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This is a "coffee-time-style" problem ( to have a taste of this style, you may like to browse the book https://www.amazon.com/Art-Mathematics-Coffee-Time-Memphis/dp/0521693950) interpreted from an anonymous problem once on the interactive whiteboard at my department, namely how to prove $e<\pi$ without much numerical computation like Taylor expansion or so. I once tried to use some "intrinsic connection" between $e$ and $\pi$ like $\sqrt{\pi}=\int_{-\infty}^{+\infty}e^{-x^2}\mathrm{d}x$ ( you can even find it in this movie http://www.imdb.com/title/tt4481414/ for testing children) and one possible way of reducing the problem is in the next paragraph. However it seems to be not that easy, any suggestion or new ideas?

A stronger version of this question is : can we construct an explicit function $f(x)$ on $\mathbb{R}$ so that $f(x)\leq e^{-x^2}$ for all $x\in\mathbb{R}$ with $f(x)< e^{-x^2}$ on an open interval, and that $\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\sqrt{e}$ ? We know from standard measure theory that there are $\beth_2$ such kind of Lebesgue-integrable functions, but this is the thing: how simple and explicit can what we're looking for be? Examples of very simple and explicit functions include but are not limited to piecewise elementary functions (https://en.wikipedia.org/wiki/Elementary_function). Unfortunately a function $f(x)$ defined piecewisely by $$f(x)|_{(-1,1)}=e^{-|x|^r}\ \text{where}\ r\in\mathbb{Q}\cap(-\infty,2)\ \text{or}\ \mathbb{Q}\cap (-\infty,2]\ \text{respectively}$$ and $$f(x)|_{(-\infty,-1]\cup[1,\infty)}=e^{-|x|^s}\ \text{where}\ s\in\mathbb{Q}\cap [2,\infty)\ \text{or}\ \mathbb{Q}\cap(2,\infty)$$ would NOT satisfy $\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\sqrt{e}$, if the values of the Gamma function at rational points are linearly (or even algebraically) independent with $\sqrt{e}$ (https://en.wikipedia.org/wiki/Particular_values_of_the_gamma_function). The question is then how to move on from this first failure to search other explicit functions.

I am aware that it is probably hard to ask such a question as solid as "can we prove that CH is independent from ZFC"; after all, one can argue that any numerical inequality essentially also comes from some intrinsic inequality and hence not numerical at all. However one might try to ask in a relatively sloppy way: is there something that is at least seemingly simpler or less numerical, if not completely non-numerical ?

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    $\begingroup$ It is not too difficult to show that $e<3$ and $\pi>3$. $\endgroup$ – CY Aries Apr 2 '18 at 9:47
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    $\begingroup$ Thank you, still a bit "numerical" though. $\endgroup$ – Philimathmuse Apr 2 '18 at 9:47
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    $\begingroup$ What do you mean by numerical precisely? $\endgroup$ – ChemiCalChems Apr 2 '18 at 11:35
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    $\begingroup$ Can you provide definitions of $e$ and $\pi$ that don't themselves fall prey to your objection of 'still a bit "numerical" though'? $\endgroup$ – Rahul Apr 2 '18 at 11:39
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    $\begingroup$ @ChemiCalChems: I cannot define it precisely, but at least it includes Taylor expansions- which means if a proof calculates a few terms then it is already "numerical". $\endgroup$ – Philimathmuse Apr 2 '18 at 11:40
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Just a suggestion.

$$\int_{-\infty}^\infty\frac{\cos x}{x^2+1}\operatorname d\!x=\frac\pi e$$

If you can prove that the above integral is $>1$ you know $\pi > e$.

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  • $\begingroup$ Thank you, it is still a bit numerical in order to show the integral is larger than one. $\endgroup$ – Philimathmuse Apr 2 '18 at 10:50
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If we define $e$ as $\lim_{n\to\infty}\left(1+\frac1n\right)^n$, then $e<3$ because, for each $n\in\mathbb{N}$,\begin{align}\left(1+\frac1n\right)^n&\leqslant1+1+\frac1{2!}+\cdots+\frac1{n!}\\&<1+1+\frac12+\frac1{2^2}+\cdots+\frac1{2^{n-1}}\\&<3.\end{align}On the other hand, $2\pi$ is greater than the perimeter of a regular hexagon inscribed in a circle with radius $1$, which is $6$. Therefore, $\pi>3>e$.

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    $\begingroup$ Thank you, although I like it very much it is still a bit "numerical" comparing with the Gaussian integral for example. $\endgroup$ – Philimathmuse Apr 2 '18 at 9:58
  • $\begingroup$ Is not this problematic since strict inequality is not preserved after taking limits? Now you have $a_n<b_n<3$ but what guarantees $\lim a_n\neq 3$ in this argumentation? $\endgroup$ – Shashi Apr 2 '18 at 10:59
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    $\begingroup$ @Shashi You are right. What I did only proves that $e\leqslant3$. But, of course, this is enough to prove that $e<\pi$. But it is easy to prove that $e<3$ indeed. It follows from the fact that\begin{align}e-\pi&=\left(1+1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots\right)-\left(1+1+1\frac{2!}+\frac1{3!}+\cdots\right)\\&=\left(\frac1{2^2}-\frac1{3!}\right)+\left(\frac1{2^3}-\frac1{4!}\right)+\cdots\\&>0.\end{align} $\endgroup$ – José Carlos Santos Apr 2 '18 at 11:07
  • $\begingroup$ Yes, indeed, so actually you could easily make your last sentence $\pi>3\geq e$ without any trouble. $\endgroup$ – Shashi Apr 2 '18 at 11:11
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Incoming overkilll! The error function has both a simple series representation and a simple continued fraction representation, allowing to produce nice algebraic approximations for the Mills ratio.
This answer of mine on MO proves the inequality $$ \sqrt{\frac{\pi}{2}}\,e^{k^2/2}\, \text{Erfc}\left(\tfrac{k}{\sqrt{2}}\right) \geq \frac{2}{k+\sqrt{k^2+4}} \tag{1}$$ by only using Fubini's theorem and the elementary (convexity) inequality $\frac{2}{\pi}x<\sin(x)<x$ for $x\in\left(0,\frac{\pi}{2}\right)$. By considering $(1)$ at $k=1$ we have $$ \sqrt{e}\left(\sqrt{\frac{\pi}{2}}-\sum_{n\geq 0}\frac{(-1)^n}{2^n(2n+1)n!}\right)\geq \frac{1}{\varphi} \tag{2}$$ relating $e,\pi$ and the golden ratio $\varphi$ (and allowing to prove $\pi>e$, of course). This is pretty much in the spirit of the "natural" relation between $\pi$ and $e$ given by $\pi=\Gamma\left(\frac{1}{2}\right)^2$ and $\Gamma(s)=\int_{0}^{+\infty}x^{s-1} e^{-x}\,dx.$

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  • $\begingroup$ This is awesome, to make things even better is it possible to directly prove $e<\pi$ only "conceptually" through the gamma function or with some extra property of it (like log-convex property, but probably overkill)? To me $\pi=\Gamma(1/2)^2$ is conceptual enough but how to establish $e$ conceptually also through $Gamma$ seems to be more difficult. $\endgroup$ – Philimathmuse Apr 3 '18 at 2:57
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There's not a lot of computation here, just some simple arithmetic and easy upper bounds (plus one important identity):

$$\begin{align} e^2&=1+2+{2^2\over2}+{2^3\over6}+{2^4\over24}+{2^5\over120}+\cdots\\ &=1+2+2+{4\over3}+{2\over3}+{32\over120}\left(1+{2\over6}+{4\over42}+\cdots \right)\\ &\lt7+{1\over2}\left(1+{1\over2}+{1\over4}+\cdots\right)\\ &=8\\ &\lt6+{3\over2}+{2\over3}\\ &\lt6+{6\over4}+{6\over9}+{6\over16}+{6\over25}+\cdots\\ &=6\sum_{n=1}^\infty{1\over n^2}\\ &=\pi^2 \end{align}$$

Added later: Here's an alternative, which uses the "easy" geometric inequality $\pi\gt3$ (comparing the circumference of a circle to the perimeter of an inscribed hexagon, as in José Carlos Santos's answer) and a small amount of computation:

$$\ln\pi\gt\ln3=-\ln(1/3)=-\int_1^{1/3}{dx\over x}=\int_0^{2/3}{du\over1-u}=\int_0^{2/3}\left(1+u+u^2+\cdots\right)du\\ =\left(2\over3\right)+{1\over2}\left(2\over3\right)^2+{1\over3}\left(2\over3\right)^3+{1\over4}\left(2\over3\right)^4+\cdots\\ \gt{2\over3}+{2\over9}+{8\over81}+{4\over81}={54+18+12\over81}={84\over81}\gt1=\ln e$$

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  • $\begingroup$ +1. It's pretty nice. $\endgroup$ – Felix Marin Apr 7 '18 at 17:08
  • $\begingroup$ Although this does not cater for my purpose I would say that it is a really beautiful proof, thank you for sharing this to everybody! $\endgroup$ – Philimathmuse Apr 7 '18 at 17:28
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    $\begingroup$ @Philimathmuse, thank you, and you are welcome. I was composing an addition when you posted your comment. It doesn't cater to your purpose either, but I hope you like it as well. $\endgroup$ – Barry Cipra Apr 7 '18 at 17:36
  • $\begingroup$ @Barry: I do enjoy it as well, thank you! $\endgroup$ – Philimathmuse Apr 7 '18 at 17:56
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By the Cauchy-Schwarz inequality together with the equality condition, we have

$$ 1 = \left( \int_{0}^{1} dx \right)^2 < \left( \int_{0}^{1} \frac{dx}{1+x^2} \right)\left( \int_{0}^{1} (1+x^2) \, dx \right) = \frac{\pi}{3}. $$

Now utilizing the inequality $e^{-x} \geq 1 - x$ which is true for all $x \in \mathbb{R}$,

$$ \frac{1}{e} = \int_{0}^{1} (1-x)e^{-x} \, dx \geq \int_{0}^{1} (1-x)^2 \, dx = \frac{1}{3}. $$

Combining two inequality yields $\pi > e$.

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  • $\begingroup$ Great job on the definite integrals Sangchul! Really nice to combine them all together. From your answer I think I can work out a proof without touching $3$ explicitly. $\endgroup$ – Philimathmuse Apr 7 '18 at 21:19
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Here's my contribution: We need to show that $\ln \pi >1$. So we need to show that $\int_1^\pi \frac{1}{x} \; dx >1.$ The graph of $y=\frac{1}{x}$ is concave up, so any tangent line lies below the curve. Find the tangent line half way through the interval:

$$y = \frac{4}{(\pi+1)^2}(\pi +1 -x).$$

The area under that line and between $x=1$ and $x=\pi$ is less than the area represented by the integral. So we need to show the area of this trapezoid is greater than $1$. After some algebra that area turns out to be

$$2\left(\frac{\pi-1}{\pi+1}\right).$$

So we need to show this expression is greater than $1$. I finally have to stoop to "numerics" and use the fact that $\pi$ is greater than $3$. Either note that $(x-1)/(x+1)$ is an increasing function and hence the last expression is greater than $2(3-1)(3+1) = 1.$ Or calculate as follows:

$$2\left(\frac{\pi-1}{\pi+1}\right) = 2 - \frac{4}{\pi+1} > 2-\frac{4}{4} = 1.$$

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  • $\begingroup$ Thank you for sharing this wonderful proof and so far it is the closet one for my purpose: up to the final part everything is perfect (actually you can put that as the first part since you have implicitly used the fact that $\pi>1$ for your wonderful contribution to treat $e$ anyway, and @José gave a really nice proof for $\pi>3$. ) I upvoted it but did not accepted it as the "final answer" since in some sense it did not provide a direct answer to my "stronger version" of the question in the second paragraph of the question. $\endgroup$ – Philimathmuse Apr 7 '18 at 17:36
  • $\begingroup$ Well, the shortest distance between antipodal points on a circle is the diameter. If the diameter is 1, then 1 is less than the distance around the semi-circle, which is $\pi/2.$ So we get $\pi > 2$ with no numerics, just pure geometry. $\endgroup$ – B. Goddard Apr 7 '18 at 17:45
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    $\begingroup$ There something else interesting here. The tangent line that gives the maximum area is the one at the point where $x = (1+\pi)/2.$ Exactly half way. I didn't check, but this is probably a property of the curve $1/x$. $\endgroup$ – B. Goddard Apr 7 '18 at 17:47
  • $\begingroup$ Yes exactly and that's the reason that I personally like your answer the best. But right now I have no clue about my stronger version question, do you have any idea about it or would like to try? $\endgroup$ – Philimathmuse Apr 7 '18 at 17:55
  • $\begingroup$ I have checked it is correct that for any interval [a,b] in the positive real line, the max area under the tangent line of the curve is taken exactly at the mid point of the interval. $\endgroup$ – Philimathmuse Apr 17 '18 at 21:51

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