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Certain formulas I really enjoy looking at like the Euler-Maclaurin formula or the Leibniz integral rule. What's your favorite equation, formula, identity or inequality?

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    $\begingroup$ Voting to close. People are at the repeating-other-people's-answers stage now. $\endgroup$ Commented Aug 21, 2010 at 18:23
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    $\begingroup$ The question has been closed as no longer relevant. It had a long and healthy life, but the large number of answers has become unwieldy. If the question had been asked more recently, it would probably have been closed sooner as being "overly broad". I encourage people who are interested in following up issues raised in the question or the answers with further questions. Please be specific! $\endgroup$ Commented Aug 22, 2010 at 9:02
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    $\begingroup$ Sadly my favourite $\sum \frac{1}{n^2 +a^2} = \frac{\pi}{a} cth(\pi a)$ wasn't listed $\endgroup$ Commented May 1, 2011 at 16:57
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    $\begingroup$ $$Rf_*R\hbox{Hom}(F,f^!G)\approx R\hbox{Hom}(Rf_!F,G)$$ $\endgroup$ Commented Oct 22, 2016 at 22:01
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    $\begingroup$ Mine is Poisson formula which can take the form $\int_H f = \int_{H^{\perp}} \hat f$. $\endgroup$
    – Watson
    Commented Dec 21, 2019 at 9:36

62 Answers 62

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$e^{\pi i} + 1 = 0$

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    $\begingroup$ I never liked writing it in the form e^(i pi) + 1 = 0, it's not a way I'd ever write anything. I think it looks a lot nicer when written e^(i pi) = -1. $\endgroup$ Commented Oct 28, 2009 at 22:45
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    $\begingroup$ More interesting and useful (and less mysterious) is: exp(i t) = cos(t) + i sin(t). $\endgroup$ Commented Oct 29, 2009 at 14:49
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    $\begingroup$ @Gerald: I certainly agree. But I think this particular equation is so pretty I once had it henna tattoo-ed on my hand. @Sam: It's just a matter of eye, beholder etc. I don't like minus signs, but do like 0 on the RHS. $\endgroup$ Commented Oct 29, 2009 at 15:29
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    $\begingroup$ I think people ascribe to this formula excessive mystery. The mystery disappears if one replaces the usual exponential with the matrix exponential on M_2(R) and then this is just the statement that the only trajectory of a particle whose velocity is always perpendicular to its displacement (and in the same proportion) is a circle. $\endgroup$ Commented Dec 8, 2009 at 18:04
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    $\begingroup$ I think the people who ascribe to this formula excessive mystery might not find matrix exponentials particularly less mysterious. $\endgroup$ Commented Mar 4, 2010 at 18:16
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Stokes' Theorem $$\int_M\mathrm{d}\omega=\oint_{\partial M}\omega.$$

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Trivial as this is, it has amazed me for decades:

$(1+2+3+\cdots+n)^2=1^3+2^3+3^3+\cdots+n^3.$

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    $\begingroup$ I wouldn't call it trivial. There is also a nice combinatorial proof for it. $\endgroup$ Commented Jun 3, 2010 at 14:12
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    $\begingroup$ I am a bit late in seeing this post, but I completely agree with you, Yaakov. The equality is a bit related to the following pattern which I discovered as a child but continues to amaze me to this day: $1=1^3$; $3+5=2^3$; $7+9+11=3^3$; $13+15+17+19=4^3$;... Many mathematicians know that the sum of the first n odd numbers is n2, but I think very few are aware of this trivial yet incredible pattern. I actually wrote a little post about this on a math blog (the Everything Seminar): $\endgroup$ Commented Nov 3, 2010 at 5:34
  • $\begingroup$ oops, here is the link: cornellmath.wordpress.com/2008/02/15/… $\endgroup$ Commented Nov 3, 2010 at 5:34
  • $\begingroup$ Also see mathoverflow.net/questions/67117/… and in particular Vladimir Dotsenko's answer, where other similar relations are listed, including $2P_3^2=P_7+P_5$. $\endgroup$ Commented Jun 10, 2011 at 12:13
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    $\begingroup$ @DenisSerre, it is an interesting position that belonging to an infinite family makes a formula not beautiful …. $\endgroup$
    – LSpice
    Commented Feb 13, 2020 at 11:19
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There are many, but here is one.

$d^2=0$

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    $\begingroup$ que?``````````` $\endgroup$
    – BlueRaja
    Commented Jul 15, 2010 at 16:24
  • $\begingroup$ I'm sorry, but I don't get it. Could someone please give more information about what this equation is all about? $\endgroup$
    – JRN
    Commented Oct 27, 2011 at 13:35
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    $\begingroup$ en.wikipedia.org/wiki/Exterior_derivative $\endgroup$ Commented Mar 3, 2012 at 16:22
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$$ \frac{24}{7\sqrt{7}} \int_{\pi/3}^{\pi/2} \log \left| \frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}\right| dt\\ = \sum_{n\geq 1} \left(\frac n7\right)\frac{1}{n^2}, $$ where $\left(\frac n7\right)$ denotes the Legendre symbol. Not really my favorite identity, but it has the interesting feature that it is a conjecture! [Update: the conjecture was proved in 2010, https://arxiv.org/abs/1005.0414.] It is a rare example of a conjectured explicit identity between real numbers that can be checked to arbitrary accuracy. This identity has been verified to over 20,000 decimal places. See J. M. Borwein and D. H. Bailey, Mathematics by Experiment: Plausible Reasoning in the 21st Century, A K Peters, Natick, MA, 2004 (pages 90-91).

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    $\begingroup$ @Chandrasekhar: I don't know whether it has a name. $\endgroup$ Commented Jun 20, 2012 at 20:55
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    $\begingroup$ It may be formulated with integrals only, and quite similar: \begin{align*} 7\int_0^1 \log x\cdot \frac{1+x-x^2+x^3-x^4-x^5}{1-x^7}dx=\\ 12 \int_{\frac{\sqrt{21}-5}2}^1 \frac{\log |x|}{2+3x+2x^2}dx \end{align*} $\endgroup$ Commented Mar 9, 2015 at 0:07
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    $\begingroup$ As mentioned in this question, a proof of this identity can be found in Section 5 of arxiv.org/abs/1005.0414 $\endgroup$
    – j.c.
    Commented May 7, 2019 at 19:19
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    $\begingroup$ For anyone else who was confused by the notation, the notation $(\frac{n}{7})$ is something called a Legendre symbol. It's not $n/7$ in parens. $\endgroup$
    – user21349
    Commented May 8, 2019 at 22:50
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    $\begingroup$ The Legendre-symbol notation is standard in number theory, at least enough that LaTeX has the fancy command \genfrac devoted to producing it and its variants; e.g., $\genfrac(){}{}n 7$ \genfrac(){}{}n 7. $\endgroup$
    – LSpice
    Commented May 25, 2019 at 23:44
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Mine is definitely $$1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}+\cdots=\frac{\pi^2}{6},$$ an amazing relation between integers and pi.

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    $\begingroup$ I honestly find the infinite product expression of this formula to be more intriguing - this now relates pi to just the primes. $\endgroup$ Commented Dec 18, 2009 at 18:47
  • $\begingroup$ pi also appears in other values of $\zeta(2n)$, for positive integers $n$. $\endgroup$
    – Regenbogen
    Commented Mar 4, 2010 at 18:05
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    $\begingroup$ @ Regenbogen: It appears in ALL values of $\zeta(2n)$. $\endgroup$
    – Max Muller
    Commented Jul 8, 2010 at 15:32
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    $\begingroup$ This identity is the subject of one of my favourite proofs: For $0 < x < \pi/2$ we have $0 < \sin x < x < \tan x$, thus $1/\sin^2 x -1 < 1/x^2 < 1/\sin^2 x$. Split the interval $(0,\pi/2)$ into $2^n$ equal parts, and sum the inequality over the interior grid points $x_k=(k/2^n)(\pi/2)$, $k=1,\ldots,2^n-1$. This gives $S_n - (2^n-1) < (4^{n+1}/\pi^2) \sum_{k=1}^{2^n-1} k^{-2} < S_n$, where $S_n=\sum_k 1/\sin^2 x_k$ satisfies $S_1=2$ and $S_n = 2+4 S_{n-1}$ (since $1/\sin^2 x+1/\sin^2(\pi/2-x) = 4/\sin^2 2x$), so that $S_n=2(4^n-1)/3$. Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. Voilà! $\endgroup$ Commented Jul 15, 2010 at 14:08
  • $\begingroup$ I should also mention that variants of the proof above can be found in Aigner & Ziegler's "Proofs from the book" and in a paper by Josef Hofbauer (Amer. Math. Monthly, February 2002). $\endgroup$ Commented Jul 15, 2010 at 14:10
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There's lots to choose from. Riemann-Roch and various other formulas from cohomology are pretty neat. But I think I'll go with

$$\sum\limits_{n=1}^{\infty} n^{-s} = \prod\limits_{p \text{ prime}} \left( 1 - p^{-s}\right)^{-1}$$

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$$1+2+3+4+5+\cdots = -\frac1{12}\,,$$ once suitably regularised of course :-)

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    $\begingroup$ Ramanujan got rejected by many famous mathematicians (looking for a scholarship outside of India) because they did not understand this.. $\endgroup$
    – BlueRaja
    Commented Jul 15, 2010 at 5:26
  • $\begingroup$ @BlueRaja What do you mean by that? Do you mean that his work, without this regularization, was insufficient for a scholarship? $\endgroup$ Commented May 7, 2019 at 21:35
  • $\begingroup$ @Acccumulation: He sent his results to multiple professors before Hardy, with a universally negative response. The response from MHM Hill about this very equation: "[Ramanujan] does not understand the precautions which have to be taken in dealing with divergent series, otherwise he could not have obtained the erroneous results you send me" $\endgroup$
    – BlueRaja
    Commented May 26, 2020 at 22:46
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$$\frac{1}{1-z} = (1+z)(1+z^2)(1+z^4)(1+z^8)...$$

Both sides as formal power series work out to $1 + z + z^2 + z^3 + ...$, where all the coefficients are 1. This is an analytic version of the fact that every positive integer can be written in exactly one way as a sum of distinct powers of two, i. e. that binary expansions are unique.

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    $\begingroup$ That's simple, but quite neat too. Haven't thought about that expansion before. Does it have any applications? $\endgroup$ Commented Oct 29, 2009 at 23:25
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    $\begingroup$ Well, it says binary expansions are unique, which is kind of a nice fact if you want to make computers that depend on that. As for applications within mathematics itself, I don't really know of any. $\endgroup$ Commented Oct 30, 2009 at 0:15
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    $\begingroup$ multiplying the RHS by $(1-z)$ starts an impressive chain reaction $\endgroup$ Commented May 25, 2011 at 6:33
  • $\begingroup$ @MichaelLugo Sorry, undergrad here. Could you please explain how this says that binary expansions are unique? $\endgroup$
    – Ovi
    Commented May 7, 2019 at 22:51
  • $\begingroup$ @Ovi The coefficient of $z^n$ in the Taylor series of $\frac1{1-z}$ counts how many subsets of the powers of $2$ sum up to $n$. Picking a term from each factor in Lugo's product picks out a subset of the powers of $2$ $\endgroup$
    – wlad
    Commented May 8, 2019 at 12:10
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I'm currently obsessed with the identity $\det (\mathbf{I} - \mathbf{A}t)^{-1} = \exp \text{tr } \log (\mathbf{I} - \mathbf{A}t)^{-1}$. It's straightforward to prove algebraically, but its combinatorial meaning is very interesting.

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  • $\begingroup$ So what is its combinatorial meaning, anyway? $\endgroup$ Commented Oct 29, 2009 at 4:17
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    $\begingroup$ If A is the adjacency matrix of a finite graph G, then the coefficient of t^k counts the number of non-negative integer linear combination of aperiodic closed walks on G (without a distinguished vertex) with a total of k vertices. This is an equivalent to an Euler product for the RHS which is again straightforward to prove algebraically and very interesting combinatorially. $\endgroup$ Commented Oct 29, 2009 at 4:24
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    $\begingroup$ That's awesome! $\endgroup$ Commented Oct 29, 2009 at 10:47
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    $\begingroup$ I should also note that as a special case of the Euler product one gets the cyclotomic identity: en.wikipedia.org/wiki/Cyclotomic_identity $\endgroup$ Commented Oct 29, 2009 at 14:40
  • $\begingroup$ Funny. I got obsessed with that one for a while. I guess you know it's connected with my answer to this question about 1+2+3+...=-1/12 because you combine both when you compute regularised determinants of Laplacians. $\endgroup$
    – Dan Piponi
    Commented May 14, 2010 at 21:05
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$V - E + F = 2$

Euler's characteristic for connected planar graphs.

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$196884 = 196883 + 1$

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    $\begingroup$ Can you explain what this is about? $\endgroup$
    – M. Winter
    Commented Dec 26, 2018 at 23:06
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    $\begingroup$ This apparently references Monstrous Moonshine conjectures, which started with an observation that the coefficient $c_1=196884$ of $q$ in the $q$-expansion of the modular invariant $j(\tau)$ is the sum $1+196883$ of the dimensions of two smallest irreducible representations of the largest sporadic finite simple group $M$, known as the Monster. $\endgroup$ Commented May 8, 2019 at 14:22
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For a triangle with angles a, b, c $$\tan a + \tan b + \tan c = (\tan a) (\tan b) (\tan c)$$

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    $\begingroup$ It is very interesting. So if inverse is also true we have interesting algebraic structure connected to set of triangles with equivalence relation given by similarity of triangles. Very interesting indeed! $\endgroup$
    – kakaz
    Commented Feb 9, 2010 at 8:44
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Given a square matrix $M \in SO_n$ decomposed as illustrated with square blocks $A,D$ and rectangular blocks $B,C,$

$$M = \left( \begin{array}{cc} A & B \\\ C & D \end{array} \right) ,$$

then $\det A = \det D.$

What this says is that, in Riemannian geometry with an orientable manifold, the Hodge star operator is an isometry, a fact that has relevance for Poincare duality.

https://en.wikipedia.org/wiki/Hodge_duality

https://en.wikipedia.org/wiki/Poincar%C3%A9_duality

But the proof is a single line:

$$ \left( \begin{array}{cc} A & B \\\ 0 & I \end{array} \right) \left( \begin{array}{cc} A^t & C^t \\\ B^t & D^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\\ B^t & D^t \end{array} \right). $$

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  • $\begingroup$ Ooh... that's a cute proof. $\endgroup$ Commented Apr 25, 2010 at 21:34
  • $\begingroup$ I can't remember the guy's name who found it, he was at UCSD and did lots of stuff with matrices. He was one of many people I told this little fact, nobody had ever heard of it, although it pops out if you go through the material on the Hodge star in Frank Warner's book, "Foundations of Differentiable Manifolds and Lie Groups." I think later editions have a different title. $\endgroup$
    – Will Jagy
    Commented Apr 27, 2010 at 2:28
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I always thought this one was really funny: $1 = 0!$

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  • $\begingroup$ Yeah, so do I. It is also convention that zero product is $1$. $\endgroup$
    – Sunni
    Commented Mar 4, 2010 at 1:15
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    $\begingroup$ True, but that's not what's going on here. One can define 0! by extending the rule n!=n*(n-1)! with n=1, or by computing that \Gamma(1) in its integral form is indeed 1. And it's only very mildly a convention that we choose \Gamma as the interpolating function for the factorial function. $\endgroup$ Commented Mar 4, 2010 at 18:23
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    $\begingroup$ Or, 0! is the number of bijective maps on a set with 0 elements, since the empty set is the only such map. $\endgroup$ Commented Mar 19, 2010 at 3:52
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    $\begingroup$ Ha! Very funny! Reminded me of when one of my professors ended a proof with the statement to the effect that the Q is an element in set D. (get it?) $\endgroup$
    – Burhan
    Commented May 21, 2010 at 18:25
  • $\begingroup$ At first I thought you said $0=1$ with an exclamation point for emphasis. Maybe it was your favorite since it is a prototype of a false statement ;) $\endgroup$ Commented Jul 15, 2010 at 10:39
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It's too hard to pick just one formula, so here's another: the Cauchy-Schwarz inequality:

||x|| ||y|| >= |(x.y)|, with equality iff x&y are parallel.

Simple, yet incredibly useful. It has many nice generalizations (like Holder's inequality), but here's a cute generalization to three vectors in a real inner product space:

||x||2 ||y||2 ||z||2 + 2(x.y)(y.z)(z.x) >= ||x||2(y.z)2 + ||y||2(z.x)2 + ||z||2(x.y)2, with equality iff one of x,y,z is in the span of the others.

There are corresponding inequalities for 4 vectors, 5 vectors, etc., but they get unwieldy after this one. All of the inequalities, including Cauchy-Schwarz, are actually just generalizations of the 1-dimensional inequality:

||x|| >= 0, with equality iff x = 0,

or rather, instantiations of it in the 2nd, 3rd, etc. exterior powers of the vector space.

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I think that Weyl's character formula is pretty awesome! It's a generating function for the dimensions of the weight spaces in a finite dimensional irreducible highest weight module of a semisimple Lie algebra.

$$\operatorname{ch}(V)=\frac{\sum_{w\in W}(-1)^{\ell(w)}w\left(e^{\lambda+\rho}\right)}{e^\rho\prod_{\alpha>0}\left(1-e^{-\alpha}\right)}$$

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Gauss-Bonnet, even though I am not a geometer.

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It has to be the ergodic theorem, $$\frac{1}{n}\sum_{k=0}^{n-1}f(T^kx) \to \int f\:d\mu,\;\;\mu\text{-a.e.}\;x,$$ the central principle which holds together pretty much my entire research existence.

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$2^n>n $

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  • $\begingroup$ What is so special about that? It's pretty easy to prove. $\endgroup$ Commented Jan 10, 2010 at 18:14
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    $\begingroup$ I didn't say it was hard to prove, but you've asked a fair question. It struck me as beautiful when I first learned as an undergrad of this way to see that there are infinitely many different infinite cardinals. $\endgroup$ Commented Jan 10, 2010 at 18:40
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    $\begingroup$ Perhaps I should have written $|2^X|>|X|$ to better indicate what I had in mind. $\endgroup$ Commented Jan 10, 2010 at 18:43
  • $\begingroup$ Or $2^\kappa>\kappa$. But I like it in the way that you stated it. $\endgroup$
    – Carsten S
    Commented Jan 15, 2010 at 0:06
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The formula $\displaystyle \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2+1} dx = \frac{\pi}{e}$. It is astounding in that we can retrieve $e$ from a formula involving the cosine. It is not surprising if we know the formula $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$, yet this integral is of a purely real-valued function. It shows how complex analysis actually underlies even the real numbers.

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  • $\begingroup$ This is tantalizingly close to giving a variant of $e$ as a period. $\endgroup$
    – user44143
    Commented May 7, 2019 at 15:31
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It may be trivial, but I've always found

$\sqrt{\pi}=\int_{-\infty}^{\infty}e^{-x^{2}}dx$

to be particularly beautiful.

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    $\begingroup$ Once one decides that the "right" definition of $n!$ when $n$ isn't a natural number is $\Gamma(n+1)$, this formula becomes equivalent to one of my favorites: $(-{\frac{1}{2}})!=\sqrt{\pi}$. $\endgroup$ Commented Aug 21, 2010 at 18:55
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Euclid, Elements, Book1 Prop 47:

Ἐν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν τετραγώνοις.

That is,

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.

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$\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2} \frac{q-1}{2}}$.

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Riemann-Roch, and its generalizations:

Hirzebruch-Riemann-Roch

Grothendieck-Hirzebruch-Riemann-Roch

Atiyah-Singer (which is also a generalization of Gauss-Bonnet)

Is it cheating to put all of these in a single answer? :-)

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My favorite is the Koike-Norton-Zagier product identity for the j-function (which classifies complex elliptic curves):

j(p) - j(q) = p-1 \prodm>0,n>-1 (1-pmqn)c(mn),

where j(q)-744 = \sumn >-2 c(n) qn = q-1 + 196884q + 21493760q2 + ... The left side is a difference of power series pure in p and q, so all of the mixed terms on the right cancel out. This yields infinitely many identities relating the coefficients of j.

It is also the Weyl denominator formula for the monster Lie algebra.

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$${\mathbb E}[X+Y]={\mathbb E}[X]+{\mathbb E}[Y]$$ for any two random varibles $X$ and $Y$.

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For X a based smooth manifold, the category of finite covers over X is equivalent to the category of actions of the fundamental group of X on based finite sets:

                       \pi-sets   ===     et/X

The same statement for number fields essentially describes the Galois theory. Now the idea that those should be somehow unified was one of the reasons in the development of abstract schemes, a very fruitful topic that is studied in the amazing area of mathematics called the abstract algebraic geometry. Also, note that "actions on sets" is very close to "representations on vector spaces" and this moves us in the direction of representation theory.

Now you see, this simple line actually somehow relates number theory and representation theory. How exactly? Well, if I knew, I would write about that, but I'm just starting to learn about those things.

(Of course, one of the specific relations hinted here should be the Langlands conjectures, since we're so close to having L-functions and representations here!)

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  • $\begingroup$ This is a pretty fact, but I would say that -- despite the way it is written! -- it is not an "equation, formula, identity or inequality". Rather it is a Galois correspondence, or an equivalence of categories. (Brief justification: it's not asserting that any two particular objects are equal. The content is more functorial than that.) $\endgroup$ Commented Aug 22, 2010 at 9:06
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$ D_A\star F = 0 $

Yang-Mills

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$\prod_{n=1}^{\infty} (1-x^n) = \sum_{k=-\infty}^{\infty} (-1)^k x^{k(3k-1)/2}$

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