulting in a far more spectacular image than just the single unit cells posted above!
The images seem to be a vast interconnected network of lines once you tile them, but in fact the entire picture is actually just a single circle, which has been aliased into a tiling cell thousands of times.
Here is a video of the first couple thosand patterns:
http://www.youtube.com/watch?v=1UVbUWuyNmk
Here is the Mathematica code used to generate and save the images. There are two parameters that are adjustable: mag is the magnification and must be an integer, with 1 generating 600 by 600 images, 2 generating 1200 by 1200 images, etc. i is a parameter which can be any real number between 0 and ~1000, with values between 0 and 500 being typical (most of the preceding images used i values between 200 and 300). By varying i, thousands of unique diagrams can be created. Small values of i create simple patterns (low degree of aliasing), and large values generate complex patterns (high degree of aliasing).
$HistoryLength = 0;
p = {x, y, L};
nnn = 8;
q = 2.0 Table[{Cos[2 [Pi] j/nnn], Sin[2 [Pi] j/nnn], 0}, {j, nnn}];
k = ConstantArray[I, nnn];
n[x_] := Sqrt[x.x];
conjugate[expr_] := expr /. Complex[x_, y_] -> x - I y;
a = Table[k[[i]]/n[p - q[[i]]], {i, nnn}];
[Gamma] = Table[Exp[-I [Omega] n[p - q[[i]]]/c], {i, nnn}];
expr = [Gamma].a /. {L -> 0.1, c -> 1, [Omega] -> 100};
ff = Compile[{{x, _Real}, {y, _Real}}, Evaluate[expr],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}];
i = 250;
mag = 1;
d = 6 i mag;
[Delta] = 0.02 i;
nn = Floor[Length[Range[-d, d, [Delta]]]/2];
A = Compile[{{x, _Integer}, {y, _Integer}}, Exp[I (x + y)],
CompilationTarget -> "C", RuntimeAttributes -> {Listable}] @@
Transpose[
Outer[List, Range[Length[Range[-d, d, [Delta]]]],
Range[Length[Range[-d, d, [Delta]]]]], {2, 3, 1}];
SaveImage =
Export[CharacterRange["a", "z"][[RandomInteger[{1, 26}, 20]]] <>
".PNG", #] &;
{#, SaveImage@#} &@
Image[RotateRight[
Abs[Fourier[
1 A mag i/
nnn ff @@
Transpose[
Outer[List, Range[-d, d, [Delta]],
Range[-d, d, [Delta]]], {2, 3, 1}]]], {nn, nn}],
Magnification -> 1]
edited Apr 9 '14 at 23:25
community wiki
3 revs
DumpsterDoofus
42
$begingroup$
Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me.
$endgroup$
– gregschlom
Apr 7 '14 at 9:11
2
$begingroup$
"Visually stunning math concepts which are easy to explain"
$endgroup$
– Rahul
Apr 8 '14 at 1:06
3
$begingroup$
@Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed.
$endgroup$
– DumpsterDoofus
Apr 8 '14 at 1:17
$begingroup$
This reminds me of the Radon transform and the Hough transform.
$endgroup$
– koo
Apr 8 '14 at 23:59
1
$begingroup$
Could you post a link to the code here?
$endgroup$
– Catherine Hwang
Apr 9 '14 at 18:46
|
show 3 more comments
42
$begingroup$
Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me.
$endgroup$
– gregschlom
Apr 7 '14 at 9:11
2
$begingroup$
"Visually stunning math concepts which are easy to explain"
$endgroup$
– Rahul
Apr 8 '14 at 1:06
3
$begingroup$
@Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed.
$endgroup$
– DumpsterDoofus
Apr 8 '14 at 1:17
$begingroup$
This reminds me of the Radon transform and the Hough transform.
$endgroup$
– koo
Apr 8 '14 at 23:59
1
$begingroup$
Could you post a link to the code here?
$endgroup$
– Catherine Hwang
Apr 9 '14 at 18:46
42
42
$begingroup$
Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me.
$endgroup$
– gregschlom
Apr 7 '14 at 9:11
$begingroup$
Those images are incredibly beautiful, but can you explain what exactly they represent and how they were generated? "Fourier transform of the light intensity due to a diffraction pattern caused by light going through 8 pinholes and interfering on a wall" isn't very clear for me.
$endgroup$
– gregschlom
Apr 7 '14 at 9:11
2
2
$begingroup$
"Visually stunning math concepts which are easy to explain"
$endgroup$
– Rahul
Apr 8 '14 at 1:06
$begingroup$
"Visually stunning math concepts which are easy to explain"
$endgroup$
– Rahul
Apr 8 '14 at 1:06
3
3
$begingroup$
@Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed.
$endgroup$
– DumpsterDoofus
Apr 8 '14 at 1:17
$begingroup$
@Rahul: It's aliasing of a circle. Aliasing is easy to explain. Draw a big circle on a clear plastic sheet. Cut the image into little squares. Stack the squares on top of each other, and look at it. That's the image. The different images above were done using little squares with various side-lengths. I can post the code if you'd like, there are literally tens of thousands of visually distinct diagrams which can be formed.
$endgroup$
– DumpsterDoofus
Apr 8 '14 at 1:17
$begingroup$
This reminds me of the Radon transform and the Hough transform.
$endgroup$
– koo
Apr 8 '14 at 23:59
$begingroup$
This reminds me of the Radon transform and the Hough transform.
$endgroup$
– koo
Apr 8 '14 at 23:59
1
1
$begingroup$
Could you post a link to the code here?
$endgroup$
– Catherine Hwang
Apr 9 '14 at 18:46
$begingroup$
Could you post a link to the code here?
$endgroup$
– Catherine Hwang
Apr 9 '14 at 18:46
|
show 3 more comments
$begingroup$
A very satisfying visualization of the area of a circle.
$endgroup$
9
$begingroup$
I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious.
$endgroup$
– Lqueryvg
Sep 1 '14 at 10:52
15
$begingroup$
Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :)
$endgroup$
– msouth
Sep 26 '14 at 0:27
$begingroup$
The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD
$endgroup$
– The Great Duck
Apr 12 '16 at 18:20
add a comment |
$begingroup$
A very satisfying visualization of the area of a circle.
$endgroup$
9
$begingroup$
I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious.
$endgroup$
– Lqueryvg
Sep 1 '14 at 10:52
15
$begingroup$
Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :)
$endgroup$
– msouth
Sep 26 '14 at 0:27
$begingroup$
The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD
$endgroup$
– The Great Duck
Apr 12 '16 at 18:20
add a comment |
$begingroup$
A very satisfying visualization of the area of a circle.
$endgroup$
A very satisfying visualization of the area of a circle.
answered Aug 17 '14 at 8:02
community wiki
StopReadingThisUsername
9
$begingroup$
I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious.
$endgroup$
– Lqueryvg
Sep 1 '14 at 10:52
15
$begingroup$
Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :)
$endgroup$
– msouth
Sep 26 '14 at 0:27
$begingroup$
The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD
$endgroup$
– The Great Duck
Apr 12 '16 at 18:20
add a comment |
9
$begingroup$
I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious.
$endgroup$
– Lqueryvg
Sep 1 '14 at 10:52
15
$begingroup$
Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :)
$endgroup$
– msouth
Sep 26 '14 at 0:27
$begingroup$
The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD
$endgroup$
– The Great Duck
Apr 12 '16 at 18:20
9
9
$begingroup$
I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious.
$endgroup$
– Lqueryvg
Sep 1 '14 at 10:52
$begingroup$
I like this, but one tiny criticism is that it's not visually obvious why the unrolled rings should form a triangle. It's obvious if you can see that the unrolled length of the rings is linearly proportional to their radii, but not visually obvious.
$endgroup$
– Lqueryvg
Sep 1 '14 at 10:52
15
15
$begingroup$
Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :)
$endgroup$
– msouth
Sep 26 '14 at 0:27
$begingroup$
Is it obvious that you can straighten a ring out and get a rectangle when the ring started out bent? I think a lot of these "bang! pi r squared!" ones leave you with real analysis-induced heebie jeebies :)
$endgroup$
– msouth
Sep 26 '14 at 0:27
$begingroup$
The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD
$endgroup$
– The Great Duck
Apr 12 '16 at 18:20
$begingroup$
The rings get smaller as you move in, narrowing to a point. It doesn't matter what the proportion is. It's a simple fact Demonstated by two things. 1. The circles smoothly decrease in length. 2. They narrow to a point. A better way to do this image would be to stack the rings. Then it makes more sense.... Although one can always just rotate by 90 degrees. XD
$endgroup$
– The Great Duck
Apr 12 '16 at 18:20
add a comment |
$begingroup$
Math is always fun to learn. Here are some of the images that explain some things beautifully visually
$endgroup$
$begingroup$
In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes.
$endgroup$
– David K
Aug 28 '14 at 21:25
1
$begingroup$
@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^circ$.
$endgroup$
– Mario Carneiro
Jan 9 '15 at 1:42
$begingroup$
@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all.
$endgroup$
– David K
Jan 9 '15 at 2:56
13
$begingroup$
When I read the number142857it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!"
$endgroup$
– Qwerty
Jan 11 '15 at 4:20
2
$begingroup$
@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 cdot 12$ and see how it corresponds to the diagram.)
$endgroup$
– Martin Ender
Mar 16 '16 at 15:57
|
show 3 more comments
$begingroup$
Math is always fun to learn. Here are some of the images that explain some things beautifully visually
$endgroup$
$begingroup$
In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes.
$endgroup$
– David K
Aug 28 '14 at 21:25
1
$begingroup$
@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^circ$.
$endgroup$
– Mario Carneiro
Jan 9 '15 at 1:42
$begingroup$
@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all.
$endgroup$
– David K
Jan 9 '15 at 2:56
13
$begingroup$
When I read the number142857it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!"
$endgroup$
– Qwerty
Jan 11 '15 at 4:20
2
$begingroup$
@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 cdot 12$ and see how it corresponds to the diagram.)
$endgroup$
– Martin Ender
Mar 16 '16 at 15:57
|
show 3 more comments
$begingroup$
Math is always fun to learn. Here are some of the images that explain some things beautifully visually
$endgroup$
Math is always fun to learn. Here are some of the images that explain some things beautifully visually
edited Aug 29 '14 at 17:58
community wiki
3 revs
Raghav
$begingroup$
In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes.
$endgroup$
– David K
Aug 28 '14 at 21:25
1
$begingroup$
@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^circ$.
$endgroup$
– Mario Carneiro
Jan 9 '15 at 1:42
$begingroup$
@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all.
$endgroup$
– David K
Jan 9 '15 at 2:56
13
$begingroup$
When I read the number142857it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!"
$endgroup$
– Qwerty
Jan 11 '15 at 4:20
2
$begingroup$
@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 cdot 12$ and see how it corresponds to the diagram.)
$endgroup$
– Martin Ender
Mar 16 '16 at 15:57
|
show 3 more comments
$begingroup$
In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes.
$endgroup$
– David K
Aug 28 '14 at 21:25
1
$begingroup$
@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^circ$.
$endgroup$
– Mario Carneiro
Jan 9 '15 at 1:42
$begingroup$
@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all.
$endgroup$
– David K
Jan 9 '15 at 2:56
13
$begingroup$
When I read the number142857it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!"
$endgroup$
– Qwerty
Jan 11 '15 at 4:20
2
$begingroup$
@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 cdot 12$ and see how it corresponds to the diagram.)
$endgroup$
– Martin Ender
Mar 16 '16 at 15:57
$begingroup$
In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes.
$endgroup$
– David K
Aug 28 '14 at 21:25
$begingroup$
In E-M wave propagation, I thought the E and M components were supposed to be $90$ degrees out of phase. The figure appears to have them in phase or $180$ degrees out of phase, depending on how you interpret the directions of the axes.
$endgroup$
– David K
Aug 28 '14 at 21:25
1
1
$begingroup$
@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^circ$.
$endgroup$
– Mario Carneiro
Jan 9 '15 at 1:42
$begingroup$
@DavidK That's only for circularly polarized photons. This is illustrating a plane-polarized wave, whose components are in phase but offset $90^circ$.
$endgroup$
– Mario Carneiro
Jan 9 '15 at 1:42
$begingroup$
@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all.
$endgroup$
– David K
Jan 9 '15 at 2:56
$begingroup$
@MarioCarneiro Quite right, a quick review of wave equations verify the diagram as shown above. I have this mental image of one of my high school teachers walking across the room waving one arm up and down and one arm side to side to simulate the $E$ and $B$ components, but apparently I don't actually remember it very well after all.
$endgroup$
– David K
Jan 9 '15 at 2:56
13
13
$begingroup$
When I read the number
142857 it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!"$endgroup$
– Qwerty
Jan 11 '15 at 4:20
$begingroup$
When I read the number
142857 it was like a dejavu for me and I was like "OH MY GOED!! I'VE KNOWN THIS NUMBER!"$endgroup$
– Qwerty
Jan 11 '15 at 4:20
2
2
$begingroup$
@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 cdot 12$ and see how it corresponds to the diagram.)
$endgroup$
– Martin Ender
Mar 16 '16 at 15:57
$begingroup$
@NateGlenn It's really just a visualisation of long multiplication. (Try writing out the long multiplication of $13 cdot 12$ and see how it corresponds to the diagram.)
$endgroup$
– Martin Ender
Mar 16 '16 at 15:57
|
show 3 more comments
$begingroup$
Francis Galton's Bean machine is interesting as it demonstrates Central Limit Theorem :
$endgroup$
1
$begingroup$
It's one of the best answers.
$endgroup$
– Seifolahi
Mar 16 '16 at 3:42
12
$begingroup$
it probably demonstrates it...
$endgroup$
– enthdegree
Aug 11 '16 at 5:41
add a comment |
$begingroup$
Francis Galton's Bean machine is interesting as it demonstrates Central Limit Theorem :
$endgroup$
1
$begingroup$
It's one of the best answers.
$endgroup$
– Seifolahi
Mar 16 '16 at 3:42
12
$begingroup$
it probably demonstrates it...
$endgroup$
– enthdegree
Aug 11 '16 at 5:41
add a comment |
$begingroup$
Francis Galton's Bean machine is interesting as it demonstrates Central Limit Theorem :
$endgroup$
Francis Galton's Bean machine is interesting as it demonstrates Central Limit Theorem :
edited Mar 11 '16 at 13:38
community wiki
2 revs
Sil
1
$begingroup$
It's one of the best answers.
$endgroup$
– Seifolahi
Mar 16 '16 at 3:42
12
$begingroup$
it probably demonstrates it...
$endgroup$
– enthdegree
Aug 11 '16 at 5:41
add a comment |
1
$begingroup$
It's one of the best answers.
$endgroup$
– Seifolahi
Mar 16 '16 at 3:42
12
$begingroup$
it probably demonstrates it...
$endgroup$
– enthdegree
Aug 11 '16 at 5:41
1
1
$begingroup$
It's one of the best answers.
$endgroup$
– Seifolahi
Mar 16 '16 at 3:42
$begingroup$
It's one of the best answers.
$endgroup$
– Seifolahi
Mar 16 '16 at 3:42
12
12
$begingroup$
it probably demonstrates it...
$endgroup$
– enthdegree
Aug 11 '16 at 5:41
$begingroup$
it probably demonstrates it...
$endgroup$
– enthdegree
Aug 11 '16 at 5:41
add a comment |
$begingroup$
I do not know if this meets your criteria of "visually stunning", but nonetheless -
I like this proof of Pythagoras' Theorem (image taken from www.wisfaq.nl):
The key to understanding this is to realize that the inner quadrilateral must be a square - the sides are equal in length (obviously) and each of its angles is $90^{circ}$ because the two angles on either side sum to $90^{circ}$, and the sum of the three angles is $180^{circ}$. The area of this square is $c^2$.
The outer square's area is $(a + b)^2$, which is $c^2$ plus $2 a b$ which is the total area of the four triangles, each of area $frac{1}{2} a b$.
$(a + b)^2 = c^2 + 2 a b$
$a^2 + b^2 + 2 a b = c^2 + 2 a b$
$a^2 + b^2 = c^2$, which is Pythagoras' theorem.
$endgroup$
1
$begingroup$
+1 finally i can prove Pythagoras theorem without looking into internet or books :)
$endgroup$
– lowtech
Jul 31 '14 at 20:58
$begingroup$
yes it is the most common proof, the easiest to remember
$endgroup$
– reuns
Apr 22 '16 at 21:13
add a comment |
$begingroup$
I do not know if this meets your criteria of "visually stunning", but nonetheless -
I like this proof of Pythagoras' Theorem (image taken from www.wisfaq.nl):
The key to understanding this is to realize that the inner quadrilateral must be a square - the sides are equal in length (obviously) and each of its angles is $90^{circ}$ because the two angles on either side sum to $90^{circ}$, and the sum of the three angles is $180^{circ}$. The area of this square is $c^2$.
The outer square's area is $(a + b)^2$, which is $c^2$ plus $2 a b$ which is the total area of the four triangles, each of area $frac{1}{2} a b$.
$(a + b)^2 = c^2 + 2 a b$
$a^2 + b^2 + 2 a b = c^2 + 2 a b$
$a^2 + b^2 = c^2$, which is Pythagoras' theorem.
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+1 finally i can prove Pythagoras theorem without looking into internet or books :)
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– lowtech
Jul 31 '14 at 20:58
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yes it is the most common proof, the easiest to remember
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– reuns
Apr 22 '16 at 21:13
add a comment |
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I do not know if this meets your criteria of "visually stunning", but nonetheless -
I like this proof of Pythagoras' Theorem (image taken from www.wisfaq.nl):
The key to understanding this is to realize that the inner quadrilateral must be a square - the sides are equal in length (obviously) and each of its angles is $90^{circ}$ because the two angles on either side sum to $90^{circ}$, and the sum of the three angles is $180^{circ}$. The area of this square is $c^2$.
The outer square's area is $(a + b)^2$, which is $c^2$ plus $2 a b$ which is the total area of the four triangles, each of area $frac{1}{2} a b$.
$(a + b)^2 = c^2 + 2 a b$
$a^2 + b^2 + 2 a b = c^2 + 2 a b$
$a^2 + b^2 = c^2$, which is Pythagoras' theorem.
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I do not know if this meets your criteria of "visually stunning", but nonetheless -
I like this proof of Pythagoras' Theorem (image taken from www.wisfaq.nl):
The key to understanding this is to realize that the inner quadrilateral must be a square - the sides are equal in length (obviously) and each of its angles is $90^{circ}$ because the two angles on either side sum to $90^{circ}$, and the sum of the three angles is $180^{circ}$. The area of this square is $c^2$.
The outer square's area is $(a + b)^2$, which is $c^2$ plus $2 a b$ which is the total area of the four triangles, each of area $frac{1}{2} a b$.
$(a + b)^2 = c^2 + 2 a b$
$a^2 + b^2 + 2 a b = c^2 + 2 a b$
$a^2 + b^2 = c^2$, which is Pythagoras' theorem.
answered Apr 1 '14 at 7:25
community wiki
Aky
1
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+1 finally i can prove Pythagoras theorem without looking into internet or books :)
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– lowtech
Jul 31 '14 at 20:58
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yes it is the most common proof, the easiest to remember
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– reuns
Apr 22 '16 at 21:13
add a comment |
1
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+1 finally i can prove Pythagoras theorem without looking into internet or books :)
$endgroup$
– lowtech
Jul 31 '14 at 20:58
$begingroup$
yes it is the most common proof, the easiest to remember
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– reuns
Apr 22 '16 at 21:13
1
1
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+1 finally i can prove Pythagoras theorem without looking into internet or books :)
$endgroup$
– lowtech
Jul 31 '14 at 20:58
$begingroup$
+1 finally i can prove Pythagoras theorem without looking into internet or books :)
$endgroup$
– lowtech
Jul 31 '14 at 20:58
$begingroup$
yes it is the most common proof, the easiest to remember
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– reuns
Apr 22 '16 at 21:13
$begingroup$
yes it is the most common proof, the easiest to remember
$endgroup$
– reuns
Apr 22 '16 at 21:13
add a comment |
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protected by DeepSea Dec 23 '14 at 4:02
Thank you for your interest in this question.
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see also, fractals
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– Guy
Mar 31 '14 at 11:55
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It looks like mathpop or demand for math entertainment)
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– rook
Apr 2 '14 at 12:59
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@ColeJohnson the 'transcendentality" is the beauty of it!
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– Guy
Apr 4 '14 at 17:07
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There is a considerable overlap with mathoverflow.net/questions/8846/proofs-without-words
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– Martin Brandenburg
Jun 20 '14 at 22:13
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@TheGuywithTheHat That's the reason for the second spike of visits, on August 27. The comment by LTS is from April; back then the traffic was driven by Ycombinator. As a result, this same question made both April 7 and August 27 the two days with most visits to the site.
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– user147263
Aug 29 '14 at 18:37