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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 ?

José Carlos Santos
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Philimathmuse
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    It is not too difficult to show that $e<3$ and $\pi>3$. – CY Aries Apr 02 '18 at 09:47
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    Thank you, still a bit "numerical" though. – Philimathmuse Apr 02 '18 at 09:47
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    What do you mean by numerical precisely? – ChemiCalChems Apr 02 '18 at 11:35
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    Can you provide definitions of $e$ and $\pi$ that don't themselves fall prey to your objection of 'still a bit "numerical" though'? –  Apr 02 '18 at 11:39
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    @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". – Philimathmuse Apr 02 '18 at 11:40
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    @Rahul: In some sense we cannot provide "absolute" definitions for each of them but there are some relations between them which are less "numerical", for example the Gaussian integral $\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$ is considered intrinsic enough here since the proof of it essentially only uses the definition of $\pi$ as that of a geometric object, and the definition of $e$ as a number (existence) $a$ so that the derivative of $f(x)=a^x$ is itself. In this relative sense we see that it is not the numerical property of $\pi$ and $e$ that matter in their definitions. – Philimathmuse Apr 02 '18 at 11:44
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    And let me add one thing to explain this seemingly strange objection: we cannot define but do agree implicitly that $\pi$ is more "geometric", $3$ more "arithmetic" and $e$ more "analytic". To compare $\pi$ and $e$ without $numerical$ elements in it is a seemingly naive requirement, but if we read it from the point of view which tries to connect the "geometric" and "analytic" part of the world without touching the "arithmetic" part then it is more interesting. Hence the objection to comparing $\pi$ and $e$ with $3$ is not objection in itself, rather an invitation into the secret of "them". – Philimathmuse Apr 02 '18 at 12:08
  • How about this one ?: $\displaystyle{\pi \over \mathrm{e}} = \int_{-\infty}^{\infty}{\cos\left(x\right) \over x^{2} + 1},\mathrm{d}x$. – Felix Marin Apr 07 '18 at 17:02
  • I think of 3 very geometrically. I start with a segment of length 1. Then I extend it by another segment of the same length. Finally, I extend that by another segment of the same length. The result is a segment of length 3. – Lee Mosher Apr 07 '18 at 17:03
  • @Lee but that is exactly arithmetic: the process that you extend it is essentially induction in Peano Axioms and independent from the geometry of $\mathbb{S}^1$ or topology of $\mathbb{R}$ which guarantees the existence of a real number $a$ so that $a^x$ has derivative $a^x$. Those things won't appear directly from counting by induction. – Philimathmuse Apr 07 '18 at 17:11
  • @Felix Thank you, this formula was exactly the one suggested in the first answer to this question. But the problem is then how to show the integral is larger than one without doing it piecewisely with some numerical estimates. – Philimathmuse Apr 07 '18 at 17:13
  • @Lee What I wrote is wrong strictly speaking, but I guess that you get my point, and thank you for sharing your idea. – Philimathmuse Apr 07 '18 at 17:18
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    Another fun integral is $$\sqrt{\frac{\pi}{e}} = \int_{0}^{\infty} e^{-\frac{1}{4}(x^2 + x^{-2})} , dx$$ although I have no idea if this will ever lead to a proof of $\pi > e$. – Sangchul Lee Apr 07 '18 at 18:04
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    Just to follow up, my comment was written with Euclid's axioms in mind, not with induction in mind. I must say, though, that I am growing suspicious of a kind of unstated assumption in your question that "geometry" and "arithmetic" and "analysis" can be cleanly separated, or separated at all. – Lee Mosher Apr 07 '18 at 18:47
  • @Lee Mosher: Thank you for this important comment. I did not mean to doubt the harmony and unification of these three aspects of mathematics, so here is the story: once I heard from a public forum that a famous working number theorist commented about his area like this: even for aliens whose "space form" is essentially different from ours, one must count and the arithmetic part is more or less the same. Now what would be the "worst" situation happening for those aliens? What if they can count but very badly, say their "Peano" only allows them to count accurately until 2 and from 3 on – Philimathmuse Apr 07 '18 at 20:58
  • @Lee Moster: they can only count weakly, and what if they have more or less the same topology as ours for the reals, and the idea of a circle and measure-all this "coincidence" would still lead them to discover $\sqrt{\pi^\ast}=\int_{\mathbb{R}^\ast}{e^\ast}^{-x^2}\mathrm{d^\ast}x$ where $\cdot^\ast$ denotes a corresponding object in their "universe". Then would they ever be able to compare $e^\ast$ and $\pi^\ast$ in the same way as we do, without help from our universe? – Philimathmuse Apr 07 '18 at 21:10
  • How about solution where $e=\sum a_n < \sum b_n = \pi$ where $a_n < b_n$ for all $n$ is shown by induction?. Such solution is not that hard to get, and there is not much numerical parts in it, apart from the intial step in induction (proving for $n=0$). – Sil Apr 07 '18 at 21:45
  • @Sil: You are right but if that does not use Taylor expansion it is then less numerical. Taylor expansion is great as always, it is just not what I am looking for when comparing $e$ and $\pi$. – Philimathmuse Apr 07 '18 at 21:52

6 Answers6

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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 $\pi > e$.

orlp
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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, $\pi>3>e$.

José Carlos Santos
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    Thank you, although I like it very much it is still a bit "numerical" comparing with the Gaussian integral for example. – Philimathmuse Apr 02 '18 at 09:58
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    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? – Shashi Apr 02 '18 at 10:59
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    @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} – José Carlos Santos Apr 02 '18 at 11:07
  • Yes, indeed, so actually you could easily make your last sentence $\pi>3\geq e$ without any trouble. – Shashi Apr 02 '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 $\Gamma(s)=\int_{0}^{+\infty}x^{s-1} e^{-x}\,dx.$

Jack D'Aurizio
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  • 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. – Philimathmuse Apr 03 '18 at 02:57
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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 $\pi > e$.

Sangchul Lee
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  • 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. – Philimathmuse Apr 07 '18 at 21:19
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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:

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

Barry Cipra
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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:

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

B. Goddard
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  • 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. – Philimathmuse Apr 07 '18 at 17:36
  • 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. – B. Goddard Apr 07 '18 at 17:45
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    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 $1/x$. – B. Goddard Apr 07 '18 at 17:47
  • 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? – Philimathmuse Apr 07 '18 at 17:55
  • 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. – Philimathmuse Apr 17 '18 at 21:51