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Let's not invoke the polynomial expansion of $\arctan$ function.

I remember I saw somewhere here a very simple proof showing that $3<\pi<4$ but I don't remember where I saw it.. (I remember that this proof is also in Wikipedia)

How do I prove this inequality?

My definition for $\pi$ is the twice the first positive real number such that $\cos x= 0$ where $\cos x = \frac{e^{ix} + e^{-ix}}{2}$

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    $\begingroup$ How are you defining $e^{ix}$? $\endgroup$
    – Git Gud
    Commented Apr 5, 2014 at 8:57
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    $\begingroup$ Why not just use $\pi=\sum\limits_{i=0}^{\infty}\frac{4}{2i+1}(-1)^i$? $\endgroup$ Commented Apr 5, 2014 at 9:02
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    $\begingroup$ @GitGud I assume, $$e^x = \lim_{n\to\infty} \left( 1+\frac{x}{n}\right)^n$$ $\endgroup$
    – Guy
    Commented Apr 5, 2014 at 9:02
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    $\begingroup$ That's a terrifying definition. Define $\pi$ to be the arclength of the circle with unit diameter. For the upper bound, circumscribe a square; it has circumference 4. The lower bound will come from some other nice shape (an inscribed square is too small; that only bounds below by $2\sqrt{2}$; a hexagon or octagon would do just fine.) $\endgroup$
    – user98602
    Commented Apr 5, 2014 at 9:06
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    $\begingroup$ @Sabyasachi i don't even know if I can prove the relevant parts. I'll wait for someone else ^_^ $\endgroup$
    – Git Gud
    Commented Apr 5, 2014 at 9:53

6 Answers 6

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After all the comments it's unclear what is allowed and what is forbidden. I'd suggest the following: $${1\over 2}=\sin{\pi\over 6}<{\pi\over 6}<\tan{\pi\over 6}={1\over\sqrt{3}}\ ,$$ which leads to $3<\pi<3.5$.

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    $\begingroup$ +1 I was thinking the same thing, but using $\sin(\pi/6)=\frac12$ and $\tan(\pi/4)=1$ then using this answer to show that $\sin(x)\le x\le\tan(x)$. $\endgroup$
    – robjohn
    Commented Apr 5, 2014 at 10:18
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Starting from the antique definition of $\pi$ : The length of cirle of radius $1$ is $=2\pi$.

On the figure below, obviously the length of the circle is between $6$ and $8$ (lengths of the hexagon and square, respectively)

So, $6 < 2\pi < 8$ hence $3 < \pi < 4$

enter image description here

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    $\begingroup$ Downvoted because this doesn't answer the question. $\endgroup$
    – Guy
    Commented Apr 5, 2014 at 9:45
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    $\begingroup$ This perfectly adresses the problem. $\endgroup$
    – mookid
    Commented Apr 5, 2014 at 9:53
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    $\begingroup$ @mookid Read the definition of $\pi$ in question. $\endgroup$
    – Guy
    Commented Apr 5, 2014 at 9:53
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    $\begingroup$ @Sabyasachi Still, I think it is an useful answer. $\endgroup$
    – Pedro
    Commented Apr 5, 2014 at 11:09
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    $\begingroup$ In fact, my answer above is of no merit because it has been known for a long time. I posted this banality just to make obvious the ambiguity of the wording of the question. If we accept a simple definition of $\pi$, my answer shows how trivial can be the solution. If the definion of $\pi$ is based on imaginary power of $e$, what is the definition of $e$ that you accept ? Is it with infinite series ? Is there not a contradiction with the exclusion of polynomial expansion as mentioned in the wording ? It was already pointed out that it is not clearly stated what is allowed and what is forbidden. $\endgroup$
    – JJacquelin
    Commented Apr 5, 2014 at 12:15
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I assume that $x\to e^x$ is the unique differentiable function that is its own derivative and maps $0\mapsto 1$. From this (especially using uniqueness) one quickly establishes $e^{x+y}=e^xe^y$, $e^{\bar z}=\overline{e^z}$ etc.

With the definitions $\cos x=\frac{e^{ix}+e^{-ix}}{2}$, $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ we then see that $\cos $ and $\sin$ map reals to reals and that $\cos^2x+\sin^2x=1$, $\sin'=\cos$, $\cos'=-\sin$ etc.

Claim 1: There is no $a$ such that $\sin x>\frac12$ for all $x\in [a,a+4]$.

Proof: Otherwise, $2\ge\cos (a+2)-\cos a=\int_a^{a+4}\sin x\,\mathrm dx>2$. $_\square$

Claim 2: There exists $a>0$ such that $\cos a=0$.

Proof: Otherwise we would have $\cos x>0$ for all $x>0$, hence $\sin x$ strictly increasing towards some limit $s:=\lim_{x\to+\infty}\sin x\in(0,1]$. Then $\lim_{x\to\infty}\cos x=\sqrt{1-s^2}=0$ and hence $s=1$. On the other hand, $s\le \frac12$ by claim 1. $_\square$.

By continuity, there exists a minimal positive real $p$ such that $\cos p=0$.

By the IVT, $-\sin \xi =\frac{0-1}{p}$, i.e. $\sin\xi=\frac1p$ for some $\xi\in(0,p)$. This already shows $$p\ge1.$$ Consider the function $f(x)=\cos x-1+\frac12x^2$. We have $f'(x)=-\sin x+x$, $f''(x)=1-\cos x\ge 0$ for all $x$; hence $f'(x)\ge f'(0)$ for all $x\ge 0$, i.e. $\sin x\le x$ for all $x\ge 0$; hence $f(x)\ge f(0)=0$ for all $x\ge0$. We conclude $\cos 1\ge \frac12$. In fact the iniequality is strict, i.e. $\cos1>\frac12$ as otherwise we'd have $f''(x)=0$ for all $x\in[0,1]$. From $\cos'=-\sin\ge -1$ we then see that $\cos (1+x)>\frac12-x$ for $x> 0$. Thus $$p> \frac32.$$

Consider $g(x)=\cos x-1+\frac12x^2-\frac1{24}x^4$. Then $g'(x)=-\sin x+x-\frac16x^3$, $g''(x)=-\cos x-1-\frac12x^2=-f(x)\le 0$ for $x\ge 0$. Hence $g'(x)\le g'(0)=0$ for all $x\ge 0$ and finally $g(x)\le g(0)=0$ for all $x\ge 0$. We conclude $0\ge g(2)=\cos 2-1+\frac42-\frac{16}{24}$, i.e. $\cos 2\le-\frac13$ and hence $$ p<2.$$

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We need to show that $3/2<\pi/2<2$, so (using your definition of $\pi$) that $\cos 2<0$ and $\cos t>0$ for all $0\leq t\leq 3/2$.

The Taylor series for $\cos t$ follows immediately from the series for $e^t$ and your definition of $\cos t$. The signs alternate. The terms decrease in absolute value for $t=2$ (except for the second term), so taking the first three terms give the upper bound $\cos 2 < -\tfrac{1}{3}$.

On the other hand for $t=\tfrac{3}{2}$, the first four terms give the lower bound $\cos \tfrac{3}{2}>0.0701171875$.

Finally, $\cos' = -\sin$ (immediate from your definitions), so if $\sin t\geq 0$ for $0\leq t\leq 3/2$ we will know that $\cos$ is decreasing in that interval and so cannot be negative. For this interval, $\sin t > t-t^3/6$ (same argument about alternating terms with decreasing absolute value); $t-t^3/6$ is positive between its zeros at $t=0$ and $t=\sqrt{6}>3/2$.

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Well you can take that $$\cos 6>0\\\cos 8<0$$ Since $\cos$ is a continuos function,a zero($\pi$) has to be between $6$ and $8$.If you have to prove that $\cos 6> 0 \land \cos 8<0$,I guess you could do so with Taylor series of $e^{ix}+e^{-ix}$

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  • $\begingroup$ The OP explicitly asks for no series expansion. $\endgroup$
    – Guy
    Commented Apr 5, 2014 at 9:39
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    $\begingroup$ @Sabyasachi He asked for no arctan series expansion,anyway if he defined the $\cos x$ as a sum of $\frac{e^{ix}+e^{-ix}}{2}$ the proof probably has to include it,and the definition of $e^{ix}$ pretty much has to include Taylor series. $\endgroup$
    – kingW3
    Commented Apr 5, 2014 at 9:50
  • $\begingroup$ Or the limit definition of $e^x$ $\endgroup$
    – Guy
    Commented Apr 5, 2014 at 9:51
  • $\begingroup$ Could be but I don't know how that could be used to prove it in this case. $\endgroup$
    – kingW3
    Commented Apr 5, 2014 at 9:52
  • $\begingroup$ Agreed.${}{}{}$ $\endgroup$
    – Guy
    Commented Apr 5, 2014 at 9:53
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Similarly to

$$ \frac{22}{7}-\pi=\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx $$ Why do we need an integral to prove that $\frac{22}{7} > \pi$?

we have $$ 4-\pi=4\int_0^1\frac{x^2}{1+x^2}dx $$

and $$ \pi-3=2\int_0^1\frac{x(1-x)^2}{1+x^2}dx $$

https://math.stackexchange.com/a/1593090/134791

Combining them $$3+2\int_0^1\frac{x(1-x)^2}{1+x^2}dx=\pi=4-4\int_0^1\frac{x^2}{1+x^2}dx$$ and noting that the integrands are positive for $0<x<1$ $$3<\pi<4$$

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