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I am a high school student.
Help me prove $\pi^{e}<23<e^{\pi}$.
First, I tried it by proving $\frac{\log \pi}{\pi}<\frac{\log 23}{\pi e}<\frac{\log e}{e}$. However, I was not able to prove the magnitude relationship between $23$ and the others.
Second, I attemted to approach by quadratic curve at $x=3$.
$e^{x}=e^{3}+e^{3}(x-3)+\frac{1}{2}e^{3}(x-3)^2$.
And the third derivative of $e^{x}$ is positive.So,
$e^{\pi}>\frac{1}{2}e^{3}(\pi^{2}-4\pi +5)> \frac{1}{2}*2.717^{3}(3.14159^{2}-4*3.1416+5)>23$
This is how I can prove the supremum of the inequality, but I can't the other. I ended up stuck here.

Moveover, I'm going so far as to believe there is a more beautiful, plain, and concise proof for this...

Frenchtoast
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  • this might be helpful https://math.stackexchange.com/questions/337565/a-question-comparing-pie-to-e-pi – Hidaw Oct 21 '17 at 00:15
  • Thank you for answering. I can prove $\pi^{e}<e^{\pi}$ by proving $\frac{\log\pi}{\pi}<\frac{\log e}{e}$, but I can't prove the magnitude relationship between them and $23$. – Frenchtoast Oct 21 '17 at 00:21
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    Hi @Frenchtoast, I removed a couple of tags from your question, and I just wanted to explain why. Functional analysis is not about studying funtions, but about studying (often infinite-dimensional) topological vector spaces, often by examining their dual space of continuous linear functionals. Proof theory is about studying the nature of proofs, and is not necessary when asking for a proof. Interesting question though! – Theo Bendit Oct 21 '17 at 00:39
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    I'm sorry for my ignorance of elementary knowledges, and thank you for your kindness! – Frenchtoast Oct 21 '17 at 00:44
  • hint, it's easier to consider the more general problem: for which positive x is $e^x>x^e$ – shai horowitz Oct 21 '17 at 01:03
  • Thank you for answering. Then how can I compare $23$ with the both upper and lower side? – Frenchtoast Oct 21 '17 at 01:10
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    It should be $e^{x}>e^{3}+e^{3}(x-3)+\frac12 e^{3}(x-3)^2$, giving $e^\pi>23.098\ldots$. –  Oct 21 '17 at 06:28
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    (+1) for showing own working that are most of the times missing from questions of new users... – Soham Oct 21 '17 at 13:40
  • I'm sorry but I don't know what (+1) is. – Frenchtoast Oct 21 '17 at 23:42

3 Answers3

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For the first part we use $\pi<{22\over7}$. Furthermore the theorem on alternating series gives $${1\over e}>1-1+{1\over2}-{1\over6}+{1\over24}-{1\over120}={44\over120}={11\over30}\ .$$ It follows that $$\pi^e<\left({22\over7}\right)^{30/11}\ .$$ It is therefore sufficient to prove that $$22^{30}<7^{30}\cdot 23^{11}\ .$$ Here the LHS computes to $$18\,736\,153\,019\,903\,829\,443\,036\,278\,993\,864\,332\,673\,024\ ,$$ and the RHS to $$21\,475\,703\,365\,914\,111\,444\,329\,770\,286\,387\,088\,568\,823\ .$$

Christian Blatter
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  • Thanks for answering. This is an awesome proof, I've never hit upon. Even if I don't have any calculator, the proof is still possible by thinking $30\log_{10}{22}-40<\log_{10}{2}<30\log_{10}{7}+11\log_{10}{23}-40$ and the value of $\log_{10}{2}$ and $\log_{10}{3}$. – Frenchtoast Oct 21 '17 at 23:41
  • That's an unbelievable solution!+1). – xpaul Oct 22 '17 at 00:46
  • Christian, just to let you know, I found a way to verify the inequality $22^{30}\lt7^{30}\cdot23^{11}$ by hand. See the "Added later" portion of my answer. – Barry Cipra Oct 30 '17 at 23:46
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Both $\pi\approx\frac{22}{7}$ and $e\approx\frac{19}{7}$ can be proved through Beuker-like integrals:

$$ \underbrace{\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx}_{\in\left(0,\frac{1}{256}\right)}=\frac{22}{7}-\pi,\qquad \underbrace{\int_{0}^{1}x^2(1-x)^2 e^{-x}\,dx}_{\in\left(0,\frac{1}{16}\right)}=14-\frac{38}{e} $$

and $\pi^e < e^\pi$ is trivial from the increasing nature of $\frac{x}{\log x}$ on $[e,+\infty)$.
$\pi^e<23$ is actually a loose inequality, since $\left(\frac{22}{7}\right)^{19}\ll 23^7$.
$23<e^\pi$ is a bit tighter, since $23^7\approx 3.4\cdot 10^9$ while $\left(\frac{19}{7}\right)^{22}\approx 3.47\cdot 10^9$.
Still, $\pi^e<23<e^\pi$ is pretty straightforward to prove by approximating both $e$ and $\pi$ in sevenths.

Jack D'Aurizio
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  • Thanks for answering. I'm sorry if I'm misunderstanding, but $\frac{22}{7}^{\frac{19}{7}}<e^{\pi}$ according to my calculator. Then, I doubt actually $e^{\pi}<23$ is proved by that. – Frenchtoast Oct 21 '17 at 23:25
  • @Frenchtoast: the first integral also gives a bound for $\pi-\frac{22}{7}$. – Jack D'Aurizio Oct 22 '17 at 00:57
  • Sorry I misunderstood. By the way, how should I prove $\frac{22}{7}^{\frac{19}{7}}<e^{\pi}$? – Frenchtoast Oct 22 '17 at 01:57
  • @Frenchtoast: replace the polynomials $x^4(1-x)^4$ and $x^2(1-x)^2$ with their squares to get more accurate rational approximations of $e$ and $\pi$, then use them to prove such inequality through the arithmetic of integers. – Jack D'Aurizio Oct 22 '17 at 02:01
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Here is a proof that $23\lt e^\pi$ based on the inequalities $e\gt2.718$ and $\pi\gt3.141$ and two calculations that can arguably be done by hand:

$$2.718\cdot1.047=2.845747\gt2.845=5.69/2$$ and $$5.69^3=184.220009\gt184=8\cdot23$$

We have

$$e^{\pi/3}\gt e^{3.141/3}=e^{1.047}=e\cdot e^{.047}\gt2.718\cdot1.047\gt5.69/2$$

and thus

$$e^\pi=(e^{\pi/3})^3\gt(5.96/2)^3\gt8\cdot23/8=23$$

Remark: You can, if you like, replace the cube of $5.69$ with the easier $5.7^3=185.193\gt185$ and then write

$$5.69^3=5.7^3\left(1-{.01\over5.7}\right)^3\gt185\left(1-3\left(1\over570\right)\right)=185\left(1-{1\over190} \right)\gt185-1=184$$

Added later: Here is a proof that $\pi^e\lt23$ using a minimal amount of computation.

We begin with Christian Blatter's observation that it suffices to show $22^{30}\lt7^{30}23^{11}$, which we'll rewrite as

$$\left(22^3\over7^3\cdot23\right)^{10}\lt23$$

We'll show this in two steps:

$${22^3\over7^3\cdot23}\lt{27\over20}\quad\text{and}\quad\left(27\over20\right)^{10}\lt23$$

For the first of these, we have

$$\begin{align} {22^3\over7^3\cdot23}\lt{27\over20} &\iff20\cdot22^3\lt21^3\cdot23\\ &\iff22^2(21-1)(21+1)\lt21^2(22-1)(22+1)\\ &\iff22^2(21^2-1)\lt21^2(22^2-1)\\ &\iff-22^2\lt-21^2 \end{align}$$

For the second, observe that $(27/20)^2=1.35^2=1.8225\lt1.836=1.8(1.02)$ and

$$1.02^5=\left(1+{1\over50}\right)^5=1+{5\over50}+{10\over50^2}+{10\over50^3}+{5\over50^4}+{1\over50^5}\lt1+{1\over10}+{1\over10^2}+{1\over10^3}+\cdots={10\over9}={2\over1.8}$$

Thus

$$\left(27\over20\right)^{10}\lt1.8^5(1.02)^5\lt2\cdot1.8^4=2\cdot3.24^2\lt2\cdot3.25^2=2\left(13\over4\right)^2={169\over8}=21{1\over8}\lt23$$

and we're done!

Remark: The trickiest hand calculation here is $135^2=18225$, which isn't that hard if you know that $13\cdot14=182$ (and know the squaring trick $d5^2=d(d+1)25$). Alternatively, you can get the inequality $1.35^2\lt1.836$ from $135^2=3^2\cdot45^2=9\cdot2025\lt9\cdot2040$.

Barry Cipra
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