Mixing
Analysis with origami.
$begingroup$
I use this place for the first time. I usually had many questions in sci.math.
Hello teacher~
There is a rectangular(P1) paper.
Fold one of the four points of P1 so that it is on one of the two sides of P1 that did not meet the point. of course, two edges overlap.
(Another expression - Fold a diagonal extending from the top-left corner to the right(or lower) edge of the paper.)
Let the overlapping part be cut out so that the paper made is a rectangular(P2) paper.
Let's repeat this process. so, P3, P4, ... , Pn, ... (They are all rectangles.)
and P1, P2, ..., Pn, ... These are "not" all squares.(given condition)
Let the area of Pn be Sn.
When $$lim_{ntoinfty} frac{S_{n+2}}{S_n} = L$$
Find all possible L values.
Hm... complex.
Without loss of generality, Fix the P1 in width t, length 1.(t>1)
Since Pi is not square for any i, t is not rational number.
so, t is irrational.
Because,
Fix the P1 in width t, length 1 (t>1).
Suppose that t is positive rational, t=b/a (b>a, a and b are natural number).
Let's make a new rectangle with width$times$a, length$times$a
Namely, this rectangle in width b, length a. (b>a)
By Euclidean algorithm,
$a,; b$
=> $a,,,, b-(k_1)a=r_1 ,,,(r1 < a)$
=> $a-(k_2)(r_1)=r_2,,,, r_1 ,,,(r_2 < r_1)$
=> $r_2,,,, (r_1)-(k_3)(r_2)=r_3 ,,,(r_3 < r_2)$
=> ...
so, $ r_n=0 $ for some n.
It means that $(r_{n-1})times n = r_{n-2}$
It means that $P_m$ is a square for some m.
Thus, t is irrational.
Anyway,
If $P_n$ and $P_{n+2}$ are not similar, $lim_{ntoinfty} frac{S_{n+2}}{S_n}$ does not exist.
(I think this is right. but how do you prove it?)
Suppose that $P_n$ are $P_{n+2}$ are similar.
Without loss of generality, Fix the $P_n$ in width t, length 1 again.(t>1) (ratio)
(1) case
$P_{n+1}$ in width (t-1), length 1.
$P_{n+2}$ in width (t-1), length (2-t).
so, ratio of similarity $P_n:P_{n+2} = 1:(2-t)$ and $S_n:S_{n+2}$ = $1^{2} : (2-t)^{2}$
Since $P_n$ are $P_{n+2}$ are similar, 1:t=(2-t):t-1. so t=$frac{1+sqrt{5}}{2}$
so, $S_n:S_{n+2}$ = $1^{2} : (2-t)^{2}$ = 1 : $frac{7-3sqrt{5}}{2}$
so, $lim_{ntoinfty} frac{S_{n+2}}{S_n}$ = $frac{7-3sqrt{5}}{2}$
(2) case
$P_{n+1}$ in width (t-1), length 1.
$P_{n+2}$ in width (t-2), length 1.
so, ratio of similarity $P_n:P_{n+2}$ = 1:(t-2) and $S_n:S_{n+2}$ = $1^2 : (t-2)^2 $
Since $P_n$ are $P_{n+2}$ are similar, 1:t=(t-2):1. so $t=1+sqrt{2}$
so, $S_n:S_{n+2}$ = $1^2 : (t-2)^2$ = $1 : (-1+sqrt{2})^2 = 1:3-2sqrt{2}$
so, $lim_{ntoinfty} frac{S_{n+2}}{S_n} = 3-2sqrt{2}$
Thus, answer is $frac{7-3sqrt{5}}{2}$ , $3-2sqrt{2}$.
origami
edited Jan 20 at 18:02
saulspatz
16.1k31331
asked Jan 20 at 16:12
mina_worldmina_world
1749
$endgroup$
|
show 3 more comments
$begingroup$
I use this place for the first time. I usually had many questions in sci.math.
Hello teacher~
There is a rectangular(P1) paper.
Fold one of the four points of P1 so that it is on one of the two sides of P1 that did not meet the point. of course, two edges overlap.
(Another expression - Fold a diagonal extending from the top-left corner to the right(or lower) edge of the paper.)
Let the overlapping part be cut out so that the paper made is a rectangular(P2) paper.
Let's repeat this process. so, P3, P4, ... , Pn, ... (They are all rectangles.)
and P1, P2, ..., Pn, ... These are "not" all squares.(given condition)
Let the area of Pn be Sn.
When $$lim_{ntoinfty} frac{S_{n+2}}{S_n} = L$$
Find all possible L values.
Hm... complex.
Without loss of generality, Fix the P1 in width t, length 1.(t>1)
Since Pi is not square for any i, t is not rational number.
so, t is irrational.
Because,
Fix the P1 in width t, length 1 (t>1).
Suppose that t is positive rational, t=b/a (b>a, a and b are natural number).
Let's make a new rectangle with width$times$a, length$times$a
Namely, this rectangle in width b, length a. (b>a)
By Euclidean algorithm,
$a,; b$
=> $a,,,, b-(k_1)a=r_1 ,,,(r1 < a)$
=> $a-(k_2)(r_1)=r_2,,,, r_1 ,,,(r_2 < r_1)$
=> $r_2,,,, (r_1)-(k_3)(r_2)=r_3 ,,,(r_3 < r_2)$
=> ...
so, $ r_n=0 $ for some n.
It means that $(r_{n-1})times n = r_{n-2}$
It means that $P_m$ is a square for some m.
Thus, t is irrational.
Anyway,
If $P_n$ and $P_{n+2}$ are not similar, $lim_{ntoinfty} frac{S_{n+2}}{S_n}$ does not exist.
(I think this is right. but how do you prove it?)
Suppose that $P_n$ are $P_{n+2}$ are similar.
Without loss of generality, Fix the $P_n$ in width t, length 1 again.(t>1) (ratio)
(1) case
$P_{n+1}$ in width (t-1), length 1.
$P_{n+2}$ in width (t-1), length (2-t).
so, ratio of similarity $P_n:P_{n+2} = 1:(2-t)$ and $S_n:S_{n+2}$ = $1^{2} : (2-t)^{2}$
Since $P_n$ are $P_{n+2}$ are similar, 1:t=(2-t):t-1. so t=$frac{1+sqrt{5}}{2}$
so, $S_n:S_{n+2}$ = $1^{2} : (2-t)^{2}$ = 1 : $frac{7-3sqrt{5}}{2}$
so, $lim_{ntoinfty} frac{S_{n+2}}{S_n}$ = $frac{7-3sqrt{5}}{2}$
(2) case
$P_{n+1}$ in width (t-1), length 1.
$P_{n+2}$ in width (t-2), length 1.
so, ratio of similarity $P_n:P_{n+2}$ = 1:(t-2) and $S_n:S_{n+2}$ = $1^2 : (t-2)^2 $
Since $P_n$ are $P_{n+2}$ are similar, 1:t=(t-2):1. so $t=1+sqrt{2}$
so, $S_n:S_{n+2}$ = $1^2 : (t-2)^2$ = $1 : (-1+sqrt{2})^2 = 1:3-2sqrt{2}$
so, $lim_{ntoinfty} frac{S_{n+2}}{S_n} = 3-2sqrt{2}$
Thus, answer is $frac{7-3sqrt{5}}{2}$ , $3-2sqrt{2}$.
origami
edited Jan 20 at 18:02
saulspatz
16.1k31331
asked Jan 20 at 16:12
mina_worldmina_world
1749
$endgroup$
$begingroup$
This is not redable. Please format your question with MathJax
$endgroup$
– saulspatz
Jan 20 at 16:17
$begingroup$
Oh, my god...I did my best.
$endgroup$
– mina_world
Jan 20 at 16:56
$begingroup$
It's getting better, but I still don't understand. If I fold a sheet of typing paper so that the lower left corner meets the right side, then I get a quadrilateral, and I can cut out the overlapped triangle to get a rectangle. If I fold it so the lower left corner meets the top, then I get a non-convex figure. I'm not sure what to do now. Is this latter fold disallowed, or is there some way to cut it that I don't see?
$endgroup$
– saulspatz
Jan 20 at 17:34
$begingroup$
Ok, press here("enter image description here" ) in the text.
$endgroup$
– mina_world
Jan 20 at 17:55
$begingroup$
@saulspatz Oh, I attached a picture and it looks right. I visited for the first time and it is very good here.
$endgroup$
– mina_world
Jan 20 at 18:08
|
show 3 more comments
$begingroup$
I use this place for the first time. I usually had many questions in sci.math.
Hello teacher~
There is a rectangular(P1) paper.
Fold one of the four points of P1 so that it is on one of the two sides of P1 that did not meet the point. of course, two edges overlap.
(Another expression - Fold a diagonal extending from the top-left corner to the right(or lower) edge of the paper.)
Let the overlapping part be cut out so that the paper made is a rectangular(P2) paper.
Let's repeat this process. so, P3, P4, ... , Pn, ... (They are all rectangles.)
and P1, P2, ..., Pn, ... These are "not" all squares.(given condition)
Let the area of Pn be Sn.
When $$lim_{ntoinfty} frac{S_{n+2}}{S_n} = L$$
Find all possible L values.
Hm... complex.
Without loss of generality, Fix the P1 in width t, length 1.(t>1)
Since Pi is not square for any i, t is not rational number.
so, t is irrational.
Because,
Fix the P1 in width t, length 1 (t>1).
Suppose that t is positive rational, t=b/a (b>a, a and b are natural number).
Let's make a new rectangle with width$times$a, length$times$a
Namely, this rectangle in width b, length a. (b>a)
By Euclidean algorithm,
$a,; b$
=> $a,,,, b-(k_1)a=r_1 ,,,(r1 < a)$
=> $a-(k_2)(r_1)=r_2,,,, r_1 ,,,(r_2 < r_1)$
=> $r_2,,,, (r_1)-(k_3)(r_2)=r_3 ,,,(r_3 < r_2)$
=> ...
so, $ r_n=0 $ for some n.
It means that $(r_{n-1})times n = r_{n-2}$
It means that $P_m$ is a square for some m.
Thus, t is irrational.
Anyway,
If $P_n$ and $P_{n+2}$ are not similar, $lim_{ntoinfty} frac{S_{n+2}}{S_n}$ does not exist.
(I think this is right. but how do you prove it?)
Suppose that $P_n$ are $P_{n+2}$ are similar.
Without loss of generality, Fix the $P_n$ in width t, length 1 again.(t>1) (ratio)
(1) case
$P_{n+1}$ in width (t-1), length 1.
$P_{n+2}$ in width (t-1), length (2-t).
so, ratio of similarity $P_n:P_{n+2} = 1:(2-t)$ and $S_n:S_{n+2}$ = $1^{2} : (2-t)^{2}$
Since $P_n$ are $P_{n+2}$ are similar, 1:t=(2-t):t-1. so t=$frac{1+sqrt{5}}{2}$
so, $S_n:S_{n+2}$ = $1^{2} : (2-t)^{2}$ = 1 : $frac{7-3sqrt{5}}{2}$
so, $lim_{ntoinfty} frac{S_{n+2}}{S_n}$ = $frac{7-3sqrt{5}}{2}$
(2) case
$P_{n+1}$ in width (t-1), length 1.
$P_{n+2}$ in width (t-2), length 1.
so, ratio of similarity $P_n:P_{n+2}$ = 1:(t-2) and $S_n:S_{n+2}$ = $1^2 : (t-2)^2 $
Since $P_n$ are $P_{n+2}$ are similar, 1:t=(t-2):1. so $t=1+sqrt{2}$
so, $S_n:S_{n+2}$ = $1^2 : (t-2)^2$ = $1 : (-1+sqrt{2})^2 = 1:3-2sqrt{2}$
so, $lim_{ntoinfty} frac{S_{n+2}}{S_n} = 3-2sqrt{2}$
Thus, answer is $frac{7-3sqrt{5}}{2}$ , $3-2sqrt{2}$.
origami
edited Jan 20 at 18:02
saulspatz
16.1k31331
asked Jan 20 at 16:12
mina_worldmina_world
1749
$endgroup$
I use this place for the first time. I usually had many questions in sci.math.
Hello teacher~
There is a rectangular(P1) paper.
Fold one of the four points of P1 so that it is on one of the two sides of P1 that did not meet the point. of course, two edges overlap.
(Another expression - Fold a diagonal extending from the top-left corner to the right(or lower) edge of the paper.)
Let the overlapping part be cut out so that the paper made is a rectangular(P2) paper.
Let's repeat this process. so, P3, P4, ... , Pn, ... (They are all rectangles.)
and P1, P2, ..., Pn, ... These are "not" all squares.(given condition)
Let the area of Pn be Sn.
When $$lim_{ntoinfty} frac{S_{n+2}}{S_n} = L$$
Find all possible L values.
Hm... complex.
Without loss of generality, Fix the P1 in width t, length 1.(t>1)
Since Pi is not square for any i, t is not rational number.
so, t is irrational.
Because,
Fix the P1 in width t, length 1 (t>1).
Suppose that t is positive rational, t=b/a (b>a, a and b are natural number).
Let's make a new rectangle with width$times$a, length$times$a
Namely, this rectangle in width b, length a. (b>a)
By Euclidean algorithm,
$a,; b$
=> $a,,,, b-(k_1)a=r_1 ,,,(r1 < a)$
=> $a-(k_2)(r_1)=r_2,,,, r_1 ,,,(r_2 < r_1)$
=> $r_2,,,, (r_1)-(k_3)(r_2)=r_3 ,,,(r_3 < r_2)$
=> ...
so, $ r_n=0 $ for some n.
It means that $(r_{n-1})times n = r_{n-2}$
It means that $P_m$ is a square for some m.
Thus, t is irrational.
Anyway,
If $P_n$ and $P_{n+2}$ are not similar, $lim_{ntoinfty} frac{S_{n+2}}{S_n}$ does not exist.
(I think this is right. but how do you prove it?)
Suppose that $P_n$ are $P_{n+2}$ are similar.
Without loss of generality, Fix the $P_n$ in width t, length 1 again.(t>1) (ratio)
(1) case
$P_{n+1}$ in width (t-1), length 1.
$P_{n+2}$ in width (t-1), length (2-t).
so, ratio of similarity $P_n:P_{n+2} = 1:(2-t)$ and $S_n:S_{n+2}$ = $1^{2} : (2-t)^{2}$
Since $P_n$ are $P_{n+2}$ are similar, 1:t=(2-t):t-1. so t=$frac{1+sqrt{5}}{2}$
so, $S_n:S_{n+2}$ = $1^{2} : (2-t)^{2}$ = 1 : $frac{7-3sqrt{5}}{2}$
so, $lim_{ntoinfty} frac{S_{n+2}}{S_n}$ = $frac{7-3sqrt{5}}{2}$
(2) case
$P_{n+1}$ in width (t-1), length 1.
$P_{n+2}$ in width (t-2), length 1.
so, ratio of similarity $P_n:P_{n+2}$ = 1:(t-2) and $S_n:S_{n+2}$ = $1^2 : (t-2)^2 $
Since $P_n$ are $P_{n+2}$ are similar, 1:t=(t-2):1. so $t=1+sqrt{2}$
so, $S_n:S_{n+2}$ = $1^2 : (t-2)^2$ = $1 : (-1+sqrt{2})^2 = 1:3-2sqrt{2}$
so, $lim_{ntoinfty} frac{S_{n+2}}{S_n} = 3-2sqrt{2}$
Thus, answer is $frac{7-3sqrt{5}}{2}$ , $3-2sqrt{2}$.
origami
origami
edited Jan 20 at 18:02
saulspatz
16.1k31331
asked Jan 20 at 16:12
mina_worldmina_world
1749
edited Jan 20 at 18:02
saulspatz
16.1k31331
asked Jan 20 at 16:12
mina_worldmina_world
1749
edited Jan 20 at 18:02
saulspatz
16.1k31331
edited Jan 20 at 18:02
saulspatz
16.1k31331
edited Jan 20 at 18:02
saulspatz
16.1k31331
16.1k31331
asked Jan 20 at 16:12
mina_worldmina_world
1749
asked Jan 20 at 16:12
mina_worldmina_world
1749
asked Jan 20 at 16:12
mina_worldmina_world
1749
1749
$begingroup$
This is not redable. Please format your question with MathJax
$endgroup$
– saulspatz
Jan 20 at 16:17
$begingroup$
Oh, my god...I did my best.
$endgroup$
– mina_world
Jan 20 at 16:56
$begingroup$
It's getting better, but I still don't understand. If I fold a sheet of typing paper so that the lower left corner meets the right side, then I get a quadrilateral, and I can cut out the overlapped triangle to get a rectangle. If I fold it so the lower left corner meets the top, then I get a non-convex figure. I'm not sure what to do now. Is this latter fold disallowed, or is there some way to cut it that I don't see?
$endgroup$
– saulspatz
Jan 20 at 17:34
$begingroup$
Ok, press here("enter image description here" ) in the text.
$endgroup$
– mina_world
Jan 20 at 17:55
$begingroup$
@saulspatz Oh, I attached a picture and it looks right. I visited for the first time and it is very good here.
$endgroup$
– mina_world
Jan 20 at 18:08
|
show 3 more comments
$begingroup$
This is not redable. Please format your question with MathJax
$endgroup$
– saulspatz
Jan 20 at 16:17
$begingroup$
Oh, my god...I did my best.
$endgroup$
– mina_world
Jan 20 at 16:56
$begingroup$
It's getting better, but I still don't understand. If I fold a sheet of typing paper so that the lower left corner meets the right side, then I get a quadrilateral, and I can cut out the overlapped triangle to get a rectangle. If I fold it so the lower left corner meets the top, then I get a non-convex figure. I'm not sure what to do now. Is this latter fold disallowed, or is there some way to cut it that I don't see?
$endgroup$
– saulspatz
Jan 20 at 17:34
$begingroup$
Ok, press here("enter image description here" ) in the text.
$endgroup$
– mina_world
Jan 20 at 17:55
$begingroup$
@saulspatz Oh, I attached a picture and it looks right. I visited for the first time and it is very good here.
$endgroup$
– mina_world
Jan 20 at 18:08
$begingroup$
This is not redable. Please format your question with MathJax
$endgroup$
– saulspatz
Jan 20 at 16:17
$begingroup$
This is not redable. Please format your question with MathJax
$endgroup$
– saulspatz
Jan 20 at 16:17
$begingroup$
Oh, my god...I did my best.
$endgroup$
– mina_world
Jan 20 at 16:56
$begingroup$
Oh, my god...I did my best.
$endgroup$
– mina_world
Jan 20 at 16:56
$begingroup$
It's getting better, but I still don't understand. If I fold a sheet of typing paper so that the lower left corner meets the right side, then I get a quadrilateral, and I can cut out the overlapped triangle to get a rectangle. If I fold it so the lower left corner meets the top, then I get a non-convex figure. I'm not sure what to do now. Is this latter fold disallowed, or is there some way to cut it that I don't see?
$endgroup$
– saulspatz
Jan 20 at 17:34
$begingroup$
It's getting better, but I still don't understand. If I fold a sheet of typing paper so that the lower left corner meets the right side, then I get a quadrilateral, and I can cut out the overlapped triangle to get a rectangle. If I fold it so the lower left corner meets the top, then I get a non-convex figure. I'm not sure what to do now. Is this latter fold disallowed, or is there some way to cut it that I don't see?
$endgroup$
– saulspatz
Jan 20 at 17:34
$begingroup$
Ok, press here("enter image description here" ) in the text.
$endgroup$
– mina_world
Jan 20 at 17:55
$begingroup$
Ok, press here("enter image description here" ) in the text.
$endgroup$
– mina_world
Jan 20 at 17:55
$begingroup$
@saulspatz Oh, I attached a picture and it looks right. I visited for the first time and it is very good here.
$endgroup$
– mina_world
Jan 20 at 18:08
$begingroup$
@saulspatz Oh, I attached a picture and it looks right. I visited for the first time and it is very good here.
$endgroup$
– mina_world
Jan 20 at 18:08
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This is a comment, not an answer, but I can't fit what I want to say into a comment. I've been trying to check your calculations, and I'm confused. In case $1,$ which applies when $1<t<2,$ you have $P_n$ is $1$ by $t$ and $P{n+2}$ is $t-1$ by $2-t.$ These two figures are similar, so we have either $${1over
t}={t-1over 2-t}implies t=sqrt{2}tag{1a}$$ or $${1over t}={2-tover t-1}implies t={1+sqrt{5}over2}tag{1b}$$
You only find the the second of these. However, what I really don't follow is how you compute the ratio of $S_n$ to $S_{n+2}.$ Since $S_n$ is the area of $P_n,$ we have $S_n=t,$ and $S_{n+2}=(t-1)(2-t)$ and so $$
{S_{n+2}over S_n} = {(t-1)(t-2)over t}$$
In your calculation of the ratio, you seem to be assuming that $P_{n+1}$ and $P_n$ are similar, but this is not the case. To find the limits, substitute the appropriate values of $t$ in the last expression. (By the way, in relation to $(1a)$ note that this is the ratio for a sheet of A4 paper.)
You case $2$ applies when $t>2.$ We have that $P_n$ is $t$ by $1$ and $P_{n+2}$ is $t-2$ by $1.$ Again we have two cases, $$ {1over t}={1over 2-t}implies t=1tag{2a}$$ and $${1over t}= {2-tover 1} implies t=1tag{2b}$$
Both $(2a)$ and $(2b)$ are contradictory, so case $2$ cannot arise.
The most interesting aspect of your post is the question you raise, "Does the existence of the limit imply that $P_n$ and $P_{n+2}$ are similar, for $n$ sufficiently large?" Like you, I think that this is probably true, but I haven't been able to prove it.
I think the answer may have to do with regular continued fractions. As you pointed out, this paper-folding is related to the Euclidean algorithm as it appears in the Elements, using repeated subtraction, rather than division in the modern algorithm. The calculation of regular continued fractions is essentially the modern Euclidean algorithm, as explained here, for example.
Let me just mention the continued fraction expansions of the two ratios. In the notation of the first link cited,$${1+sqrt{5}over 2}=[1;1,1,1,dots]$$
and $$ sqrt{2}=[1;2,2,2dots]$$ When the ratio is ${1+sqrt5over2}$ we always fold alternate sides of the paper. First the top, then the left, then the top, and so on. When the ratio is $sqrt2,$ after the initial fold, we always fold the same side twice in succession. Fits the top, then the left twice, then the top twice, and so on.
While it's plain that continued fractions have something to do with this problem, I haven't been able to turn that insight to account. In particular, how do the areas of the rectangles enter into it?
If I can't figure out how to solve this, I may post a question of my own.
answered Feb 25 at 18:55
saulspatzsaulspatz
16.1k31331
$endgroup$
add a comment |
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This is a comment, not an answer, but I can't fit what I want to say into a comment. I've been trying to check your calculations, and I'm confused. In case $1,$ which applies when $1<t<2,$ you have $P_n$ is $1$ by $t$ and $P{n+2}$ is $t-1$ by $2-t.$ These two figures are similar, so we have either $${1over
t}={t-1over 2-t}implies t=sqrt{2}tag{1a}$$ or $${1over t}={2-tover t-1}implies t={1+sqrt{5}over2}tag{1b}$$
You only find the the second of these. However, what I really don't follow is how you compute the ratio of $S_n$ to $S_{n+2}.$ Since $S_n$ is the area of $P_n,$ we have $S_n=t,$ and $S_{n+2}=(t-1)(2-t)$ and so $$
{S_{n+2}over S_n} = {(t-1)(t-2)over t}$$
In your calculation of the ratio, you seem to be assuming that $P_{n+1}$ and $P_n$ are similar, but this is not the case. To find the limits, substitute the appropriate values of $t$ in the last expression. (By the way, in relation to $(1a)$ note that this is the ratio for a sheet of A4 paper.)
You case $2$ applies when $t>2.$ We have that $P_n$ is $t$ by $1$ and $P_{n+2}$ is $t-2$ by $1.$ Again we have two cases, $$ {1over t}={1over 2-t}implies t=1tag{2a}$$ and $${1over t}= {2-tover 1} implies t=1tag{2b}$$
Both $(2a)$ and $(2b)$ are contradictory, so case $2$ cannot arise.
The most interesting aspect of your post is the question you raise, "Does the existence of the limit imply that $P_n$ and $P_{n+2}$ are similar, for $n$ sufficiently large?" Like you, I think that this is probably true, but I haven't been able to prove it.
I think the answer may have to do with regular continued fractions. As you pointed out, this paper-folding is related to the Euclidean algorithm as it appears in the Elements, using repeated subtraction, rather than division in the modern algorithm. The calculation of regular continued fractions is essentially the modern Euclidean algorithm, as explained here, for example.
Let me just mention the continued fraction expansions of the two ratios. In the notation of the first link cited,$${1+sqrt{5}over 2}=[1;1,1,1,dots]$$
and $$ sqrt{2}=[1;2,2,2dots]$$ When the ratio is ${1+sqrt5over2}$ we always fold alternate sides of the paper. First the top, then the left, then the top, and so on. When the ratio is $sqrt2,$ after the initial fold, we always fold the same side twice in succession. Fits the top, then the left twice, then the top twice, and so on.
While it's plain that continued fractions have something to do with this problem, I haven't been able to turn that insight to account. In particular, how do the areas of the rectangles enter into it?
If I can't figure out how to solve this, I may post a question of my own.
answered Feb 25 at 18:55
saulspatzsaulspatz
16.1k31331
$endgroup$
add a comment |
$begingroup$
This is a comment, not an answer, but I can't fit what I want to say into a comment. I've been trying to check your calculations, and I'm confused. In case $1,$ which applies when $1<t<2,$ you have $P_n$ is $1$ by $t$ and $P{n+2}$ is $t-1$ by $2-t.$ These two figures are similar, so we have either $${1over
t}={t-1over 2-t}implies t=sqrt{2}tag{1a}$$ or $${1over t}={2-tover t-1}implies t={1+sqrt{5}over2}tag{1b}$$
You only find the the second of these. However, what I really don't follow is how you compute the ratio of $S_n$ to $S_{n+2}.$ Since $S_n$ is the area of $P_n,$ we have $S_n=t,$ and $S_{n+2}=(t-1)(2-t)$ and so $$
{S_{n+2}over S_n} = {(t-1)(t-2)over t}$$
In your calculation of the ratio, you seem to be assuming that $P_{n+1}$ and $P_n$ are similar, but this is not the case. To find the limits, substitute the appropriate values of $t$ in the last expression. (By the way, in relation to $(1a)$ note that this is the ratio for a sheet of A4 paper.)
You case $2$ applies when $t>2.$ We have that $P_n$ is $t$ by $1$ and $P_{n+2}$ is $t-2$ by $1.$ Again we have two cases, $$ {1over t}={1over 2-t}implies t=1tag{2a}$$ and $${1over t}= {2-tover 1} implies t=1tag{2b}$$
Both $(2a)$ and $(2b)$ are contradictory, so case $2$ cannot arise.
The most interesting aspect of your post is the question you raise, "Does the existence of the limit imply that $P_n$ and $P_{n+2}$ are similar, for $n$ sufficiently large?" Like you, I think that this is probably true, but I haven't been able to prove it.
I think the answer may have to do with regular continued fractions. As you pointed out, this paper-folding is related to the Euclidean algorithm as it appears in the Elements, using repeated subtraction, rather than division in the modern algorithm. The calculation of regular continued fractions is essentially the modern Euclidean algorithm, as explained here, for example.
Let me just mention the continued fraction expansions of the two ratios. In the notation of the first link cited,$${1+sqrt{5}over 2}=[1;1,1,1,dots]$$
and $$ sqrt{2}=[1;2,2,2dots]$$ When the ratio is ${1+sqrt5over2}$ we always fold alternate sides of the paper. First the top, then the left, then the top, and so on. When the ratio is $sqrt2,$ after the initial fold, we always fold the same side twice in succession. Fits the top, then the left twice, then the top twice, and so on.
While it's plain that continued fractions have something to do with this problem, I haven't been able to turn that insight to account. In particular, how do the areas of the rectangles enter into it?
If I can't figure out how to solve this, I may post a question of my own.
answered Feb 25 at 18:55
saulspatzsaulspatz
16.1k31331
$endgroup$
add a comment |
$begingroup$
This is a comment, not an answer, but I can't fit what I want to say into a comment. I've been trying to check your calculations, and I'm confused. In case $1,$ which applies when $1<t<2,$ you have $P_n$ is $1$ by $t$ and $P{n+2}$ is $t-1$ by $2-t.$ These two figures are similar, so we have either $${1over
t}={t-1over 2-t}implies t=sqrt{2}tag{1a}$$ or $${1over t}={2-tover t-1}implies t={1+sqrt{5}over2}tag{1b}$$
You only find the the second of these. However, what I really don't follow is how you compute the ratio of $S_n$ to $S_{n+2}.$ Since $S_n$ is the area of $P_n,$ we have $S_n=t,$ and $S_{n+2}=(t-1)(2-t)$ and so $$
{S_{n+2}over S_n} = {(t-1)(t-2)over t}$$
In your calculation of the ratio, you seem to be assuming that $P_{n+1}$ and $P_n$ are similar, but this is not the case. To find the limits, substitute the appropriate values of $t$ in the last expression. (By the way, in relation to $(1a)$ note that this is the ratio for a sheet of A4 paper.)
You case $2$ applies when $t>2.$ We have that $P_n$ is $t$ by $1$ and $P_{n+2}$ is $t-2$ by $1.$ Again we have two cases, $$ {1over t}={1over 2-t}implies t=1tag{2a}$$ and $${1over t}= {2-tover 1} implies t=1tag{2b}$$
Both $(2a)$ and $(2b)$ are contradictory, so case $2$ cannot arise.
The most interesting aspect of your post is the question you raise, "Does the existence of the limit imply that $P_n$ and $P_{n+2}$ are similar, for $n$ sufficiently large?" Like you, I think that this is probably true, but I haven't been able to prove it.
I think the answer may have to do with regular continued fractions. As you pointed out, this paper-folding is related to the Euclidean algorithm as it appears in the Elements, using repeated subtraction, rather than division in the modern algorithm. The calculation of regular continued fractions is essentially the modern Euclidean algorithm, as explained here, for example.
Let me just mention the continued fraction expansions of the two ratios. In the notation of the first link cited,$${1+sqrt{5}over 2}=[1;1,1,1,dots]$$
and $$ sqrt{2}=[1;2,2,2dots]$$ When the ratio is ${1+sqrt5over2}$ we always fold alternate sides of the paper. First the top, then the left, then the top, and so on. When the ratio is $sqrt2,$ after the initial fold, we always fold the same side twice in succession. Fits the top, then the left twice, then the top twice, and so on.
While it's plain that continued fractions have something to do with this problem, I haven't been able to turn that insight to account. In particular, how do the areas of the rectangles enter into it?
If I can't figure out how to solve this, I may post a question of my own.
answered Feb 25 at 18:55
saulspatzsaulspatz
16.1k31331
$endgroup$
This is a comment, not an answer, but I can't fit what I want to say into a comment. I've been trying to check your calculations, and I'm confused. In case $1,$ which applies when $1<t<2,$ you have $P_n$ is $1$ by $t$ and $P{n+2}$ is $t-1$ by $2-t.$ These two figures are similar, so we have either $${1over
t}={t-1over 2-t}implies t=sqrt{2}tag{1a}$$ or $${1over t}={2-tover t-1}implies t={1+sqrt{5}over2}tag{1b}$$
You only find the the second of these. However, what I really don't follow is how you compute the ratio of $S_n$ to $S_{n+2}.$ Since $S_n$ is the area of $P_n,$ we have $S_n=t,$ and $S_{n+2}=(t-1)(2-t)$ and so $$
{S_{n+2}over S_n} = {(t-1)(t-2)over t}$$
In your calculation of the ratio, you seem to be assuming that $P_{n+1}$ and $P_n$ are similar, but this is not the case. To find the limits, substitute the appropriate values of $t$ in the last expression. (By the way, in relation to $(1a)$ note that this is the ratio for a sheet of A4 paper.)
You case $2$ applies when $t>2.$ We have that $P_n$ is $t$ by $1$ and $P_{n+2}$ is $t-2$ by $1.$ Again we have two cases, $$ {1over t}={1over 2-t}implies t=1tag{2a}$$ and $${1over t}= {2-tover 1} implies t=1tag{2b}$$
Both $(2a)$ and $(2b)$ are contradictory, so case $2$ cannot arise.
The most interesting aspect of your post is the question you raise, "Does the existence of the limit imply that $P_n$ and $P_{n+2}$ are similar, for $n$ sufficiently large?" Like you, I think that this is probably true, but I haven't been able to prove it.
I think the answer may have to do with regular continued fractions. As you pointed out, this paper-folding is related to the Euclidean algorithm as it appears in the Elements, using repeated subtraction, rather than division in the modern algorithm. The calculation of regular continued fractions is essentially the modern Euclidean algorithm, as explained here, for example.
Let me just mention the continued fraction expansions of the two ratios. In the notation of the first link cited,$${1+sqrt{5}over 2}=[1;1,1,1,dots]$$
and $$ sqrt{2}=[1;2,2,2dots]$$ When the ratio is ${1+sqrt5over2}$ we always fold alternate sides of the paper. First the top, then the left, then the top, and so on. When the ratio is $sqrt2,$ after the initial fold, we always fold the same side twice in succession. Fits the top, then the left twice, then the top twice, and so on.
While it's plain that continued fractions have something to do with this problem, I haven't been able to turn that insight to account. In particular, how do the areas of the rectangles enter into it?
If I can't figure out how to solve this, I may post a question of my own.
answered Feb 25 at 18:55
saulspatzsaulspatz
16.1k31331
answered Feb 25 at 18:55
saulspatzsaulspatz
16.1k31331
answered Feb 25 at 18:55
saulspatzsaulspatz
16.1k31331
answered Feb 25 at 18:55
saulspatzsaulspatz
16.1k31331
16.1k31331
add a comment |
add a comment |
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$begingroup$
This is not redable. Please format your question with MathJax
$endgroup$
– saulspatz
Jan 20 at 16:17
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Oh, my god...I did my best.
$endgroup$
– mina_world
Jan 20 at 16:56
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It's getting better, but I still don't understand. If I fold a sheet of typing paper so that the lower left corner meets the right side, then I get a quadrilateral, and I can cut out the overlapped triangle to get a rectangle. If I fold it so the lower left corner meets the top, then I get a non-convex figure. I'm not sure what to do now. Is this latter fold disallowed, or is there some way to cut it that I don't see?
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– saulspatz
Jan 20 at 17:34
$begingroup$
Ok, press here("enter image description here" ) in the text.
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– mina_world
Jan 20 at 17:55
$begingroup$
@saulspatz Oh, I attached a picture and it looks right. I visited for the first time and it is very good here.
$endgroup$
– mina_world
Jan 20 at 18:08