Proof that $ pi > 2 cdot sqrt{2}$ and $ pi > 3 $
$begingroup$
I need to prove that $pi > 3$ and $ pi > 2 cdot sqrt{2}$ only in use of definition of cosine (by series) or $cos(x) = frac{e^{iz}+ e^{-iz}}{2}$and definition of $pi$ as $pi = 2cdot x_0$ where $cos(x_0) = 0$ and $x_0 in (0,2)$
what I did
I thought that I can use $frac{pi^2}{6} = sumfrac{1}{n^2}$:
$$frac{pi^2}{6} = sumfrac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = frac{205}{144}$$
$$ pi^2 > frac{205}{24} > 8 $$
$$ pi > 2 cdot sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $pi > 3$. Moreover I have just understood that I can't use $sumfrac{1}{n^2}$ because "it was not main part of my lecture"
real-analysis summation pi
$endgroup$
|
show 3 more comments
$begingroup$
I need to prove that $pi > 3$ and $ pi > 2 cdot sqrt{2}$ only in use of definition of cosine (by series) or $cos(x) = frac{e^{iz}+ e^{-iz}}{2}$and definition of $pi$ as $pi = 2cdot x_0$ where $cos(x_0) = 0$ and $x_0 in (0,2)$
what I did
I thought that I can use $frac{pi^2}{6} = sumfrac{1}{n^2}$:
$$frac{pi^2}{6} = sumfrac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = frac{205}{144}$$
$$ pi^2 > frac{205}{24} > 8 $$
$$ pi > 2 cdot sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $pi > 3$. Moreover I have just understood that I can't use $sumfrac{1}{n^2}$ because "it was not main part of my lecture"
real-analysis summation pi
$endgroup$
3
$begingroup$
Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
$endgroup$
– Winther
Dec 18 '18 at 22:48
$begingroup$
Ok, but as I have mentioned, I can't use it...
$endgroup$
– VirtualUser
Dec 18 '18 at 22:51
1
$begingroup$
Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
$endgroup$
– Winther
Dec 18 '18 at 22:52
$begingroup$
"only in use of definition of ..."
$endgroup$
– VirtualUser
Dec 18 '18 at 22:52
1
$begingroup$
The second inequality implies the first. Forget the first.
$endgroup$
– Yves Daoust
Dec 18 '18 at 23:01
|
show 3 more comments
$begingroup$
I need to prove that $pi > 3$ and $ pi > 2 cdot sqrt{2}$ only in use of definition of cosine (by series) or $cos(x) = frac{e^{iz}+ e^{-iz}}{2}$and definition of $pi$ as $pi = 2cdot x_0$ where $cos(x_0) = 0$ and $x_0 in (0,2)$
what I did
I thought that I can use $frac{pi^2}{6} = sumfrac{1}{n^2}$:
$$frac{pi^2}{6} = sumfrac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = frac{205}{144}$$
$$ pi^2 > frac{205}{24} > 8 $$
$$ pi > 2 cdot sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $pi > 3$. Moreover I have just understood that I can't use $sumfrac{1}{n^2}$ because "it was not main part of my lecture"
real-analysis summation pi
$endgroup$
I need to prove that $pi > 3$ and $ pi > 2 cdot sqrt{2}$ only in use of definition of cosine (by series) or $cos(x) = frac{e^{iz}+ e^{-iz}}{2}$and definition of $pi$ as $pi = 2cdot x_0$ where $cos(x_0) = 0$ and $x_0 in (0,2)$
what I did
I thought that I can use $frac{pi^2}{6} = sumfrac{1}{n^2}$:
$$frac{pi^2}{6} = sumfrac{1}{n^2} > 1 + 1/4 + 1/9 + 1/16 = frac{205}{144}$$
$$ pi^2 > frac{205}{24} > 8 $$
$$ pi > 2 cdot sqrt{2} $$
Fine... but there are 2 problems: firstly, in this way is hard to proof that $pi > 3$. Moreover I have just understood that I can't use $sumfrac{1}{n^2}$ because "it was not main part of my lecture"
real-analysis summation pi
real-analysis summation pi
edited Dec 18 '18 at 22:50
Bernard
121k740116
121k740116
asked Dec 18 '18 at 22:40
VirtualUserVirtualUser
899114
899114
3
$begingroup$
Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
$endgroup$
– Winther
Dec 18 '18 at 22:48
$begingroup$
Ok, but as I have mentioned, I can't use it...
$endgroup$
– VirtualUser
Dec 18 '18 at 22:51
1
$begingroup$
Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
$endgroup$
– Winther
Dec 18 '18 at 22:52
$begingroup$
"only in use of definition of ..."
$endgroup$
– VirtualUser
Dec 18 '18 at 22:52
1
$begingroup$
The second inequality implies the first. Forget the first.
$endgroup$
– Yves Daoust
Dec 18 '18 at 23:01
|
show 3 more comments
3
$begingroup$
Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
$endgroup$
– Winther
Dec 18 '18 at 22:48
$begingroup$
Ok, but as I have mentioned, I can't use it...
$endgroup$
– VirtualUser
Dec 18 '18 at 22:51
1
$begingroup$
Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
$endgroup$
– Winther
Dec 18 '18 at 22:52
$begingroup$
"only in use of definition of ..."
$endgroup$
– VirtualUser
Dec 18 '18 at 22:52
1
$begingroup$
The second inequality implies the first. Forget the first.
$endgroup$
– Yves Daoust
Dec 18 '18 at 23:01
3
3
$begingroup$
Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
$endgroup$
– Winther
Dec 18 '18 at 22:48
$begingroup$
Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
$endgroup$
– Winther
Dec 18 '18 at 22:48
$begingroup$
Ok, but as I have mentioned, I can't use it...
$endgroup$
– VirtualUser
Dec 18 '18 at 22:51
$begingroup$
Ok, but as I have mentioned, I can't use it...
$endgroup$
– VirtualUser
Dec 18 '18 at 22:51
1
1
$begingroup$
Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
$endgroup$
– Winther
Dec 18 '18 at 22:52
$begingroup$
Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
$endgroup$
– Winther
Dec 18 '18 at 22:52
$begingroup$
"only in use of definition of ..."
$endgroup$
– VirtualUser
Dec 18 '18 at 22:52
$begingroup$
"only in use of definition of ..."
$endgroup$
– VirtualUser
Dec 18 '18 at 22:52
1
1
$begingroup$
The second inequality implies the first. Forget the first.
$endgroup$
– Yves Daoust
Dec 18 '18 at 23:01
$begingroup$
The second inequality implies the first. Forget the first.
$endgroup$
– Yves Daoust
Dec 18 '18 at 23:01
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.
Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.
Approach using the series definition of cosine and pi.
We can use the fact that $cos(x)$ is continuous, and that
$$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
$$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
$$frac{3245071}{45875200}approx 0.0707$$
with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.
$endgroup$
$begingroup$
@VirtualUser I have edited to show you how to do it using the series definition of $pi$.
$endgroup$
– Frpzzd
Dec 18 '18 at 23:04
add a comment |
$begingroup$
You can show that in $[0,2]$n the cosine function is decreasing. Then
$$fracpi2>frac32iff0<cosfrac32.$$
You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).
The first partial sums are
$$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$
Below, the last partial sum above, and the exact value.
$$0.0707369341169leftrightarrow 0.0707372016677$$
$endgroup$
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%2f3045796%2fproof-that-pi-2-cdot-sqrt2-and-pi-3%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
$begingroup$
The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.
Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.
Approach using the series definition of cosine and pi.
We can use the fact that $cos(x)$ is continuous, and that
$$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
$$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
$$frac{3245071}{45875200}approx 0.0707$$
with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.
$endgroup$
$begingroup$
@VirtualUser I have edited to show you how to do it using the series definition of $pi$.
$endgroup$
– Frpzzd
Dec 18 '18 at 23:04
add a comment |
$begingroup$
The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.
Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.
Approach using the series definition of cosine and pi.
We can use the fact that $cos(x)$ is continuous, and that
$$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
$$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
$$frac{3245071}{45875200}approx 0.0707$$
with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.
$endgroup$
$begingroup$
@VirtualUser I have edited to show you how to do it using the series definition of $pi$.
$endgroup$
– Frpzzd
Dec 18 '18 at 23:04
add a comment |
$begingroup$
The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.
Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.
Approach using the series definition of cosine and pi.
We can use the fact that $cos(x)$ is continuous, and that
$$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
$$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
$$frac{3245071}{45875200}approx 0.0707$$
with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.
$endgroup$
The area of a regular octagon is $2sqrt{2}$ times its squared circumradius. Thus, considering a regular octagon inscribed in a circle is enough to prove that $pigt 2sqrt{2}$.
Similarly, the area of a dodecagon is $3$ times its squared circumradius, proving that $pigt 3$.
Approach using the series definition of cosine and pi.
We can use the fact that $cos(x)$ is continuous, and that
$$cos(x)=sum_{n=0}^infty frac{(-1)^n x^{2n}}{(2n)!}$$
We have that $cos(x)gt 0$ for $xlt pi/2$ and $cos(x)lt 0$ for $xgt pi/2$ (assuming that $xin (0,2)$). Consider the sum
$$sum_{n=0}^infty frac{(-1)^n(9/4)^{n}}{(2n)!}$$
By the alternating series test, the error of the approximation obtained by summing the first $N$ terms is less than the $(N+1)$th term. By adding the first $5$ terms, for instance, we get
$$frac{3245071}{45875200}approx 0.0707$$
with an error less than $10^{-6}$. This shows that $cos(3/2)gt 0$ and $pigt 3gt 2sqrt{2}$.
edited Dec 19 '18 at 21:16
answered Dec 18 '18 at 22:52
FrpzzdFrpzzd
23k841109
23k841109
$begingroup$
@VirtualUser I have edited to show you how to do it using the series definition of $pi$.
$endgroup$
– Frpzzd
Dec 18 '18 at 23:04
add a comment |
$begingroup$
@VirtualUser I have edited to show you how to do it using the series definition of $pi$.
$endgroup$
– Frpzzd
Dec 18 '18 at 23:04
$begingroup$
@VirtualUser I have edited to show you how to do it using the series definition of $pi$.
$endgroup$
– Frpzzd
Dec 18 '18 at 23:04
$begingroup$
@VirtualUser I have edited to show you how to do it using the series definition of $pi$.
$endgroup$
– Frpzzd
Dec 18 '18 at 23:04
add a comment |
$begingroup$
You can show that in $[0,2]$n the cosine function is decreasing. Then
$$fracpi2>frac32iff0<cosfrac32.$$
You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).
The first partial sums are
$$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$
Below, the last partial sum above, and the exact value.
$$0.0707369341169leftrightarrow 0.0707372016677$$
$endgroup$
add a comment |
$begingroup$
You can show that in $[0,2]$n the cosine function is decreasing. Then
$$fracpi2>frac32iff0<cosfrac32.$$
You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).
The first partial sums are
$$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$
Below, the last partial sum above, and the exact value.
$$0.0707369341169leftrightarrow 0.0707372016677$$
$endgroup$
add a comment |
$begingroup$
You can show that in $[0,2]$n the cosine function is decreasing. Then
$$fracpi2>frac32iff0<cosfrac32.$$
You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).
The first partial sums are
$$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$
Below, the last partial sum above, and the exact value.
$$0.0707369341169leftrightarrow 0.0707372016677$$
$endgroup$
You can show that in $[0,2]$n the cosine function is decreasing. Then
$$fracpi2>frac32iff0<cosfrac32.$$
You evaluate the cosine by Taylor's formula. As it is alternating, it is easy to know when to stop (the remainder is smaller than the first omitted term).
The first partial sums are
$$1,-frac18,frac{11}{128},frac{359}{5120},frac{16229}{229376},frac{3245071}{45875200},cdots$$
Below, the last partial sum above, and the exact value.
$$0.0707369341169leftrightarrow 0.0707372016677$$
answered Dec 18 '18 at 23:10
Yves DaoustYves Daoust
129k675227
129k675227
add a comment |
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%2f3045796%2fproof-that-pi-2-cdot-sqrt2-and-pi-3%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
3
$begingroup$
Sum up to the temr $1/7^2$ and you'll get $pi > 3$. And if you prove $pi > 3$ then $pi > 2sqrt{2}$ follows since $3 > 2sqrt{2}$.
$endgroup$
– Winther
Dec 18 '18 at 22:48
$begingroup$
Ok, but as I have mentioned, I can't use it...
$endgroup$
– VirtualUser
Dec 18 '18 at 22:51
1
$begingroup$
Prove that $pi>3$ using geometry ; Can someone give an analytical proof that $pi$ is greater than 3?
$endgroup$
– Winther
Dec 18 '18 at 22:52
$begingroup$
"only in use of definition of ..."
$endgroup$
– VirtualUser
Dec 18 '18 at 22:52
1
$begingroup$
The second inequality implies the first. Forget the first.
$endgroup$
– Yves Daoust
Dec 18 '18 at 23:01