Using Wallis' product to derive $sqrtpi$
$begingroup$
Recall Wallis' product:
$$lim_{ntoinfty}Big(frac{2}{1}cdotfrac{2}{3}cdotfrac{4}{3}cdotfrac{4}{5}cdotfrac{6}{5}cdotsfrac{2n}{2n-1}cdotfrac{2n}{2n+1}Big)=frac{pi}{2}$$
We have to show that $$lim_{ntoinfty}frac{(n!)^22^{2n}}{(2n)!sqrt n}=sqrtpi$$
The hint I got was to use $$P_n=frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2cdot2cdot4cdot4cdots(2n)cdot(2n)$ alternatively. How do I use $P_n$ to derive $sqrtpi:$?
sequences-and-series analysis
$endgroup$
add a comment |
$begingroup$
Recall Wallis' product:
$$lim_{ntoinfty}Big(frac{2}{1}cdotfrac{2}{3}cdotfrac{4}{3}cdotfrac{4}{5}cdotfrac{6}{5}cdotsfrac{2n}{2n-1}cdotfrac{2n}{2n+1}Big)=frac{pi}{2}$$
We have to show that $$lim_{ntoinfty}frac{(n!)^22^{2n}}{(2n)!sqrt n}=sqrtpi$$
The hint I got was to use $$P_n=frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2cdot2cdot4cdot4cdots(2n)cdot(2n)$ alternatively. How do I use $P_n$ to derive $sqrtpi:$?
sequences-and-series analysis
$endgroup$
$begingroup$
I think, you should need Stirling formula.
$endgroup$
– hamam_Abdallah
Dec 8 '18 at 19:58
$begingroup$
@hamam_Abdallah that comes later, so I can't use it.
$endgroup$
– Tomás Palamás
Dec 8 '18 at 19:59
add a comment |
$begingroup$
Recall Wallis' product:
$$lim_{ntoinfty}Big(frac{2}{1}cdotfrac{2}{3}cdotfrac{4}{3}cdotfrac{4}{5}cdotfrac{6}{5}cdotsfrac{2n}{2n-1}cdotfrac{2n}{2n+1}Big)=frac{pi}{2}$$
We have to show that $$lim_{ntoinfty}frac{(n!)^22^{2n}}{(2n)!sqrt n}=sqrtpi$$
The hint I got was to use $$P_n=frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2cdot2cdot4cdot4cdots(2n)cdot(2n)$ alternatively. How do I use $P_n$ to derive $sqrtpi:$?
sequences-and-series analysis
$endgroup$
Recall Wallis' product:
$$lim_{ntoinfty}Big(frac{2}{1}cdotfrac{2}{3}cdotfrac{4}{3}cdotfrac{4}{5}cdotfrac{6}{5}cdotsfrac{2n}{2n-1}cdotfrac{2n}{2n+1}Big)=frac{pi}{2}$$
We have to show that $$lim_{ntoinfty}frac{(n!)^22^{2n}}{(2n)!sqrt n}=sqrtpi$$
The hint I got was to use $$P_n=frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2cdot2cdot4cdot4cdots(2n)cdot(2n)$ alternatively. How do I use $P_n$ to derive $sqrtpi:$?
sequences-and-series analysis
sequences-and-series analysis
asked Dec 8 '18 at 19:44
Tomás PalamásTomás Palamás
363211
363211
$begingroup$
I think, you should need Stirling formula.
$endgroup$
– hamam_Abdallah
Dec 8 '18 at 19:58
$begingroup$
@hamam_Abdallah that comes later, so I can't use it.
$endgroup$
– Tomás Palamás
Dec 8 '18 at 19:59
add a comment |
$begingroup$
I think, you should need Stirling formula.
$endgroup$
– hamam_Abdallah
Dec 8 '18 at 19:58
$begingroup$
@hamam_Abdallah that comes later, so I can't use it.
$endgroup$
– Tomás Palamás
Dec 8 '18 at 19:59
$begingroup$
I think, you should need Stirling formula.
$endgroup$
– hamam_Abdallah
Dec 8 '18 at 19:58
$begingroup$
I think, you should need Stirling formula.
$endgroup$
– hamam_Abdallah
Dec 8 '18 at 19:58
$begingroup$
@hamam_Abdallah that comes later, so I can't use it.
$endgroup$
– Tomás Palamás
Dec 8 '18 at 19:59
$begingroup$
@hamam_Abdallah that comes later, so I can't use it.
$endgroup$
– Tomás Palamás
Dec 8 '18 at 19:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that
$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$
Where the last equality was got by translation invariance.
$endgroup$
$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10
$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12
$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13
$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15
$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24
|
show 4 more comments
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%2f3031546%2fusing-wallis-product-to-derive-sqrt-pi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that
$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$
Where the last equality was got by translation invariance.
$endgroup$
$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10
$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12
$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13
$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15
$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24
|
show 4 more comments
$begingroup$
As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that
$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$
Where the last equality was got by translation invariance.
$endgroup$
$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10
$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12
$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13
$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15
$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24
|
show 4 more comments
$begingroup$
As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that
$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$
Where the last equality was got by translation invariance.
$endgroup$
As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that
$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$
Where the last equality was got by translation invariance.
edited Dec 9 '18 at 3:47
answered Dec 8 '18 at 20:03
Dionel JaimeDionel Jaime
1,748514
1,748514
$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10
$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12
$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13
$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15
$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24
|
show 4 more comments
$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10
$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12
$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13
$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15
$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24
$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10
$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10
$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12
$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12
$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13
$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13
$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15
$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15
$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24
$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24
|
show 4 more comments
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.
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%2f3031546%2fusing-wallis-product-to-derive-sqrt-pi%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
$begingroup$
I think, you should need Stirling formula.
$endgroup$
– hamam_Abdallah
Dec 8 '18 at 19:58
$begingroup$
@hamam_Abdallah that comes later, so I can't use it.
$endgroup$
– Tomás Palamás
Dec 8 '18 at 19:59