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

$\begingroup$ There are numerous significant formula and laws from our mathematical legacy. It should be noticed while we are now enjoying them freely, there are still quite a lot of basic problems that we are not able to solve using the techniques we know. $\endgroup$– SunniMar 4, 2010 at 1:20

2$\begingroup$ Voting to close. People are at the repeatingotherpeople'sanswers stage now. $\endgroup$– Qiaochu YuanAug 21, 2010 at 18:23

3$\begingroup$ 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! $\endgroup$– Pete L. ClarkAug 22, 2010 at 9:02

1$\begingroup$ Sadly my favourite $\sum \frac{1}{n^2 +a^2} = \frac{\pi}{a} cth(\pi a)$ wasn't listed $\endgroup$– Ostap ChervakMay 1, 2011 at 16:57

$\begingroup$ $$Rf_*R\hbox{Hom}(F,f^!G)\approx R\hbox{Hom}(Rf_!F,G)$$ $\endgroup$– Steven LandsburgOct 22, 2016 at 22:01
62 Answers
$e^{\pi i} + 1 = 0$

24$\begingroup$ 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. $\endgroup$ Oct 28, 2009 at 22:45

22$\begingroup$ More interesting and useful (and less mysterious) is: exp(i t) = cos(t) + i sin(t). $\endgroup$ Oct 29, 2009 at 14:49

10$\begingroup$ @Gerald: I certainly agree. But I think this particular equation is so pretty I once had it henna tattooed 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. $\endgroup$ Oct 29, 2009 at 15:29

12$\begingroup$ 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. $\endgroup$ Dec 8, 2009 at 18:04

37$\begingroup$ I think the people who ascribe to this formula excessive mystery might not find matrix exponentials particularly less mysterious. $\endgroup$ Mar 4, 2010 at 18:16
Trivial as this is, it has amazed me for decades:
$(1+2+3+...+n)^2=(1^3+2^3+3^3+...+n^3)$

1$\begingroup$ I wouldn't call it trivial. There is also a nice combinatorial proof for it. $\endgroup$ Jun 3, 2010 at 14:12

12$\begingroup$ 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): $\endgroup$ Nov 3, 2010 at 5:34

$\begingroup$ oops, here is the link: cornellmath.wordpress.com/2008/02/15/… $\endgroup$ 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 $2P_3^2=P_7+P_5$. $\endgroup$ Jun 10, 2011 at 12:13

$\begingroup$ Actually, I asked a question about this formula (mathoverflow.net/questions/67117). It turns out that it is only the first one in an infinite list. Thus it is not so beautiful, at least not by itself. $\endgroup$ Dec 17, 2011 at 11:18
$$ \frac{24}{7\sqrt{7}} \int_{\pi/3}^{\pi/2} \log \left \frac{\tan t+\sqrt{7}}{\tan t\sqrt{7}}\right dt\\ = \sum_{n\geq 1} \left(\frac n7\right)\frac{1}{n^2}, $$ 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! 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 9091).

$\begingroup$ @Richard: Could you tell me the name of this conjecture. $\endgroup$– C.S.Jun 20, 2012 at 8:01

1$\begingroup$ @Chandrasekhar: I don't know whether it has a name. $\endgroup$ Jun 20, 2012 at 20:55

13$\begingroup$ 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.May 7, 2019 at 19:19

1$\begingroup$ 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. $\endgroup$– user21349May 8, 2019 at 22:50

1$\begingroup$ The Legendresymbol 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
. $\endgroup$– LSpiceMay 25, 2019 at 23:44
There are many, but here is one.
$d^2=0$

2

$\begingroup$ I'm sorry, but I don't get it. Could someone please give more information about what this equation is all about? $\endgroup$ Oct 27, 2011 at 13:35

6
There's lots to choose from. RiemannRoch and various other formulas from cohomology are pretty neat. But I think I'll go with
$$\sum\limits_{n=1}^{\infty} n^{s} = \prod\limits_{p \text{ prime}} \left( 1  p^{s}\right)^{1}$$
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.

4$\begingroup$ I honestly find the infinite product expression of this formula to be more intriguing  this now relates pi to just the primes. $\endgroup$ Dec 18, 2009 at 18:47

$\begingroup$ pi also appears in other values of $\zeta(2n)$, for positive integers $n$. $\endgroup$ Mar 4, 2010 at 18:05

2$\begingroup$ @ Regenbogen: It appears in ALL values of $\zeta(2n)$. $\endgroup$ Jul 8, 2010 at 15:32

4$\begingroup$ 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 $x_k=(k/2^n)(\pi/2)$, $k=1,\ldots,2^n1$. This gives $S_n  (2^n1) < (4^{n+1}/\pi^2) \sum_{k=1}^{2^n1} k^{2} < S_n$, where $S_n=\sum_k 1/\sin^2 x_k$ satisfies $S_1=2$ and $S_n = 2+4 S_{n1}$ (since $1/\sin^2 x+1/\sin^2(\pi/2x) = 4/\sin^2 2x$), so that $S_n=2(4^n1)/3$. Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. Voilà! $\endgroup$ 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). $\endgroup$ Jul 15, 2010 at 14:10

9$\begingroup$ Ramanujan got rejected by many famous mathematicians (looking for a scholarship outside of India) because they did not understand this.. $\endgroup$– BlueRajaJul 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? $\endgroup$ 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" $\endgroup$– BlueRajaMay 26, 2020 at 22:46
$$\frac{1}{1z} = (1+z)(1+z^2)(1+z^4)(1+z^8)...$$
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.

4$\begingroup$ That's simple, but quite neat too. Haven't thought about that expansion before. Does it have any applications? $\endgroup$ Oct 29, 2009 at 23:25

2$\begingroup$ 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. $\endgroup$ Oct 30, 2009 at 0:15

18$\begingroup$ multiplying the RHS by $(1z)$ starts an impressive chain reaction $\endgroup$ May 25, 2011 at 6:33

$\begingroup$ @MichaelLugo Sorry, undergrad here. Could you please explain how this says that binary expansions are unique? $\endgroup$– OviMay 7, 2019 at 22:51

$\begingroup$ @Ovi The coefficient of $z^n$ in the Taylor series of $\frac1{1z}$ 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 $2$ $\endgroup$– wladMay 8, 2019 at 12:10
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.

$\begingroup$ So what is its combinatorial meaning, anyway? $\endgroup$ Oct 29, 2009 at 4:17

8$\begingroup$ If A is the adjacency matrix of a finite graph G, then the coefficient of t^k counts the number of nonnegative 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. $\endgroup$ Oct 29, 2009 at 4:24

3

5$\begingroup$ I should also note that as a special case of the Euler product one gets the cyclotomic identity: en.wikipedia.org/wiki/Cyclotomic_identity $\endgroup$ 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. $\endgroup$ May 14, 2010 at 21:05
$V  E + F = 2$
Euler's characteristic for connected planar graphs.
$196884 = 196883 + 1$

2

2$\begingroup$ 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. $\endgroup$ May 8, 2019 at 14:22
For a triangle with angles a, b, c $$\tan a + \tan b + \tan c = (\tan a) (\tan b) (\tan c)$$

2$\begingroup$ 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! $\endgroup$– kakazFeb 9, 2010 at 8:44
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.
http://en.wikipedia.org/wiki/Hodge_duality
http://en.wikipedia.org/wiki/Poincar%C3%A9_duality
But the proof is a single line:
$$ \left( \begin{array}{cc} A & B \\\ 0 & I \end{array} \right) \left( \begin{array}{cc} A^t & C^t \\\ B^t & D^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\\ B^t & D^t \end{array} \right). $$


$\begingroup$ 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. $\endgroup$ Apr 27, 2010 at 2:28
It's too hard to pick just one formula, so here's another: the CauchySchwarz 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 CauchySchwarz, are actually just generalizations of the 1dimensional inequality:
x >= 0, with equality iff x = 0,
or rather, instantiations of it in the 2^{nd}, 3^{rd}, etc. exterior powers of the vector space.
I always thought this one was really funny: $1 = 0!$

$\begingroup$ Yeah, so do I. It is also convention that zero product is $1$. $\endgroup$– SunniMar 4, 2010 at 1:15

3$\begingroup$ True, but that's not what's going on here. One can define 0! by extending the rule n!=n*(n1)! 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. $\endgroup$ Mar 4, 2010 at 18:23

6$\begingroup$ Or, 0! is the number of bijective maps on a set with 0 elements, since the empty set is the only such map. $\endgroup$ Mar 19, 2010 at 3:52

5$\begingroup$ 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?) $\endgroup$– BurhanMay 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 ;) $\endgroup$ Jul 15, 2010 at 10:39
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.
$2^n>n $

$\begingroup$ What is so special about that? It's pretty easy to prove. $\endgroup$ Jan 10, 2010 at 18:14

6$\begingroup$ 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. $\endgroup$ Jan 10, 2010 at 18:40

$\begingroup$ Perhaps I should have written $2^X>X$ to better indicate what I had in mind. $\endgroup$ Jan 10, 2010 at 18:43

$\begingroup$ Or $2^\kappa>\kappa$. But I like it in the way that you stated it. $\endgroup$ Jan 15, 2010 at 0:06
It has to be the ergodic theorem, $$\frac{1}{n}\sum_{k=0}^{n1}f(T^kx) \to \int f\:d\mu,\;\;\mu\text{a.e.}\;x,$$ the central principle which holds together pretty much my entire research existence.
It may be trivial, but I've always found
$\sqrt{\pi}=\int_{\infty}^{\infty}e^{x^{2}}dx$
to be particularly beautiful.

3$\begingroup$ 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:
$({\frac{1}{2}})!=\sqrt{\pi}$
. $\endgroup$ Aug 21, 2010 at 18:55
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 realvalued function. It shows how complex analysis actually underlies even the real numbers.

$\begingroup$ This is tantalizingly close to giving a variant of $e$ as a period. $\endgroup$– Matt F.May 7, 2019 at 15:31
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 rightangled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.
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:
\pisets === 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 Lfunctions and representations here!)

$\begingroup$ 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.) $\endgroup$ Aug 22, 2010 at 9:06
$\prod_{n=1}^{\infty} (1x^n) = \sum_{k=\infty}^{\infty} (1)^k x^{k(3k1)/2}$
$\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (1)^{\frac{p1}{2} \frac{q1}{2}}$.
RiemannRoch, and its generalizations:
GrothendieckHirzebruchRiemannRoch
AtiyahSinger (which is also a generalization of GaussBonnet)
Is it cheating to put all of these in a single answer? :)
My favorite is the KoikeNortonZagier product identity for the jfunction (which classifies complex elliptic curves):
j(p)  j(q) = p^{1} \prod_{m>0,n>1} (1p^{m}q^{n})^{c(mn)},
where j(q)744 = \sum_{n >2} c(n) q^{n} = q^{1} + 196884q + 21493760q^{2} + ... 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.