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I would like to use mathematical tools to prove that $$9.9998\lt \frac{\pi^9}{e^8}\lt 10$$

With an on-line calculator I got

$$ \frac{\pi^9}{e^8}\approx 9.9998387978$$ But I do not know any good way to prove this. I failed to use Taylor expansion $$e^8 =\sum_{n=0}^{\infty}\frac{8^n}{n!}$$

Any idea?

Guy Fsone
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  • Well, it looks like taking natural logs helps. You want $\ln(9.9998)<9\ln (\pi)-8<\ln (10)$. Still numerical. – lulu Oct 25 '17 at 16:01
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    It's looks nicer to proof $,2980909<100 \cdot \pi^9<2980910,$ and $,298095<100 \cdot e^8<298096,$ . :-) – user90369 Oct 25 '17 at 16:37
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    That's funny. How did you discover such inequality? – Jack D'Aurizio Oct 25 '17 at 16:42
  • A natural question might be the following: can you show two rational numbers $r,s$ such that there are no integers $i,k,j$ such that $r^i \sim 10^k s^j$? – Ivan Di Liberti Oct 25 '17 at 17:39
  • @IvanDiLiberti: how is that possible, if $\mathbb{Z}\log(r)+\mathbb{Z}\log(s)$ is dense in $\mathbb{R}$? – Jack D'Aurizio Oct 25 '17 at 17:46
  • But than this inequality was expected! We couldn't know that $i=9, j=8, k=1$, but such numbers had to exist. – Ivan Di Liberti Oct 25 '17 at 17:48
  • @IvanDiLiberti: of course, I get your point, there is nothing really special here, but I still find such kind of numerical coincidences amusing. – Jack D'Aurizio Oct 25 '17 at 17:50
  • @JackD'Aurizio Please here is another one. I found your answer here clear and direct. But for this I haven,t found a satisfactory answer: https://math.stackexchange.com/questions/2482717/how-to-prove-19-999e-pi-pi20?rq=1 – Guy Fsone Oct 25 '17 at 17:57
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    @GuyFsone: nbubis' answer to the linked question has a structure that is similar to my answer here. The point is that the constant $e^{\pi}$ appears in the context of j-invariants and modular forms, hence it can be computed through fast-convergent numerical algorithms, recalling the AGM mean for the computation of complete elliptic integrals of the first kind. nbubis' answer is very satisfactory by my point of view, but I will take some time to think about an alternative approach. I am not sure to be able to do better. – Jack D'Aurizio Oct 25 '17 at 18:09
  • Thanks in advance. I am not so familiar with the notion of "j-invariants and modular forms" :) that is why I got lost at some point. – Guy Fsone Oct 25 '17 at 18:14
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    Guy: Please stop making very trivial edits to get you post bumped to the front page. This example here, among many others: indicate very inappropriate behavior. – amWhy Nov 03 '17 at 18:30

1 Answers1

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My approach can be summarized as follows:

  1. By a well-known fact about Euler numbers, $\pi^9$ is given by a rational multiple of the fast-converging series $\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^9}$;
  2. By the generalized continued fraction for the hyperbolic cotangent function, it is pretty simple to produce accurate rational approximations of $e^8$;
  3. The given inequality can be proved by producing accurate rational approximations for both $\pi^9$ and $e^8$ through the previous points, then by comparing them.
  1. $$\pi^9 = \frac{203325460470370265464832}{6820919298826171875}\pm 7\cdot 10^{-5} $$
  2. $$ e^8 = \frac{6456755}{2166}\pm 4\cdot 10^{-8}$$
  3. Done.
Jack D'Aurizio
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