What are your favorite relations between e and pi?

Question

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.

The power series of $\frac1{e^x+1}$ has a radius of convergence of $\pi$. — Akiva Weinberger, Nov 22 '15 at 23:27
@AkivaWeinberger Do you mean the classic $\int_0^1 \frac{\pi}{e}\ dx$? — pjs36, Nov 22 '15 at 23:28
dig in http://math.stackexchange.com/questions/689315/interesting-and-unexpected-applications-of-pi/689369#689369 for a few more — janmarqz, Nov 22 '15 at 23:29
Of course I don't know your friend nor why he's losing interest in mathematics, but I don't see how a sheet of formulas might help. He will likely be delighted that you care enough about him to actually try to do something about this, but wouldn't a fun problem or an inspiring lecture be better? — A.P., Nov 23 '15 at 00:08
@A.P. He would like it... Just trust me on that one. The other idea you had was pretty interesting too — , Nov 23 '15 at 00:10
Not only do they both appear in the word pie but i appears in it too,and $e^{\pi i}+1=0.$ — DanielWainfleet, Nov 23 '15 at 01:59
Two links explaining why such connections are there in the first place. — Lucian, Nov 23 '15 at 02:13
Interestingly, $\pi^x$, $\pi^e$, and $\pi^{-1}$ don't occur in any of these. $e$ works well as a base, but $\pi$ doesn't really know how to be one. EDIT: Normal distribution has $\frac1{2\pi}$. But that's the only one. — Akiva Weinberger, Nov 23 '15 at 12:09
My favorite:
$$\int_{0}^{1} \left(\frac{5}{2} \left((x - \sqrt{x^2 - 1})^{2i} + x^4\right) - 1\right) , dx = e^{\pi}.$$ — Emmanuel José García, Feb 11 '24 at 21:04

score 29 · Answer 1 · answered Nov 22 '15 at 23:28

29

I'm a fan of $$e^\pi-\pi=20{}$$(Well... almost...)

answered Nov 22 '15 at 23:28

Arthur

199,419

@Justin Yes, that is where my link takes you (note that the equation is blue; try clicking on it). – Arthur Nov 23 '15 at 06:53
1

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. – Justin Nov 23 '15 at 07:14

Akiva Weinberger · Answer 2 · 2015-11-22T23:35:03.417

24

$$\int_{-\infty}^\infty\frac{\cos x}{x^2+1}\operatorname d\!x=\frac\pi e$$ EDIT: Also: $$\int_{-\infty}^\infty e^{-x^2}\operatorname d\!x=\sqrt\pi$$

edited Nov 22 '15 at 23:35

answered Nov 22 '15 at 23:31

Akiva Weinberger

23,062

I guess that's more interesting than mine :) – pjs36 Nov 22 '15 at 23:33
1

@pjs36 I added the Gaussian integral. – Akiva Weinberger Nov 22 '15 at 23:36

Ben Longo · Accepted Answer · 2015-12-01T03:58:19.773

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.

extremely creative findings! Where did you find these? – Nov 23 '15 at 07:00 — , Nov 23 '15 at 07:00

score 16 · Answer 4 · answered Nov 23 '15 at 01:16

16

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?"

answered Nov 23 '15 at 01:16

Tito Piezas III

54,565

6

Ramanujan?${}{}{}$ – YoTengoUnLCD Nov 23 '15 at 02:13
2

Well the question is now in the hot network questions list, and not everyone reading it will know mathematicians work by sight... – curiousdannii Nov 23 '15 at 02:14
1

For a proof, see Mathologer's video on the subject: https://www.youtube.com/watch?v=6iTdNmDHfV0 – Akiva Weinberger Oct 20 '23 at 19:11
@AkivaWeinberger Thanks for referring the video. You know, I never did consider if the two components of the identity had closed-forms. A quick check with Mathematica showed that $$\sqrt{\frac{\pi,e}{2}}\operatorname{erf}\left(\tfrac1{\sqrt2}\right)+\sqrt{\frac{\pi,e}{2}}\left(1-\operatorname{erf}\left(\tfrac1{\sqrt2}\right)\right)=\sqrt{\frac{\pi,e}{2}}$$ where the first addend is the series and the second is the cfrac. It must be nice for people like Ramanujan to have Mathematica built in one's brain. :) – Tito Piezas III Oct 21 '23 at 02:49
@AkivaWeinberger I investigated further. It turns out the series can also be expressed as a nice continued fraction. Kindly see this new post. I assume Mathologer is not aware of this fact. – Tito Piezas III Oct 22 '23 at 07:08

score 12 · Answer 5 · edited Nov 23 '15 at 01:29

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).

score 11 · Answer 6 · answered Nov 22 '15 at 23:52

11

$$e^{\pi\sqrt{163}} =262537412640768743.99999999999925...$$

answered Nov 22 '15 at 23:52

Empy2

50,853

1

+1 for reminding me of the Feynman point and making me crack a smile. – Nov 22 '15 at 23:55
2

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. – Empy2 Nov 23 '15 at 00:22
This was observed in the 19th century by a mathematician who was a performing calculating prodigy when young. – DanielWainfleet Nov 23 '15 at 01:55
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. – Tito Piezas III Nov 23 '15 at 04:49

nicole · Answer 7 · 2015-11-22T23:32:52.487

5

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

edited Nov 22 '15 at 23:32

answered Nov 22 '15 at 23:32

nicole

744

I already had that, please read the question before answering – Nov 22 '15 at 23:32
3

I think you should read the answer before commenting: the answer is suggesting an alternative to what you wrote. – Rob Arthan Nov 22 '15 at 23:37
1

@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?! ;) – Nov 22 '15 at 23:37
4

@Bye_World: Nicole's formula is the standard way to express Euler's identity. I don't understand the point of your criticism. – Rob Arthan Nov 22 '15 at 23:42
1

@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 – Nov 22 '15 at 23:43
@RobArthan I understand that it's standard. My (very mild) criticism is with people who say that this identity is one that links together "the $5$ most important constants in mathematics". $1$ and $0$ only show up because we choose to write it that way instead of the simpler equation $e^{i\pi}=-1$ which is profound enough on its own IMO. – Nov 22 '15 at 23:44
@Bye_World: OK. I can see why you like $e^{i\pi} = -1$. $1$ still shows up in there though $\ddot{\smile}$. – Rob Arthan Nov 22 '15 at 23:52
3

Wow! I am a complete idiot! Sorry all, I misread the answer. – Nov 22 '15 at 23:56

What are your favorite relations between e and pi?

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