Sum to Infinity of Trigonometry to $pi$
$begingroup$
For
$$y=sum_{n=0}^a2cdot2^ncdottanleft(frac{45}{2^n}right)cdotsinleft(frac{90}{2^n}right)^2$$
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
4-sided regular→ 8-sided regular→ 16-sided regular→ 32-sided regular→... →n-sided regular
(When n tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
We have already tried Geometric Sequence Infinite Sum, but there does not seem to have a common ratio.
Moreover, we have used our calculator to input the numbers up to 128[(sin ...] and we get the value of 3.140... which is very close to π But we can't be completely sure that the infinity sum really equals to π.
That is why we really need your knowledge of Maths to solve this.
Is there a way to prove that when $a$ tends to infinity, $y$ tends to $pi$?
Thanks in advance and Happy Holidays Everyone! :D
summation infinity pi
$endgroup$
|
show 1 more comment
$begingroup$
For
$$y=sum_{n=0}^a2cdot2^ncdottanleft(frac{45}{2^n}right)cdotsinleft(frac{90}{2^n}right)^2$$
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
4-sided regular→ 8-sided regular→ 16-sided regular→ 32-sided regular→... →n-sided regular
(When n tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
We have already tried Geometric Sequence Infinite Sum, but there does not seem to have a common ratio.
Moreover, we have used our calculator to input the numbers up to 128[(sin ...] and we get the value of 3.140... which is very close to π But we can't be completely sure that the infinity sum really equals to π.
That is why we really need your knowledge of Maths to solve this.
Is there a way to prove that when $a$ tends to infinity, $y$ tends to $pi$?
Thanks in advance and Happy Holidays Everyone! :D
summation infinity pi
$endgroup$
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 13:24
2
$begingroup$
Thanks for the tip :D I will edit this right away!
$endgroup$
– user629248
Dec 25 '18 at 13:25
1
$begingroup$
Jolly Christmas and Happy Holidays to you, My Good Fellow! :)
$endgroup$
– user629248
Dec 25 '18 at 13:26
$begingroup$
Thank you, and Happy Holidays to you too! Let me know once you've edited the question, then I'll probably upvote :)
$endgroup$
– Shaun
Dec 25 '18 at 13:32
1
$begingroup$
Hey by the way, Shaun, can you tell me how can I type Equations Symbols when I submit my Question?
$endgroup$
– user629248
Dec 25 '18 at 13:43
|
show 1 more comment
$begingroup$
For
$$y=sum_{n=0}^a2cdot2^ncdottanleft(frac{45}{2^n}right)cdotsinleft(frac{90}{2^n}right)^2$$
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
4-sided regular→ 8-sided regular→ 16-sided regular→ 32-sided regular→... →n-sided regular
(When n tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
We have already tried Geometric Sequence Infinite Sum, but there does not seem to have a common ratio.
Moreover, we have used our calculator to input the numbers up to 128[(sin ...] and we get the value of 3.140... which is very close to π But we can't be completely sure that the infinity sum really equals to π.
That is why we really need your knowledge of Maths to solve this.
Is there a way to prove that when $a$ tends to infinity, $y$ tends to $pi$?
Thanks in advance and Happy Holidays Everyone! :D
summation infinity pi
$endgroup$
For
$$y=sum_{n=0}^a2cdot2^ncdottanleft(frac{45}{2^n}right)cdotsinleft(frac{90}{2^n}right)^2$$
I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
4-sided regular→ 8-sided regular→ 16-sided regular→ 32-sided regular→... →n-sided regular
(When n tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
We have already tried Geometric Sequence Infinite Sum, but there does not seem to have a common ratio.
Moreover, we have used our calculator to input the numbers up to 128[(sin ...] and we get the value of 3.140... which is very close to π But we can't be completely sure that the infinity sum really equals to π.
That is why we really need your knowledge of Maths to solve this.
Is there a way to prove that when $a$ tends to infinity, $y$ tends to $pi$?
Thanks in advance and Happy Holidays Everyone! :D
summation infinity pi
summation infinity pi
edited Dec 25 '18 at 13:45
asked Dec 25 '18 at 13:17
user629248
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 13:24
2
$begingroup$
Thanks for the tip :D I will edit this right away!
$endgroup$
– user629248
Dec 25 '18 at 13:25
1
$begingroup$
Jolly Christmas and Happy Holidays to you, My Good Fellow! :)
$endgroup$
– user629248
Dec 25 '18 at 13:26
$begingroup$
Thank you, and Happy Holidays to you too! Let me know once you've edited the question, then I'll probably upvote :)
$endgroup$
– Shaun
Dec 25 '18 at 13:32
1
$begingroup$
Hey by the way, Shaun, can you tell me how can I type Equations Symbols when I submit my Question?
$endgroup$
– user629248
Dec 25 '18 at 13:43
|
show 1 more comment
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 13:24
2
$begingroup$
Thanks for the tip :D I will edit this right away!
$endgroup$
– user629248
Dec 25 '18 at 13:25
1
$begingroup$
Jolly Christmas and Happy Holidays to you, My Good Fellow! :)
$endgroup$
– user629248
Dec 25 '18 at 13:26
$begingroup$
Thank you, and Happy Holidays to you too! Let me know once you've edited the question, then I'll probably upvote :)
$endgroup$
– Shaun
Dec 25 '18 at 13:32
1
$begingroup$
Hey by the way, Shaun, can you tell me how can I type Equations Symbols when I submit my Question?
$endgroup$
– user629248
Dec 25 '18 at 13:43
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 13:24
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 13:24
2
2
$begingroup$
Thanks for the tip :D I will edit this right away!
$endgroup$
– user629248
Dec 25 '18 at 13:25
$begingroup$
Thanks for the tip :D I will edit this right away!
$endgroup$
– user629248
Dec 25 '18 at 13:25
1
1
$begingroup$
Jolly Christmas and Happy Holidays to you, My Good Fellow! :)
$endgroup$
– user629248
Dec 25 '18 at 13:26
$begingroup$
Jolly Christmas and Happy Holidays to you, My Good Fellow! :)
$endgroup$
– user629248
Dec 25 '18 at 13:26
$begingroup$
Thank you, and Happy Holidays to you too! Let me know once you've edited the question, then I'll probably upvote :)
$endgroup$
– Shaun
Dec 25 '18 at 13:32
$begingroup$
Thank you, and Happy Holidays to you too! Let me know once you've edited the question, then I'll probably upvote :)
$endgroup$
– Shaun
Dec 25 '18 at 13:32
1
1
$begingroup$
Hey by the way, Shaun, can you tell me how can I type Equations Symbols when I submit my Question?
$endgroup$
– user629248
Dec 25 '18 at 13:43
$begingroup$
Hey by the way, Shaun, can you tell me how can I type Equations Symbols when I submit my Question?
$endgroup$
– user629248
Dec 25 '18 at 13:43
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
$$2tan Acdotsin^22A=2(4sin^3Acos A)=(3sin A-sin3A)2cos A$$
$$=3sin2A-(sin4A+sin2A)=2sin2A-sin4A$$
which clearly shows Telescopic form
answered Dec 25 '18 at 13:26
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
When I come across things like these: (if you do not need to prove it manually)
Wolfram Alpha computation
answered Dec 25 '18 at 13:57
yigoliyigoli
177
$endgroup$
$begingroup$
Thank you! I have heard of Wolfram before, but is there a manually proof for this question?
$endgroup$
– user629248
Dec 25 '18 at 14:02
$begingroup$
Easy to fool yourself here. If the sum happened to be $pi + 10^{-16}$ then WA would still say True to your query. Just try this (it's $pi$ up to 15 digits). In this case it seems it evaluates both sides numerically (to floating point accuracy) and checks if they agree. But it doesn't tell you that this is what it does.
$endgroup$
– Winther
Dec 25 '18 at 14:43
$begingroup$
To check this try to ask WA to just evaluate the sum, it doesn't evaluate it to $pi$ symbolically.
$endgroup$
– Winther
Dec 25 '18 at 14:44
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%2f3052096%2fsum-to-infinity-of-trigonometry-to-pi%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
$begingroup$
Hint:
$$2tan Acdotsin^22A=2(4sin^3Acos A)=(3sin A-sin3A)2cos A$$
$$=3sin2A-(sin4A+sin2A)=2sin2A-sin4A$$
which clearly shows Telescopic form
answered Dec 25 '18 at 13:26
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
$$2tan Acdotsin^22A=2(4sin^3Acos A)=(3sin A-sin3A)2cos A$$
$$=3sin2A-(sin4A+sin2A)=2sin2A-sin4A$$
which clearly shows Telescopic form
answered Dec 25 '18 at 13:26
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
$$2tan Acdotsin^22A=2(4sin^3Acos A)=(3sin A-sin3A)2cos A$$
$$=3sin2A-(sin4A+sin2A)=2sin2A-sin4A$$
which clearly shows Telescopic form
answered Dec 25 '18 at 13:26
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
Hint:
$$2tan Acdotsin^22A=2(4sin^3Acos A)=(3sin A-sin3A)2cos A$$
$$=3sin2A-(sin4A+sin2A)=2sin2A-sin4A$$
which clearly shows Telescopic form
answered Dec 25 '18 at 13:26
lab bhattacharjeelab bhattacharjee
226k15158275
answered Dec 25 '18 at 13:26
lab bhattacharjeelab bhattacharjee
226k15158275
answered Dec 25 '18 at 13:26
lab bhattacharjeelab bhattacharjee
226k15158275
answered Dec 25 '18 at 13:26
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
add a comment |
add a comment |
$begingroup$
When I come across things like these: (if you do not need to prove it manually)
Wolfram Alpha computation
answered Dec 25 '18 at 13:57
yigoliyigoli
177
$endgroup$
$begingroup$
Thank you! I have heard of Wolfram before, but is there a manually proof for this question?
$endgroup$
– user629248
Dec 25 '18 at 14:02
$begingroup$
Easy to fool yourself here. If the sum happened to be $pi + 10^{-16}$ then WA would still say True to your query. Just try this (it's $pi$ up to 15 digits). In this case it seems it evaluates both sides numerically (to floating point accuracy) and checks if they agree. But it doesn't tell you that this is what it does.
$endgroup$
– Winther
Dec 25 '18 at 14:43
$begingroup$
To check this try to ask WA to just evaluate the sum, it doesn't evaluate it to $pi$ symbolically.
$endgroup$
– Winther
Dec 25 '18 at 14:44
add a comment |
$begingroup$
When I come across things like these: (if you do not need to prove it manually)
Wolfram Alpha computation
answered Dec 25 '18 at 13:57
yigoliyigoli
177
$endgroup$
$begingroup$
Thank you! I have heard of Wolfram before, but is there a manually proof for this question?
$endgroup$
– user629248
Dec 25 '18 at 14:02
$begingroup$
Easy to fool yourself here. If the sum happened to be $pi + 10^{-16}$ then WA would still say True to your query. Just try this (it's $pi$ up to 15 digits). In this case it seems it evaluates both sides numerically (to floating point accuracy) and checks if they agree. But it doesn't tell you that this is what it does.
$endgroup$
– Winther
Dec 25 '18 at 14:43
$begingroup$
To check this try to ask WA to just evaluate the sum, it doesn't evaluate it to $pi$ symbolically.
$endgroup$
– Winther
Dec 25 '18 at 14:44
add a comment |
$begingroup$
When I come across things like these: (if you do not need to prove it manually)
Wolfram Alpha computation
answered Dec 25 '18 at 13:57
yigoliyigoli
177
$endgroup$
When I come across things like these: (if you do not need to prove it manually)
Wolfram Alpha computation
answered Dec 25 '18 at 13:57
yigoliyigoli
177
answered Dec 25 '18 at 13:57
yigoliyigoli
177
answered Dec 25 '18 at 13:57
yigoliyigoli
177
answered Dec 25 '18 at 13:57
yigoliyigoli
177
177
$begingroup$
Thank you! I have heard of Wolfram before, but is there a manually proof for this question?
$endgroup$
– user629248
Dec 25 '18 at 14:02
$begingroup$
Easy to fool yourself here. If the sum happened to be $pi + 10^{-16}$ then WA would still say True to your query. Just try this (it's $pi$ up to 15 digits). In this case it seems it evaluates both sides numerically (to floating point accuracy) and checks if they agree. But it doesn't tell you that this is what it does.
$endgroup$
– Winther
Dec 25 '18 at 14:43
$begingroup$
To check this try to ask WA to just evaluate the sum, it doesn't evaluate it to $pi$ symbolically.
$endgroup$
– Winther
Dec 25 '18 at 14:44
add a comment |
$begingroup$
Thank you! I have heard of Wolfram before, but is there a manually proof for this question?
$endgroup$
– user629248
Dec 25 '18 at 14:02
$begingroup$
Easy to fool yourself here. If the sum happened to be $pi + 10^{-16}$ then WA would still say True to your query. Just try this (it's $pi$ up to 15 digits). In this case it seems it evaluates both sides numerically (to floating point accuracy) and checks if they agree. But it doesn't tell you that this is what it does.
$endgroup$
– Winther
Dec 25 '18 at 14:43
$begingroup$
To check this try to ask WA to just evaluate the sum, it doesn't evaluate it to $pi$ symbolically.
$endgroup$
– Winther
Dec 25 '18 at 14:44
$begingroup$
Thank you! I have heard of Wolfram before, but is there a manually proof for this question?
$endgroup$
– user629248
Dec 25 '18 at 14:02
$begingroup$
Thank you! I have heard of Wolfram before, but is there a manually proof for this question?
$endgroup$
– user629248
Dec 25 '18 at 14:02
$begingroup$
Easy to fool yourself here. If the sum happened to be $pi + 10^{-16}$ then WA would still say True to your query. Just try this (it's $pi$ up to 15 digits). In this case it seems it evaluates both sides numerically (to floating point accuracy) and checks if they agree. But it doesn't tell you that this is what it does.
$endgroup$
– Winther
Dec 25 '18 at 14:43
$begingroup$
Easy to fool yourself here. If the sum happened to be $pi + 10^{-16}$ then WA would still say True to your query. Just try this (it's $pi$ up to 15 digits). In this case it seems it evaluates both sides numerically (to floating point accuracy) and checks if they agree. But it doesn't tell you that this is what it does.
$endgroup$
– Winther
Dec 25 '18 at 14:43
$begingroup$
To check this try to ask WA to just evaluate the sum, it doesn't evaluate it to $pi$ symbolically.
$endgroup$
– Winther
Dec 25 '18 at 14:44
$begingroup$
To check this try to ask WA to just evaluate the sum, it doesn't evaluate it to $pi$ symbolically.
$endgroup$
– Winther
Dec 25 '18 at 14:44
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.
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%2f3052096%2fsum-to-infinity-of-trigonometry-to-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
$begingroup$
You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– Shaun
Dec 25 '18 at 13:24
2
$begingroup$
Thanks for the tip :D I will edit this right away!
$endgroup$
– user629248
Dec 25 '18 at 13:25
1
$begingroup$
Jolly Christmas and Happy Holidays to you, My Good Fellow! :)
$endgroup$
– user629248
Dec 25 '18 at 13:26
$begingroup$
Thank you, and Happy Holidays to you too! Let me know once you've edited the question, then I'll probably upvote :)
$endgroup$
– Shaun
Dec 25 '18 at 13:32
1
$begingroup$
Hey by the way, Shaun, can you tell me how can I type Equations Symbols when I submit my Question?
$endgroup$
– user629248
Dec 25 '18 at 13:43