Anybody know a proof of $prod_{n=1}^inftycos(x/2^n)=sin x/x$.
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This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values $x$ for which $prod_{n=1}^infty cosleft(largefrac{x}{2^n}right)$ converges. I've shown that the product converges for all $x$. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that
$$largeprod_{n=1}^inftycosleft(frac{x}{2^n}right)=frac{sin (x)}{x}.$$
I can't figure out how to prove this.
real-analysis infinite-product
edited Dec 10 at 11:11
Gaby Alfonso
670315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
add a comment |
up vote
12
down vote
favorite
This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values $x$ for which $prod_{n=1}^infty cosleft(largefrac{x}{2^n}right)$ converges. I've shown that the product converges for all $x$. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that
$$largeprod_{n=1}^inftycosleft(frac{x}{2^n}right)=frac{sin (x)}{x}.$$
I can't figure out how to prove this.
real-analysis infinite-product
edited Dec 10 at 11:11
Gaby Alfonso
670315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
2
It has the correct set of zeroes, hence the quotient is an entire analytic function
– Hagen von Eitzen
Mar 18 '15 at 17:49
add a comment |
up vote
12
down vote
favorite
up vote
12
down vote
favorite
This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values $x$ for which $prod_{n=1}^infty cosleft(largefrac{x}{2^n}right)$ converges. I've shown that the product converges for all $x$. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that
$$largeprod_{n=1}^inftycosleft(frac{x}{2^n}right)=frac{sin (x)}{x}.$$
I can't figure out how to prove this.
real-analysis infinite-product
edited Dec 10 at 11:11
Gaby Alfonso
670315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values $x$ for which $prod_{n=1}^infty cosleft(largefrac{x}{2^n}right)$ converges. I've shown that the product converges for all $x$. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that
$$largeprod_{n=1}^inftycosleft(frac{x}{2^n}right)=frac{sin (x)}{x}.$$
I can't figure out how to prove this.
real-analysis infinite-product
real-analysis infinite-product
edited Dec 10 at 11:11
Gaby Alfonso
670315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
edited Dec 10 at 11:11
Gaby Alfonso
670315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
edited Dec 10 at 11:11
Gaby Alfonso
670315
edited Dec 10 at 11:11
Gaby Alfonso
670315
edited Dec 10 at 11:11
Gaby Alfonso
670315
670315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
17.3k21242
2
It has the correct set of zeroes, hence the quotient is an entire analytic function
– Hagen von Eitzen
Mar 18 '15 at 17:49
add a comment |
2
It has the correct set of zeroes, hence the quotient is an entire analytic function
– Hagen von Eitzen
Mar 18 '15 at 17:49
2
2
It has the correct set of zeroes, hence the quotient is an entire analytic function
– Hagen von Eitzen
Mar 18 '15 at 17:49
It has the correct set of zeroes, hence the quotient is an entire analytic function
– Hagen von Eitzen
Mar 18 '15 at 17:49
add a comment |
2 Answers
2
active
oldest
votes
up vote
15
down vote
accepted
Using the trig identity
$$sin (2t) = 2sin (t) cos (t),$$
we have
$$prod_{n = 1}^N cos(x/2^n) = prod_{n = 1}^N frac{sin(x/2^{n-1})}{2sin(x/2^n)} = frac{sin(x)}{2^Nsin(x/2^N)} = frac{sin x}{x}cdot frac{x/2^N}{sin(x/2^N)}$$
Take the limit as $N to infty$ and use the fact $lim_{tto 0} frac{sin t}{t} = 1$ to obtain the result.
answered Mar 18 '15 at 17:50
kobe
34.6k22247
I'm not the OP, so feel free to ignore this request, but I'm lost on the argument that $lim_{Ntoinfty} 2^N sin(2^{-N}x) = x$
– jameselmore
Mar 18 '15 at 17:59
@jameselmore By the Taylor expansion we have $sin(2^{-N}x)sim_{Ntoinfty}2^{-N}x$.
– user63181
Mar 18 '15 at 18:02
@jameselmore I've added more details. Since $(sin t)/t to 1$ as $t to 0$, $(sin x/2^N)/(x/2^N) to 1$ as $N to infty$. Therefore $(x/2^N)/sin(x/2^N) to 1$ as $N to infty$.
– kobe
Mar 18 '15 at 18:04
@kobe, Thanks! Makes a lot of sense
– jameselmore
Mar 18 '15 at 18:06
add a comment |
up vote
5
down vote
Hint
$$cos(x/2^n)=frac12frac{sin(x/2^{n-1})}{sin(x/2^n)}$$
and telescope.
How do you incorporate the denominator $x$ into your proof?
– DeepSea
Mar 18 '15 at 17:51
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
Using the trig identity
$$sin (2t) = 2sin (t) cos (t),$$
we have
$$prod_{n = 1}^N cos(x/2^n) = prod_{n = 1}^N frac{sin(x/2^{n-1})}{2sin(x/2^n)} = frac{sin(x)}{2^Nsin(x/2^N)} = frac{sin x}{x}cdot frac{x/2^N}{sin(x/2^N)}$$
Take the limit as $N to infty$ and use the fact $lim_{tto 0} frac{sin t}{t} = 1$ to obtain the result.
answered Mar 18 '15 at 17:50
kobe
34.6k22247
I'm not the OP, so feel free to ignore this request, but I'm lost on the argument that $lim_{Ntoinfty} 2^N sin(2^{-N}x) = x$
– jameselmore
Mar 18 '15 at 17:59
@jameselmore By the Taylor expansion we have $sin(2^{-N}x)sim_{Ntoinfty}2^{-N}x$.
– user63181
Mar 18 '15 at 18:02
@jameselmore I've added more details. Since $(sin t)/t to 1$ as $t to 0$, $(sin x/2^N)/(x/2^N) to 1$ as $N to infty$. Therefore $(x/2^N)/sin(x/2^N) to 1$ as $N to infty$.
– kobe
Mar 18 '15 at 18:04
@kobe, Thanks! Makes a lot of sense
– jameselmore
Mar 18 '15 at 18:06
add a comment |
up vote
15
down vote
accepted
Using the trig identity
$$sin (2t) = 2sin (t) cos (t),$$
we have
$$prod_{n = 1}^N cos(x/2^n) = prod_{n = 1}^N frac{sin(x/2^{n-1})}{2sin(x/2^n)} = frac{sin(x)}{2^Nsin(x/2^N)} = frac{sin x}{x}cdot frac{x/2^N}{sin(x/2^N)}$$
Take the limit as $N to infty$ and use the fact $lim_{tto 0} frac{sin t}{t} = 1$ to obtain the result.
answered Mar 18 '15 at 17:50
kobe
34.6k22247
I'm not the OP, so feel free to ignore this request, but I'm lost on the argument that $lim_{Ntoinfty} 2^N sin(2^{-N}x) = x$
– jameselmore
Mar 18 '15 at 17:59
@jameselmore By the Taylor expansion we have $sin(2^{-N}x)sim_{Ntoinfty}2^{-N}x$.
– user63181
Mar 18 '15 at 18:02
@jameselmore I've added more details. Since $(sin t)/t to 1$ as $t to 0$, $(sin x/2^N)/(x/2^N) to 1$ as $N to infty$. Therefore $(x/2^N)/sin(x/2^N) to 1$ as $N to infty$.
– kobe
Mar 18 '15 at 18:04
@kobe, Thanks! Makes a lot of sense
– jameselmore
Mar 18 '15 at 18:06
add a comment |
up vote
15
down vote
accepted
up vote
15
down vote
accepted
Using the trig identity
$$sin (2t) = 2sin (t) cos (t),$$
we have
$$prod_{n = 1}^N cos(x/2^n) = prod_{n = 1}^N frac{sin(x/2^{n-1})}{2sin(x/2^n)} = frac{sin(x)}{2^Nsin(x/2^N)} = frac{sin x}{x}cdot frac{x/2^N}{sin(x/2^N)}$$
Take the limit as $N to infty$ and use the fact $lim_{tto 0} frac{sin t}{t} = 1$ to obtain the result.
answered Mar 18 '15 at 17:50
kobe
34.6k22247
Using the trig identity
$$sin (2t) = 2sin (t) cos (t),$$
we have
$$prod_{n = 1}^N cos(x/2^n) = prod_{n = 1}^N frac{sin(x/2^{n-1})}{2sin(x/2^n)} = frac{sin(x)}{2^Nsin(x/2^N)} = frac{sin x}{x}cdot frac{x/2^N}{sin(x/2^N)}$$
Take the limit as $N to infty$ and use the fact $lim_{tto 0} frac{sin t}{t} = 1$ to obtain the result.
answered Mar 18 '15 at 17:50
kobe
34.6k22247
edited Mar 18 '15 at 18:02
answered Mar 18 '15 at 17:50
kobe
34.6k22247
answered Mar 18 '15 at 17:50
kobe
34.6k22247
answered Mar 18 '15 at 17:50
kobe
34.6k22247
34.6k22247
I'm not the OP, so feel free to ignore this request, but I'm lost on the argument that $lim_{Ntoinfty} 2^N sin(2^{-N}x) = x$
– jameselmore
Mar 18 '15 at 17:59
@jameselmore By the Taylor expansion we have $sin(2^{-N}x)sim_{Ntoinfty}2^{-N}x$.
– user63181
Mar 18 '15 at 18:02
@jameselmore I've added more details. Since $(sin t)/t to 1$ as $t to 0$, $(sin x/2^N)/(x/2^N) to 1$ as $N to infty$. Therefore $(x/2^N)/sin(x/2^N) to 1$ as $N to infty$.
– kobe
Mar 18 '15 at 18:04
@kobe, Thanks! Makes a lot of sense
– jameselmore
Mar 18 '15 at 18:06
add a comment |
I'm not the OP, so feel free to ignore this request, but I'm lost on the argument that $lim_{Ntoinfty} 2^N sin(2^{-N}x) = x$
– jameselmore
Mar 18 '15 at 17:59
@jameselmore By the Taylor expansion we have $sin(2^{-N}x)sim_{Ntoinfty}2^{-N}x$.
– user63181
Mar 18 '15 at 18:02
@jameselmore I've added more details. Since $(sin t)/t to 1$ as $t to 0$, $(sin x/2^N)/(x/2^N) to 1$ as $N to infty$. Therefore $(x/2^N)/sin(x/2^N) to 1$ as $N to infty$.
– kobe
Mar 18 '15 at 18:04
@kobe, Thanks! Makes a lot of sense
– jameselmore
Mar 18 '15 at 18:06
I'm not the OP, so feel free to ignore this request, but I'm lost on the argument that $lim_{Ntoinfty} 2^N sin(2^{-N}x) = x$
– jameselmore
Mar 18 '15 at 17:59
I'm not the OP, so feel free to ignore this request, but I'm lost on the argument that $lim_{Ntoinfty} 2^N sin(2^{-N}x) = x$
– jameselmore
Mar 18 '15 at 17:59
@jameselmore By the Taylor expansion we have $sin(2^{-N}x)sim_{Ntoinfty}2^{-N}x$.
– user63181
Mar 18 '15 at 18:02
@jameselmore By the Taylor expansion we have $sin(2^{-N}x)sim_{Ntoinfty}2^{-N}x$.
– user63181
Mar 18 '15 at 18:02
@jameselmore I've added more details. Since $(sin t)/t to 1$ as $t to 0$, $(sin x/2^N)/(x/2^N) to 1$ as $N to infty$. Therefore $(x/2^N)/sin(x/2^N) to 1$ as $N to infty$.
– kobe
Mar 18 '15 at 18:04
@jameselmore I've added more details. Since $(sin t)/t to 1$ as $t to 0$, $(sin x/2^N)/(x/2^N) to 1$ as $N to infty$. Therefore $(x/2^N)/sin(x/2^N) to 1$ as $N to infty$.
– kobe
Mar 18 '15 at 18:04
@kobe, Thanks! Makes a lot of sense
– jameselmore
Mar 18 '15 at 18:06
@kobe, Thanks! Makes a lot of sense
– jameselmore
Mar 18 '15 at 18:06
add a comment |
up vote
5
down vote
Hint
$$cos(x/2^n)=frac12frac{sin(x/2^{n-1})}{sin(x/2^n)}$$
and telescope.
How do you incorporate the denominator $x$ into your proof?
– DeepSea
Mar 18 '15 at 17:51
add a comment |
up vote
5
down vote
Hint
$$cos(x/2^n)=frac12frac{sin(x/2^{n-1})}{sin(x/2^n)}$$
and telescope.
How do you incorporate the denominator $x$ into your proof?
– DeepSea
Mar 18 '15 at 17:51
add a comment |
up vote
5
down vote
up vote
5
down vote
Hint
$$cos(x/2^n)=frac12frac{sin(x/2^{n-1})}{sin(x/2^n)}$$
and telescope.
Hint
$$cos(x/2^n)=frac12frac{sin(x/2^{n-1})}{sin(x/2^n)}$$
and telescope.
answered Mar 18 '15 at 17:49
user63181
How do you incorporate the denominator $x$ into your proof?
– DeepSea
Mar 18 '15 at 17:51
add a comment |
How do you incorporate the denominator $x$ into your proof?
– DeepSea
Mar 18 '15 at 17:51
How do you incorporate the denominator $x$ into your proof?
– DeepSea
Mar 18 '15 at 17:51
How do you incorporate the denominator $x$ into your proof?
– DeepSea
Mar 18 '15 at 17:51
add a comment |
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2
It has the correct set of zeroes, hence the quotient is an entire analytic function
– Hagen von Eitzen
Mar 18 '15 at 17:49