26

Prove that:

$$e^\pi+\frac{1}{\pi}< \pi^{e}+1$$

Using Wolfram Alpha $\pi e^{\pi}+1 \approx 73.698\ldots$ and $\pi(\pi^{e}+1) \approx 73.699\ldots$

Can this inequality be proven without brute-force estimations (anything of the sort $e\approx 2.7182...$ or $\pi \approx 3.1415...$)? I've just seen this and I remembered I've seen the question asked here in an older paper, but I don't remember the details.

Note that this is sharper because it can be written as:

$$e^{\pi}-\pi^e<1-\frac{1}{\pi}<1$$

I've tried, but none of the methods in the linked question (which study the function $x^\frac{1}{x}$) can be applied here.

LHF
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    What about the function $f(x)=e^x-x^e$? – Weierstraß Ramirez Feb 24 '20 at 20:50
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    @WeierstraßRamirez, That is essentially the same as studying $x^{\frac{1}{x}}$. It has two critical points at $1$ (maximum) and $e$ (minimum). I think it's only enough to show $e^{\pi} > \pi^e$. Or is there something I'm missing? – LHF Feb 24 '20 at 20:58
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    Maybe useful the next link! – user0410 Feb 24 '20 at 21:38
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    What exactly are "estimations"? What exactly may the solution involve? – joriki Feb 24 '20 at 22:19
  • Mmm, perhaps $f(x)=e^x(1-\frac{1}{\pi})+x^e(1+\frac{e^\pi}{\pi^{e+1}})$? – Weierstraß Ramirez Feb 24 '20 at 22:22
  • @joriki, I was thinking nothing of the sort $e\approx 2.7182...,\ \pi\approx 3.1415...$. – LHF Feb 24 '20 at 22:23
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    Maybe interesting. Another sharp bound for the expression $e^\pi-\pi^e$ is given by $$ e^\pi-\pi^e \approx \frac{1}{6},\sqrt [3]{75+7,\sqrt {449}}-,{\frac {2}{\sqrt [3]{75+7,\sqrt {449}}}}<1-\frac{1}{\pi} $$ In fact, I changed the real root of the polynomial $x^3+x-1$, slightly! – Amin235 Feb 26 '20 at 18:13

1 Answers1

14

From the continued fraction expansion of $\pi$, we have

$$ \frac{333}{106}\lt\frac{103993}{33102}\lt\pi\lt\frac{355}{113}\;. $$

There are various ways of proving these inequalities without using decimal approximations:

In the case of $\mathrm e$, the continued fraction expansion is regular and can be systematically derived (see e.g. A Short Proof of the Simple Continued Fraction Expansion of e by Henry Cohn, The American Mathematical Monthly, $113(1)$, $57$–$62$, The Simple Continued Fraction Expansion of e by C. D. Olds, The American Mathematical Monthly, $77(9)$, $968$–$974$, or Continued fraction for e at Topological Musings); it yields

$$ \frac{1264}{465}\lt\mathrm e\lt\frac{1457}{536}\;. $$

Thus it suffices to show that

$$ \left(\frac{1457}{536}\right)^\frac{355}{113}+\frac1{\frac{333}{106}}\lt\left(\frac{103993}{33102}\right)^\frac{1264}{465} + 1\;, $$

or

$$ \left(\frac{1457}{536}\right)^\frac{355}{113}\lt\left(\frac{103993}{33102}\right)^\frac{1264}{465} + \frac{227}{333}\;. $$

Since both sides contain fractional exponents, it’s hard to compare them directly; but we can find a fraction that lies between them and compare them to it separately. Among the suitable fractions, the one with the lowest denominator is $\frac{4767}{206}$. The rational inequalities

$$ \left(\frac{1457}{536}\right)^{355}\lt\left(\frac{4767}{206}\right)^{113} $$

and

$$ \left(\frac{4767}{206}-\frac{227}{333}\right)^{465}\lt\left(\frac{103993}{33102}\right)^{1264} $$

are readily checked with integer arithmetic, and thus with

$$ \left(\frac{1457}{536}\right)^\frac{355}{113}\lt\frac{4767}{206}\lt\left(\frac{103993}{33102}\right)^\frac{1264}{465} + \frac{227}{333} $$

the result follows.

joriki
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    I appreciate the great work and references. Given the sharpness of the inequality I doubt a better approach can be found, but I'll wait a little more time (a day or two) before accepting the answer. – LHF Feb 25 '20 at 13:19
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    With the utmost respect for you, I appreciate your intervention to this post but how to deny that underlying calculations of it are far more complicated than simply calculating both sides of the inequality. Impossible that I put you a downvote but in no way, due to the type of problem posed, would you put an upvote. Best regards. – Piquito Feb 25 '20 at 14:07
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    @Piquito: I think you'd need to explicate your concept of "simply calculating". The question asked for a proof, so you'd need to be able to control the errors in this "simple calculation". Your own answer works with approximations ($\approx$) without specifying any error bounds for them; thus it doesn't constitute a proof. Moreover, the question explicitly asked for answers that don't use this sort of approximation. – joriki Feb 25 '20 at 14:11
  • Kind of response to problems with parallel thinking. In this case, calculate everything you want out of the problem and apply it here. My best wishes for you. – Piquito Feb 25 '20 at 15:51
  • Perhaps the downvoter could explain the downvote? – joriki Feb 25 '20 at 17:11
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    @joriki, it seems like an epidemy of mysterious senseless downvotes on MSE nowadays. – LHF Feb 25 '20 at 17:14
  • I think that "calculate" means using a calculator. Try to use one that has double precision, just to be on the safe side. The WP-34s emulator on my iPhone is good to 33 digits. – richard1941 Mar 09 '20 at 20:29
  • @joriki But I still don't understand why it isn't easier to let a computer compute the difference of the two sides numerically. I mean, over the course of 30 years, people came up with various smart algorithms that allow us to control exactly how many significant digits we have i.e. how small the error is. It seems a bit strange to me that we wouldn't resort to these powerful tools... – Maximilian Janisch Mar 17 '20 at 16:32