Mixing
Is there a naive proof that $x - frac{x^3}{3!} + frac{x^5}{5!} - ldots$ has period $2pi$?
$begingroup$
I recently visited the far away land of Polynomia. The mathematicians in Polynomia are quite sophisticated algebraists: they know a lot about polynomials and their associated machinery - rings, fields, algebraic geometry, etc. But they aren't very good at analysis; they don't know much about differential equations and don't like sophisticated estimates. They're pretty good with the theory of power series because it involves taking limits of polynomials (which they love), and so they've managed to figure out at least some complex analysis.
In my recent visit I got into a discussion about the power series
$$f(x) = x - frac{x^3}{3!} + frac{x^5}{5!} - ldots$$
They knew how to prove that this power series converges everywhere on the complex plane, but they were astonished when I told them that $f$ is periodic with period $2pi$. (They are aware that the polynomial equation $x^2 + y^2 = 1$ defines a curve in $mathbb{R}^2$, and they define $2pi$ to be its arclength.) You see, since they don't really like differential equations they don't know about functions like $sin x$, $e^x$, etc.
So the Polynomians were pretty incredulous about my claim and they demanded that I prove it. The proofs I know rely heavily on methods like path integrals of transcendental functions, and their eyes just glazed over. They're looking for some property of the partial sums of $f$ which, in the limit, guarantees that $f$ is periodic with period $2 pi$. Circles are almost certainly going to have to enter into it and I can probably convince to accept path integrals of polynomials along a circle, but the more algebraic the argument the better. Can anyone help?
calculus sequences-and-series
edited Jan 11 at 7:30
Jyrki Lahtonen
109k13169372
asked May 29 '18 at 19:26
Paul SiegelPaul Siegel
5,8261942
$endgroup$
|
show 5 more comments
$begingroup$
I recently visited the far away land of Polynomia. The mathematicians in Polynomia are quite sophisticated algebraists: they know a lot about polynomials and their associated machinery - rings, fields, algebraic geometry, etc. But they aren't very good at analysis; they don't know much about differential equations and don't like sophisticated estimates. They're pretty good with the theory of power series because it involves taking limits of polynomials (which they love), and so they've managed to figure out at least some complex analysis.
In my recent visit I got into a discussion about the power series
$$f(x) = x - frac{x^3}{3!} + frac{x^5}{5!} - ldots$$
They knew how to prove that this power series converges everywhere on the complex plane, but they were astonished when I told them that $f$ is periodic with period $2pi$. (They are aware that the polynomial equation $x^2 + y^2 = 1$ defines a curve in $mathbb{R}^2$, and they define $2pi$ to be its arclength.) You see, since they don't really like differential equations they don't know about functions like $sin x$, $e^x$, etc.
So the Polynomians were pretty incredulous about my claim and they demanded that I prove it. The proofs I know rely heavily on methods like path integrals of transcendental functions, and their eyes just glazed over. They're looking for some property of the partial sums of $f$ which, in the limit, guarantees that $f$ is periodic with period $2 pi$. Circles are almost certainly going to have to enter into it and I can probably convince to accept path integrals of polynomials along a circle, but the more algebraic the argument the better. Can anyone help?
calculus sequences-and-series
edited Jan 11 at 7:30
Jyrki Lahtonen
109k13169372
asked May 29 '18 at 19:26
Paul SiegelPaul Siegel
5,8261942
$endgroup$
5
$begingroup$
If they don't know about differentiation, how do they define arc length? Or more generally, path integrals?
$endgroup$
– Thomas Andrews
May 29 '18 at 19:33
5
$begingroup$
The formal derivative is pretty well-defined in Polynomia and you can use its properties to show that $f()$ and its formal derivative $dot{f}$ satisfy $f(x)^2 +dot{f}(x)^2=1$ and that in fact they parametrize the circle with constant (formal) speed.
$endgroup$
– Steven Stadnicki
May 29 '18 at 19:35
4
$begingroup$
With the companion series of $g(x)=1-x^2/2!+cdots$ their methods seem to allow proofs for $$f(x)^2+g(x)^2=1$$ and $$f(x+y)=f(x)g(y)+f(y)g(x).$$ So if you can convince them of the fact that there exists a positive number $2pi$ such that $f(2pi)=0$, $g(2pi)=1$ you are in business. Proving that $2pi$ is related to the arc length of the circle OTOH...
$endgroup$
– Jyrki Lahtonen
May 29 '18 at 19:37
1
$begingroup$
Give them a book about differential equation and you solved the issue.
$endgroup$
– Zacky
May 29 '18 at 19:41
4
$begingroup$
This question would make more sense if you gave an algebraic definition of $pi$. You gave a definition requiring calculus and geometry and then insisted that we can't use calculus or geometry.
$endgroup$
– DanielV
May 29 '18 at 20:15
|
show 5 more comments
$begingroup$
I recently visited the far away land of Polynomia. The mathematicians in Polynomia are quite sophisticated algebraists: they know a lot about polynomials and their associated machinery - rings, fields, algebraic geometry, etc. But they aren't very good at analysis; they don't know much about differential equations and don't like sophisticated estimates. They're pretty good with the theory of power series because it involves taking limits of polynomials (which they love), and so they've managed to figure out at least some complex analysis.
In my recent visit I got into a discussion about the power series
$$f(x) = x - frac{x^3}{3!} + frac{x^5}{5!} - ldots$$
They knew how to prove that this power series converges everywhere on the complex plane, but they were astonished when I told them that $f$ is periodic with period $2pi$. (They are aware that the polynomial equation $x^2 + y^2 = 1$ defines a curve in $mathbb{R}^2$, and they define $2pi$ to be its arclength.) You see, since they don't really like differential equations they don't know about functions like $sin x$, $e^x$, etc.
So the Polynomians were pretty incredulous about my claim and they demanded that I prove it. The proofs I know rely heavily on methods like path integrals of transcendental functions, and their eyes just glazed over. They're looking for some property of the partial sums of $f$ which, in the limit, guarantees that $f$ is periodic with period $2 pi$. Circles are almost certainly going to have to enter into it and I can probably convince to accept path integrals of polynomials along a circle, but the more algebraic the argument the better. Can anyone help?
calculus sequences-and-series
edited Jan 11 at 7:30
Jyrki Lahtonen
109k13169372
asked May 29 '18 at 19:26
Paul SiegelPaul Siegel
5,8261942
$endgroup$
I recently visited the far away land of Polynomia. The mathematicians in Polynomia are quite sophisticated algebraists: they know a lot about polynomials and their associated machinery - rings, fields, algebraic geometry, etc. But they aren't very good at analysis; they don't know much about differential equations and don't like sophisticated estimates. They're pretty good with the theory of power series because it involves taking limits of polynomials (which they love), and so they've managed to figure out at least some complex analysis.
In my recent visit I got into a discussion about the power series
$$f(x) = x - frac{x^3}{3!} + frac{x^5}{5!} - ldots$$
They knew how to prove that this power series converges everywhere on the complex plane, but they were astonished when I told them that $f$ is periodic with period $2pi$. (They are aware that the polynomial equation $x^2 + y^2 = 1$ defines a curve in $mathbb{R}^2$, and they define $2pi$ to be its arclength.) You see, since they don't really like differential equations they don't know about functions like $sin x$, $e^x$, etc.
So the Polynomians were pretty incredulous about my claim and they demanded that I prove it. The proofs I know rely heavily on methods like path integrals of transcendental functions, and their eyes just glazed over. They're looking for some property of the partial sums of $f$ which, in the limit, guarantees that $f$ is periodic with period $2 pi$. Circles are almost certainly going to have to enter into it and I can probably convince to accept path integrals of polynomials along a circle, but the more algebraic the argument the better. Can anyone help?
calculus sequences-and-series
calculus sequences-and-series
edited Jan 11 at 7:30
Jyrki Lahtonen
109k13169372
asked May 29 '18 at 19:26
Paul SiegelPaul Siegel
5,8261942
edited Jan 11 at 7:30
Jyrki Lahtonen
109k13169372
asked May 29 '18 at 19:26
Paul SiegelPaul Siegel
5,8261942
edited Jan 11 at 7:30
Jyrki Lahtonen
109k13169372
edited Jan 11 at 7:30
Jyrki Lahtonen
109k13169372
edited Jan 11 at 7:30
Jyrki Lahtonen
109k13169372
109k13169372
asked May 29 '18 at 19:26
Paul SiegelPaul Siegel
5,8261942
asked May 29 '18 at 19:26
Paul SiegelPaul Siegel
5,8261942
asked May 29 '18 at 19:26
Paul SiegelPaul Siegel
5,8261942
5,8261942
5
$begingroup$
If they don't know about differentiation, how do they define arc length? Or more generally, path integrals?
$endgroup$
– Thomas Andrews
May 29 '18 at 19:33
5
$begingroup$
The formal derivative is pretty well-defined in Polynomia and you can use its properties to show that $f()$ and its formal derivative $dot{f}$ satisfy $f(x)^2 +dot{f}(x)^2=1$ and that in fact they parametrize the circle with constant (formal) speed.
$endgroup$
– Steven Stadnicki
May 29 '18 at 19:35
4
$begingroup$
With the companion series of $g(x)=1-x^2/2!+cdots$ their methods seem to allow proofs for $$f(x)^2+g(x)^2=1$$ and $$f(x+y)=f(x)g(y)+f(y)g(x).$$ So if you can convince them of the fact that there exists a positive number $2pi$ such that $f(2pi)=0$, $g(2pi)=1$ you are in business. Proving that $2pi$ is related to the arc length of the circle OTOH...
$endgroup$
– Jyrki Lahtonen
May 29 '18 at 19:37
1
$begingroup$
Give them a book about differential equation and you solved the issue.
$endgroup$
– Zacky
May 29 '18 at 19:41
4
$begingroup$
This question would make more sense if you gave an algebraic definition of $pi$. You gave a definition requiring calculus and geometry and then insisted that we can't use calculus or geometry.
$endgroup$
– DanielV
May 29 '18 at 20:15
|
show 5 more comments
5
$begingroup$
If they don't know about differentiation, how do they define arc length? Or more generally, path integrals?
$endgroup$
– Thomas Andrews
May 29 '18 at 19:33
5
$begingroup$
The formal derivative is pretty well-defined in Polynomia and you can use its properties to show that $f()$ and its formal derivative $dot{f}$ satisfy $f(x)^2 +dot{f}(x)^2=1$ and that in fact they parametrize the circle with constant (formal) speed.
$endgroup$
– Steven Stadnicki
May 29 '18 at 19:35
4
$begingroup$
With the companion series of $g(x)=1-x^2/2!+cdots$ their methods seem to allow proofs for $$f(x)^2+g(x)^2=1$$ and $$f(x+y)=f(x)g(y)+f(y)g(x).$$ So if you can convince them of the fact that there exists a positive number $2pi$ such that $f(2pi)=0$, $g(2pi)=1$ you are in business. Proving that $2pi$ is related to the arc length of the circle OTOH...
$endgroup$
– Jyrki Lahtonen
May 29 '18 at 19:37
1
$begingroup$
Give them a book about differential equation and you solved the issue.
$endgroup$
– Zacky
May 29 '18 at 19:41
4
$begingroup$
This question would make more sense if you gave an algebraic definition of $pi$. You gave a definition requiring calculus and geometry and then insisted that we can't use calculus or geometry.
$endgroup$
– DanielV
May 29 '18 at 20:15
5
5
$begingroup$
If they don't know about differentiation, how do they define arc length? Or more generally, path integrals?
$endgroup$
– Thomas Andrews
May 29 '18 at 19:33
$begingroup$
If they don't know about differentiation, how do they define arc length? Or more generally, path integrals?
$endgroup$
– Thomas Andrews
May 29 '18 at 19:33
5
5
$begingroup$
The formal derivative is pretty well-defined in Polynomia and you can use its properties to show that $f()$ and its formal derivative $dot{f}$ satisfy $f(x)^2 +dot{f}(x)^2=1$ and that in fact they parametrize the circle with constant (formal) speed.
$endgroup$
– Steven Stadnicki
May 29 '18 at 19:35
$begingroup$
The formal derivative is pretty well-defined in Polynomia and you can use its properties to show that $f()$ and its formal derivative $dot{f}$ satisfy $f(x)^2 +dot{f}(x)^2=1$ and that in fact they parametrize the circle with constant (formal) speed.
$endgroup$
– Steven Stadnicki
May 29 '18 at 19:35
4
4
$begingroup$
With the companion series of $g(x)=1-x^2/2!+cdots$ their methods seem to allow proofs for $$f(x)^2+g(x)^2=1$$ and $$f(x+y)=f(x)g(y)+f(y)g(x).$$ So if you can convince them of the fact that there exists a positive number $2pi$ such that $f(2pi)=0$, $g(2pi)=1$ you are in business. Proving that $2pi$ is related to the arc length of the circle OTOH...
$endgroup$
– Jyrki Lahtonen
May 29 '18 at 19:37
$begingroup$
With the companion series of $g(x)=1-x^2/2!+cdots$ their methods seem to allow proofs for $$f(x)^2+g(x)^2=1$$ and $$f(x+y)=f(x)g(y)+f(y)g(x).$$ So if you can convince them of the fact that there exists a positive number $2pi$ such that $f(2pi)=0$, $g(2pi)=1$ you are in business. Proving that $2pi$ is related to the arc length of the circle OTOH...
$endgroup$
– Jyrki Lahtonen
May 29 '18 at 19:37
1
1
$begingroup$
Give them a book about differential equation and you solved the issue.
$endgroup$
– Zacky
May 29 '18 at 19:41
$begingroup$
Give them a book about differential equation and you solved the issue.
$endgroup$
– Zacky
May 29 '18 at 19:41
4
4
$begingroup$
This question would make more sense if you gave an algebraic definition of $pi$. You gave a definition requiring calculus and geometry and then insisted that we can't use calculus or geometry.
$endgroup$
– DanielV
May 29 '18 at 20:15
$begingroup$
This question would make more sense if you gave an algebraic definition of $pi$. You gave a definition requiring calculus and geometry and then insisted that we can't use calculus or geometry.
$endgroup$
– DanielV
May 29 '18 at 20:15
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
There is a companion to that series,
$$g(x)=1-frac{x^2}2+frac{x^4}{4!}+cdots$$
and together they have a funny property: $$h(x+y):=g(x+y)+if(x+y)=(g(x)+if(x))(g(y)+if(y))=h(x)h(y).$$
This can be shown in an elementary way by developing the powers of $x+y$ using the binomial theorem and identifying to the products of the partial sums.
Then assuming that by some magic (such as the intermediate value theorem) we can show that there is a solution to
$$h(2pi)=1,$$
where $pi$ is the unknown, then for all $x$
$$h(x+2pi)=h(x).$$
answered May 29 '18 at 19:44
Yves DaoustYves Daoust
127k673226
$endgroup$
4
$begingroup$
They are great at algebraic manipulations, so I suspect that they will have no trouble with the identity $h(x+y) = h(x)h(y)$. I might be able to convince them that $h(x) = 1$ has a positive real solution, but I'm most worried about showing that the smallest positive real solution is related somehow to the unit circle. Maybe there's a way to show that the sequence of lengths of regular polygons with radius 1 converge to a solution, or something?
$endgroup$
– Paul Siegel
May 29 '18 at 19:52
$begingroup$
Using polygonal approximations for the circle to somehow do this would be great... maybe that's the real question you should ask (separately)...
$endgroup$
– T_M
May 29 '18 at 20:19
1
$begingroup$
Rolling the sleeves a little more, you can expand $g^2(x)+f^2(x)$ and observe the nice cancellation of the coefficients...
$endgroup$
– Yves Daoust
May 29 '18 at 20:21
add a comment |
$begingroup$
To follow up on Yves Daoust's answer, you could try explaining with some algebraic manipulation that $h(x)=lim_{m to infty}(1+frac{x}{m})^m$. Then you could argue $h(2pi i)=1$ by pointing out that multiplying a complex number $z$ by $(1+delta i)$, where $delta$ is a small positive real number is approximately the same thing as rotating $z$ by $delta$ radians anticlockwise.
answered Jun 10 '18 at 13:41
user142857user142857
32
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is a companion to that series,
$$g(x)=1-frac{x^2}2+frac{x^4}{4!}+cdots$$
and together they have a funny property: $$h(x+y):=g(x+y)+if(x+y)=(g(x)+if(x))(g(y)+if(y))=h(x)h(y).$$
This can be shown in an elementary way by developing the powers of $x+y$ using the binomial theorem and identifying to the products of the partial sums.
Then assuming that by some magic (such as the intermediate value theorem) we can show that there is a solution to
$$h(2pi)=1,$$
where $pi$ is the unknown, then for all $x$
$$h(x+2pi)=h(x).$$
answered May 29 '18 at 19:44
Yves DaoustYves Daoust
127k673226
$endgroup$
4
$begingroup$
They are great at algebraic manipulations, so I suspect that they will have no trouble with the identity $h(x+y) = h(x)h(y)$. I might be able to convince them that $h(x) = 1$ has a positive real solution, but I'm most worried about showing that the smallest positive real solution is related somehow to the unit circle. Maybe there's a way to show that the sequence of lengths of regular polygons with radius 1 converge to a solution, or something?
$endgroup$
– Paul Siegel
May 29 '18 at 19:52
$begingroup$
Using polygonal approximations for the circle to somehow do this would be great... maybe that's the real question you should ask (separately)...
$endgroup$
– T_M
May 29 '18 at 20:19
1
$begingroup$
Rolling the sleeves a little more, you can expand $g^2(x)+f^2(x)$ and observe the nice cancellation of the coefficients...
$endgroup$
– Yves Daoust
May 29 '18 at 20:21
add a comment |
$begingroup$
There is a companion to that series,
$$g(x)=1-frac{x^2}2+frac{x^4}{4!}+cdots$$
and together they have a funny property: $$h(x+y):=g(x+y)+if(x+y)=(g(x)+if(x))(g(y)+if(y))=h(x)h(y).$$
This can be shown in an elementary way by developing the powers of $x+y$ using the binomial theorem and identifying to the products of the partial sums.
Then assuming that by some magic (such as the intermediate value theorem) we can show that there is a solution to
$$h(2pi)=1,$$
where $pi$ is the unknown, then for all $x$
$$h(x+2pi)=h(x).$$
answered May 29 '18 at 19:44
Yves DaoustYves Daoust
127k673226
$endgroup$
4
$begingroup$
They are great at algebraic manipulations, so I suspect that they will have no trouble with the identity $h(x+y) = h(x)h(y)$. I might be able to convince them that $h(x) = 1$ has a positive real solution, but I'm most worried about showing that the smallest positive real solution is related somehow to the unit circle. Maybe there's a way to show that the sequence of lengths of regular polygons with radius 1 converge to a solution, or something?
$endgroup$
– Paul Siegel
May 29 '18 at 19:52
$begingroup$
Using polygonal approximations for the circle to somehow do this would be great... maybe that's the real question you should ask (separately)...
$endgroup$
– T_M
May 29 '18 at 20:19
1
$begingroup$
Rolling the sleeves a little more, you can expand $g^2(x)+f^2(x)$ and observe the nice cancellation of the coefficients...
$endgroup$
– Yves Daoust
May 29 '18 at 20:21
add a comment |
$begingroup$
There is a companion to that series,
$$g(x)=1-frac{x^2}2+frac{x^4}{4!}+cdots$$
and together they have a funny property: $$h(x+y):=g(x+y)+if(x+y)=(g(x)+if(x))(g(y)+if(y))=h(x)h(y).$$
This can be shown in an elementary way by developing the powers of $x+y$ using the binomial theorem and identifying to the products of the partial sums.
Then assuming that by some magic (such as the intermediate value theorem) we can show that there is a solution to
$$h(2pi)=1,$$
where $pi$ is the unknown, then for all $x$
$$h(x+2pi)=h(x).$$
answered May 29 '18 at 19:44
Yves DaoustYves Daoust
127k673226
$endgroup$
There is a companion to that series,
$$g(x)=1-frac{x^2}2+frac{x^4}{4!}+cdots$$
and together they have a funny property: $$h(x+y):=g(x+y)+if(x+y)=(g(x)+if(x))(g(y)+if(y))=h(x)h(y).$$
This can be shown in an elementary way by developing the powers of $x+y$ using the binomial theorem and identifying to the products of the partial sums.
Then assuming that by some magic (such as the intermediate value theorem) we can show that there is a solution to
$$h(2pi)=1,$$
where $pi$ is the unknown, then for all $x$
$$h(x+2pi)=h(x).$$
answered May 29 '18 at 19:44
Yves DaoustYves Daoust
127k673226
answered May 29 '18 at 19:44
Yves DaoustYves Daoust
127k673226
answered May 29 '18 at 19:44
Yves DaoustYves Daoust
127k673226
answered May 29 '18 at 19:44
Yves DaoustYves Daoust
127k673226
127k673226
4
$begingroup$
They are great at algebraic manipulations, so I suspect that they will have no trouble with the identity $h(x+y) = h(x)h(y)$. I might be able to convince them that $h(x) = 1$ has a positive real solution, but I'm most worried about showing that the smallest positive real solution is related somehow to the unit circle. Maybe there's a way to show that the sequence of lengths of regular polygons with radius 1 converge to a solution, or something?
$endgroup$
– Paul Siegel
May 29 '18 at 19:52
$begingroup$
Using polygonal approximations for the circle to somehow do this would be great... maybe that's the real question you should ask (separately)...
$endgroup$
– T_M
May 29 '18 at 20:19
1
$begingroup$
Rolling the sleeves a little more, you can expand $g^2(x)+f^2(x)$ and observe the nice cancellation of the coefficients...
$endgroup$
– Yves Daoust
May 29 '18 at 20:21
add a comment |
4
$begingroup$
They are great at algebraic manipulations, so I suspect that they will have no trouble with the identity $h(x+y) = h(x)h(y)$. I might be able to convince them that $h(x) = 1$ has a positive real solution, but I'm most worried about showing that the smallest positive real solution is related somehow to the unit circle. Maybe there's a way to show that the sequence of lengths of regular polygons with radius 1 converge to a solution, or something?
$endgroup$
– Paul Siegel
May 29 '18 at 19:52
$begingroup$
Using polygonal approximations for the circle to somehow do this would be great... maybe that's the real question you should ask (separately)...
$endgroup$
– T_M
May 29 '18 at 20:19
1
$begingroup$
Rolling the sleeves a little more, you can expand $g^2(x)+f^2(x)$ and observe the nice cancellation of the coefficients...
$endgroup$
– Yves Daoust
May 29 '18 at 20:21
4
4
$begingroup$
They are great at algebraic manipulations, so I suspect that they will have no trouble with the identity $h(x+y) = h(x)h(y)$. I might be able to convince them that $h(x) = 1$ has a positive real solution, but I'm most worried about showing that the smallest positive real solution is related somehow to the unit circle. Maybe there's a way to show that the sequence of lengths of regular polygons with radius 1 converge to a solution, or something?
$endgroup$
– Paul Siegel
May 29 '18 at 19:52
$begingroup$
They are great at algebraic manipulations, so I suspect that they will have no trouble with the identity $h(x+y) = h(x)h(y)$. I might be able to convince them that $h(x) = 1$ has a positive real solution, but I'm most worried about showing that the smallest positive real solution is related somehow to the unit circle. Maybe there's a way to show that the sequence of lengths of regular polygons with radius 1 converge to a solution, or something?
$endgroup$
– Paul Siegel
May 29 '18 at 19:52
$begingroup$
Using polygonal approximations for the circle to somehow do this would be great... maybe that's the real question you should ask (separately)...
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– T_M
May 29 '18 at 20:19
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Using polygonal approximations for the circle to somehow do this would be great... maybe that's the real question you should ask (separately)...
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– T_M
May 29 '18 at 20:19
1
1
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Rolling the sleeves a little more, you can expand $g^2(x)+f^2(x)$ and observe the nice cancellation of the coefficients...
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– Yves Daoust
May 29 '18 at 20:21
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Rolling the sleeves a little more, you can expand $g^2(x)+f^2(x)$ and observe the nice cancellation of the coefficients...
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– Yves Daoust
May 29 '18 at 20:21
add a comment |
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To follow up on Yves Daoust's answer, you could try explaining with some algebraic manipulation that $h(x)=lim_{m to infty}(1+frac{x}{m})^m$. Then you could argue $h(2pi i)=1$ by pointing out that multiplying a complex number $z$ by $(1+delta i)$, where $delta$ is a small positive real number is approximately the same thing as rotating $z$ by $delta$ radians anticlockwise.
answered Jun 10 '18 at 13:41
user142857user142857
32
$endgroup$
add a comment |
$begingroup$
To follow up on Yves Daoust's answer, you could try explaining with some algebraic manipulation that $h(x)=lim_{m to infty}(1+frac{x}{m})^m$. Then you could argue $h(2pi i)=1$ by pointing out that multiplying a complex number $z$ by $(1+delta i)$, where $delta$ is a small positive real number is approximately the same thing as rotating $z$ by $delta$ radians anticlockwise.
answered Jun 10 '18 at 13:41
user142857user142857
32
$endgroup$
add a comment |
$begingroup$
To follow up on Yves Daoust's answer, you could try explaining with some algebraic manipulation that $h(x)=lim_{m to infty}(1+frac{x}{m})^m$. Then you could argue $h(2pi i)=1$ by pointing out that multiplying a complex number $z$ by $(1+delta i)$, where $delta$ is a small positive real number is approximately the same thing as rotating $z$ by $delta$ radians anticlockwise.
answered Jun 10 '18 at 13:41
user142857user142857
32
$endgroup$
To follow up on Yves Daoust's answer, you could try explaining with some algebraic manipulation that $h(x)=lim_{m to infty}(1+frac{x}{m})^m$. Then you could argue $h(2pi i)=1$ by pointing out that multiplying a complex number $z$ by $(1+delta i)$, where $delta$ is a small positive real number is approximately the same thing as rotating $z$ by $delta$ radians anticlockwise.
answered Jun 10 '18 at 13:41
user142857user142857
32
answered Jun 10 '18 at 13:41
user142857user142857
32
answered Jun 10 '18 at 13:41
user142857user142857
32
answered Jun 10 '18 at 13:41
user142857user142857
32
32
add a comment |
add a comment |
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If they don't know about differentiation, how do they define arc length? Or more generally, path integrals?
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– Thomas Andrews
May 29 '18 at 19:33
5
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The formal derivative is pretty well-defined in Polynomia and you can use its properties to show that $f()$ and its formal derivative $dot{f}$ satisfy $f(x)^2 +dot{f}(x)^2=1$ and that in fact they parametrize the circle with constant (formal) speed.
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– Steven Stadnicki
May 29 '18 at 19:35
4
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With the companion series of $g(x)=1-x^2/2!+cdots$ their methods seem to allow proofs for $$f(x)^2+g(x)^2=1$$ and $$f(x+y)=f(x)g(y)+f(y)g(x).$$ So if you can convince them of the fact that there exists a positive number $2pi$ such that $f(2pi)=0$, $g(2pi)=1$ you are in business. Proving that $2pi$ is related to the arc length of the circle OTOH...
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– Jyrki Lahtonen
May 29 '18 at 19:37
1
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Give them a book about differential equation and you solved the issue.
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– Zacky
May 29 '18 at 19:41
4
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This question would make more sense if you gave an algebraic definition of $pi$. You gave a definition requiring calculus and geometry and then insisted that we can't use calculus or geometry.
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– DanielV
May 29 '18 at 20:15