13
$\begingroup$

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 ?

$\endgroup$
18
  • 4
    $\begingroup$ It is not too difficult to show that $e<3$ and $\pi>3$. $\endgroup$
    – CY Aries
    Apr 2, 2018 at 9:47
  • 4
    $\begingroup$ What do you mean by numerical precisely? $\endgroup$ Apr 2, 2018 at 11:35
  • 3
    $\begingroup$ Can you provide definitions of $e$ and $\pi$ that don't themselves fall prey to your objection of 'still a bit "numerical" though'? $\endgroup$
    – user856
    Apr 2, 2018 at 11:39
  • 2
    $\begingroup$ @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. $\endgroup$ Apr 2, 2018 at 11:44
  • 2
    $\begingroup$ 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". $\endgroup$ Apr 2, 2018 at 12:08

6 Answers 6

15
$\begingroup$

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 $\pi > e$.

$\endgroup$
1
  • $\begingroup$ Thank you, it is still a bit numerical in order to show the integral is larger than one. $\endgroup$ Apr 2, 2018 at 10:50
15
$\begingroup$

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, $\pi>3>e$.

$\endgroup$
4
  • 1
    $\begingroup$ Thank you, although I like it very much it is still a bit "numerical" comparing with the Gaussian integral for example. $\endgroup$ Apr 2, 2018 at 9:58
  • 1
    $\begingroup$ 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, 2018 at 10:59
  • 1
    $\begingroup$ @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} $\endgroup$ Apr 2, 2018 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, 2018 at 11:11
8
$\begingroup$

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 $\Gamma(s)=\int_{0}^{+\infty}x^{s-1} e^{-x}\,dx.$

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Apr 3, 2018 at 2:57
5
$\begingroup$

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 $\pi > e$.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Apr 7, 2018 at 21:19
4
$\begingroup$

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:

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

$\endgroup$
4
  • $\begingroup$ +1. It's pretty nice. $\endgroup$ Apr 7, 2018 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! $\endgroup$ Apr 7, 2018 at 17:28
  • 1
    $\begingroup$ @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. $\endgroup$ Apr 7, 2018 at 17:36
  • $\begingroup$ @Barry: I do enjoy it as well, thank you! $\endgroup$ Apr 7, 2018 at 17:56
2
$\begingroup$

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:

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

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ Apr 7, 2018 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, 2018 at 17:45
  • 1
    $\begingroup$ 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, 2018 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? $\endgroup$ Apr 7, 2018 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. $\endgroup$ Apr 17, 2018 at 21:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .