I would like to know how to prove $$e^\pi-\pi\sim 20.$$ More precisely, I want to show by using only mathematical tools that, $$19.999<e^\pi-\pi<20$$

I have checked with online calculator and I got $$e^\pi-\pi\approx19.9990999792\sim 20.$$

I tried to use the Tyalor expansion for exponential $$ e^\pi =\sum_{n=0}^{\infty} \frac{\pi^n}{n!} = \pi +1+\sum_{n=2}^{\infty} \frac{\pi^n}{n!}$$ then, $$e^\pi -\pi =1+\sum_{n=2}^{\infty} \frac{\pi^n}{n!}$$ which is not easy to continue from here, since the factor $\pi^n$ is involved. Any idea?

  • $\begingroup$ Hello , see here .sites.google.com/site/tpiezas/0016. Have a good day $\endgroup$ – max8128 Oct 21 '17 at 12:50
  • $\begingroup$ @max8128 What does this have to do with the question? $\endgroup$ – Wojowu Oct 21 '17 at 13:01
  • $\begingroup$ Seriously I went through that link I felt stupid $\endgroup$ – Guy Fsone Oct 21 '17 at 13:02
  • $\begingroup$ There is some information here: mathworld.wolfram.com/AlmostInteger.html $\endgroup$ – Ethan Bolker Oct 21 '17 at 13:05
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    $\begingroup$ For the lower bound, you need to approximate $\pi$ to at least $5$ decimal digits and then you need to sum terms of the exponential series at least to $n=14$. This is, in theory possible, but is going to be plain tedious... $\endgroup$ – Wojowu Oct 21 '17 at 13:07

Use the iteration to calculate Gelfond's constant: $$k_0 = 1/\sqrt{2},\quad k_{n+1}={\frac {1-{\sqrt {1-k_{n}^{2}}}} {1+{\sqrt {1-k_{n}^{2}}}}}$$ $$e^\pi =\lim_{n\to\infty} \left(\frac{4}{k_{n+1}}\right)^{2^{-n}}$$

Within two iterations you should have (assuming you know $\pi$ to 5 or more digits and that you're willing to compute three square roots), that $e^\pi-\pi\approx 19.999$:

$$k_1=3-2\sqrt{2}$$ $$k_2=33+24\sqrt{2}-4\sqrt{140+99\sqrt{2}}$$ And: $$\sqrt{4\over k_2} \approx 19.99926 + \pi$$

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    $\begingroup$ (+1) It might be interesting to expand this answer and explain why we may compute $e^\pi$ through a numerical algorithm that is reminiscent of the AGM mean for the computation of complete elliptic integral of the first kind. $\endgroup$ – Jack D'Aurizio Oct 25 '17 at 18:12
  • $\begingroup$ @JackD'Aurizio - I've actually been looking for the source of this series, and don't have access to the book mentioned as a source. Do you know how to derive it? If so, I'll ask a dedicated question $\endgroup$ – nbubis Oct 25 '17 at 18:44
  • $\begingroup$ @JackD'Aurizio - Here goes: math.stackexchange.com/questions/2489586/… $\endgroup$ – nbubis Oct 25 '17 at 18:53

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