Mixing
Anybody know a proof of $prod_{n=1}^inftycos(x/2^n)=sin x/x$.
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 '18 at 11:11
Gaby Alfonso
682315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
add a comment |
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 '18 at 11:11
Gaby Alfonso
682315
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 |
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 '18 at 11:11
Gaby Alfonso
682315
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 '18 at 11:11
Gaby Alfonso
682315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
edited Dec 10 '18 at 11:11
Gaby Alfonso
682315
asked Mar 18 '15 at 17:46
Tim Raczkowski
17.3k21242
edited Dec 10 '18 at 11:11
Gaby Alfonso
682315
edited Dec 10 '18 at 11:11
Gaby Alfonso
682315
edited Dec 10 '18 at 11:11
Gaby Alfonso
682315
682315
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
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
34.7k22247
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 |
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
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1195825%2fanybody-know-a-proof-of-prod-n-1-infty-cosx-2n-sin-x-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
34.7k22247
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 |
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
34.7k22247
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 |
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
34.7k22247
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
34.7k22247
edited Mar 18 '15 at 18:02
answered Mar 18 '15 at 17:50
kobe
34.7k22247
answered Mar 18 '15 at 17:50
kobe
34.7k22247
answered Mar 18 '15 at 17:50
kobe
34.7k22247
34.7k22247
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 |
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 |
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 |
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1195825%2fanybody-know-a-proof-of-prod-n-1-infty-cosx-2n-sin-x-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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