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This question is for a very cool friend of mine. See, he really is interested on how seemingly separate concepts can be connected in such nice ways. He told me that he was losing his love for mathematics due to some personal stuff in his life. (I will NOT discuss his personal life, but it has nothing to do with me).

To make him feel better about math (or love of math for that matter), I was planning on giving him a sheet of paper with quite a few relations of $e$ and $\pi$

The ones I were going to give him were:

$$e^{i\pi}=-1$$

and how $e$ and $\pi$ both occur in the distribution of primes (bounds on primes has to do with $\ln$ and the regularization of the product of primes are $4\pi^2$)

Can I have a few more examples of any relations please? I feel it could mean a lot to him. I'm sorry if this is too soft for this site, but I didn't quite know where to ask it.

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closed as primarily opinion-based by A.P., user223391, Eric Wofsey, Michael Albanese, 6005 Nov 23 '15 at 5:23

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ The power series of $\frac1{e^x+1}$ has a radius of convergence of $\pi$. $\endgroup$ – Akiva Weinberger Nov 22 '15 at 23:27
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    $\begingroup$ @AkivaWeinberger Do you mean the classic $\int_0^1 \frac{\pi}{e}\ dx$? $\endgroup$ – pjs36 Nov 22 '15 at 23:28
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    $\begingroup$ @pjs36 This is the mathematical equivalent of a bad pun. :P $\endgroup$ – Akiva Weinberger Nov 22 '15 at 23:29
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    $\begingroup$ Another connection: They both appear in the word "pie." $\endgroup$ – Akiva Weinberger Nov 22 '15 at 23:43
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    $\begingroup$ @A.P. He would like it... Just trust me on that one. The other idea you had was pretty interesting too $\endgroup$ – user285523 Nov 23 '15 at 0:10
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Some additional possibilities

$$n!\sim\sqrt{2\pi n}\left(\frac {n}{e}\right)^n$$

The normal distribution is given by

$$\phi(x) = \frac{1}{2 \pi}e^{(-1/2)x^2}$$

$$\int_{-\infty}^\infty\phi(x)dx=1$$

A personal favorite involving the Euler–Mascheroni constant.

$$\int_0^\infty e^{-x}\ln^2x \,dx=\gamma^2+\frac{\pi^2}{6}$$

Also

$$\frac{e^{3+2\gamma}}{2\pi}=\prod_{n=1}^\infty e^{-2+2/n}\left(1+\frac{2}{n}\right)^n$$

Some rather unusual ones

$$\sum_{n=1}^\infty \frac 1 {n^2}\cos\left(\frac 9{n\pi+\sqrt{n^2\pi^2-9}}\right) =-\frac{\pi^2}{12e^3}$$

$$\pi=72\sum_{n=1}^\infty \frac{1}{n(e^{n\pi}-1)}-96\sum_{n=1}^\infty \frac{1}{n(e^{2n\pi}-1)}+24\sum_{n=1}^\infty \frac 1 {n(e^{4n\pi}-1)}$$

Edit:

$$\zeta(3)=\frac{7}{180}\pi^3-2\sum_{k=1}^\infty \frac 1 {k^3(e^{2\pi k}-1)}$$

An approximation for the Feigenbaum constant is given by

$$\pi+\tan^{-1}\left(e^\pi\right)$$

$$\int_0^\infty \Gamma(x)\,dx=e+\int_0^\infty\frac{e^{-x}}{\pi^2+\ln^2x}\,dx$$

$$\prod_{k=0}^\infty \frac{1}{k^{1/k^2}}=\left(\frac{A^{12}}{2\pi e^\gamma}\right)^{\pi^2/6}$$

where $A$ is the Glaisher–Kinkelin constant.

The denominators $q_n$ of the convergents of the continued fraction expansions of almost all real numbers satisfy

$$\lim_{n\to\infty}q_n^{1/n}=e^{\pi^2/(12\ln 2)}$$

This is known as Lévy's constant.

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  • $\begingroup$ extremely creative findings! Where did you find these? $\endgroup$ – user285523 Nov 23 '15 at 7:00
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$$\int_{-\infty}^\infty\frac{\cos x}{x^2+1}\operatorname d\!x=\frac\pi e$$ EDIT: Also: $$\int_{-\infty}^\infty e^{-x^2}\operatorname d\!x=\sqrt\pi$$

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  • $\begingroup$ I guess that's more interesting than mine :) $\endgroup$ – pjs36 Nov 22 '15 at 23:33
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    $\begingroup$ @pjs36 I added the Gaussian integral. $\endgroup$ – Akiva Weinberger Nov 22 '15 at 23:36
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I'm a fan of $$e^\pi-\pi=20{}$$(Well... almost...)

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  • $\begingroup$ @Justin Yes, that is where my link takes you (note that the equation is blue; try clicking on it). $\endgroup$ – Arthur Nov 23 '15 at 6:53
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    $\begingroup$ Sorry, I didn't notice the equation was a link. I'm viewing on a dark theme which got rid of the blue on the equation. $\endgroup$ – Justin Nov 23 '15 at 7:14
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Here's a beautiful relation between $\pi$ and $e$,

$$\sqrt{\frac{\pi\,e}{2}}=1+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot5}+\frac{1}{1\cdot3\cdot5\cdot7}+\dots+\cfrac1{1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\ddots}}}}$$

It shouldn't be hard to guess who found this. As Kevin Brown of Mathpages remarked, "Is there any other mathematician whose work is instantly recognizable?"

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    $\begingroup$ Ramanujan?${}{}{}$ $\endgroup$ – YoTengoUnLCD Nov 23 '15 at 2:13
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    $\begingroup$ Well the question is now in the hot network questions list, and not everyone reading it will know mathematicians work by sight... $\endgroup$ – curiousdannii Nov 23 '15 at 2:14
  • $\begingroup$ @YoTengoUnLCD: Correct. :) $\endgroup$ – Tito Piezas III Nov 23 '15 at 4:39
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$$e^{\pi\sqrt{163}} =262537412640768743.99999999999925...$$

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    $\begingroup$ +1 for reminding me of the Feynman point and making me crack a smile. $\endgroup$ – user137731 Nov 22 '15 at 23:55
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    $\begingroup$ I don't understand any of the maths, but I gather this one is called Ramanujan's constant and related to a bunch of deep theory. $\endgroup$ – Empy2 Nov 23 '15 at 0:22
  • $\begingroup$ This was observed in the 19th century by a mathematician who was a performing calculating prodigy when young. $\endgroup$ – DanielWainfleet Nov 23 '15 at 1:55
  • $\begingroup$ Or better: $$e^{\pi\sqrt{163}} \approx 12^3(231^2-1)^3 +743.99999999999925\dots\\ e^{\pi\sqrt{427}} \approx 12^3\big((7215 + 924\sqrt{61})^2-1\big)^3+743.999999999999999999999987\dots$$ Notice the common form. $\endgroup$ – Tito Piezas III Nov 23 '15 at 4:49
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I honestly just like the fact that $e + \pi$ might be rational. This is the most embarrassing unsolved problem in mathematics in my opinion. It's clearly transcendental and we have no idea how to prove that it's even irrational.

They're so unrelated additively that we can't prove anything about how unrelated they are. Uh-huh.

$e\pi$ might also be rational. But we do know they can't both be (easy proof).

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You may consider expressing Euler's identity as $e^{i\pi}+1 = 0$ instead of the way you have it, because 0 shows up.

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    $\begingroup$ I think you should read the answer before commenting: the answer is suggesting an alternative to what you wrote. $\endgroup$ – Rob Arthan Nov 22 '15 at 23:37
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    $\begingroup$ @nicole Yeah, but that's really just a trick to make $1$ and $0$ show up. In fact you can make $0$ show up in any equation. For instance: $G_{\mu\nu} = 8\pi G(T_{\mu\nu}+\rho_\Lambda g_{\mu\nu}) \mathbf{+0}$. There, now I've shown that the additive identity shows up in the Einstein field equations! What could be the physical significance?! ;) $\endgroup$ – user137731 Nov 22 '15 at 23:37
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    $\begingroup$ @Bye_World: Nicole's formula is the standard way to express Euler's identity. I don't understand the point of your criticism. $\endgroup$ – Rob Arthan Nov 22 '15 at 23:42
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    $\begingroup$ @RobArthan The reason I don't like this answer is because it is emphasizes addition and not $\pi$ or $e$. I can see your point though $\endgroup$ – user285523 Nov 22 '15 at 23:43
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    $\begingroup$ Wow! I am a complete idiot! Sorry all, I misread the answer. $\endgroup$ – user285523 Nov 22 '15 at 23:56