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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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closed as no longer relevant by Robin Chapman, Akhil Mathew, Yemon Choi, Qiaochu Yuan, Pete L. Clark Aug 22 '10 at 9:00

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Voting to close. People are at the repeating-other-people's-answers stage now. – Qiaochu Yuan Aug 21 '10 at 18:23
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! – Pete L. Clark Aug 22 '10 at 9:02
Sadly my favourite $\sum \frac{1}{n^2 +a^2} = \frac{\pi}{a} cth(\pi a)$ wasn't listed – Ostap Chervak May 1 '11 at 16:57

63 Answers 63

up vote 35 down vote accepted

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

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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. – Sam Derbyshire Oct 28 '09 at 22:45
More interesting and useful (and less mysterious) is: exp(i t) = cos(t) + i sin(t). – Gerald Edgar Oct 29 '09 at 14:49
@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. – Sonia Balagopalan Oct 29 '09 at 15:29
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. – Qiaochu Yuan Dec 8 '09 at 18:04
I think the people who ascribe to this formula excessive mystery might not find matrix exponentials particularly less mysterious. – Cam McLeman Mar 4 '10 at 18:16

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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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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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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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.

alt text

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Riemann-Roch, and its generalizations:



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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$$\frac{1}{1-z} = (1+z)(1+z^2)(1+z^4)(1+z^8)...$$

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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That's simple, but quite neat too. Haven't thought about that expansion before. Does it have any applications? – George Lowther Oct 29 '09 at 23:25
multiplying the RHS by $(1-z)$ starts an impressive chain reaction – Pietro Majer May 25 '11 at 6:33

E[X+Y]=E[X]+E[Y] for any 2 random varibles X and Y

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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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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. – Qiaochu Yuan Oct 29 '09 at 4:24
That's awesome! – Sonia Balagopalan Oct 29 '09 at 10:47
I should also note that as a special case of the Euler product one gets the cyclotomic identity: – Qiaochu Yuan Oct 29 '09 at 14:40

$\pi = 2 \times 1/\sqrt(1/2) \times 1/\sqrt((1+\sqrt(1/2))/2) \times 1/\sqrt((1+\sqrt((1+\sqrt(1/2))/2))/2) \times \ldots $

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Var[X+Y]=Var[X]+Var[Y] for any two independent random variables X and Y, which is the statistics equivalent of the Pythagorean Theorem.

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1+2+3+4+5+... = -1/12

Once suitably regularised of course :-)

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Ramanujan got rejected by many famous mathematicians (looking for a scholarship outside of India) because they did not understand this.. – BlueRaja Jul 15 '10 at 5:26

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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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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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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I honestly find the infinite product expression of this formula to be more intriguing - this now relates pi to just the primes. – Nick Salter Dec 18 '09 at 18:47
@ Regenbogen: It appears in ALL values of $\zeta(2n)$. – Max Muller Jul 8 '10 at 15:32
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! – Hans Lundmark Jul 15 '10 at 14:08

There are many, but here is one.


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que?``````````` – BlueRaja Jul 15 '10 at 16:24

The Newton iteration for finding the inverse, X, of a matrix A:

Xi+1 = 2 * Xi - Xi * A * Xi

Completely impractical and yet so beautiful. The first time I saw a Newton iteration working I thought it was "magical".

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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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Addendum to $e^{i \pi}$

Benjamin Peirce apparently liked this mathematical synonym for the additive inverse of $1$ so much that he introduced three special symbols for $e, i, \pi$ — ones that enable $e^{i \pi}$ to be written in a single cursive ligature, like so:

Benjamin Peirce's script for e^(i pi)

Benjamin Peirce, LAA, § 15, p. 5.

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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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polynomially convex hull of K = plurisubharmonic hull of K , where K is compact subset of C^n. For n>1, the equality is very interesting.

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$V - E + F = 2$

Euler's characteristic for connected planar graphs.

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$2^n>n $

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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. – Jonas Meyer Jan 10 '10 at 18:40

The isogeny theorem: $\mathrm{Hom}_K(A,A')$

$ = \mathrm{Hom}_{G_K}(T_\ell(A),T_\ell(A'))$

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

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$ D_A\star F = 0 $


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I always thought this one was really funny: $1 = 0!$

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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. – Cam McLeman Mar 4 '10 at 18:23
Or, 0! is the number of bijective maps on a set with 0 elements, since the empty set is the only such map. – Jonas Meyer Mar 19 '10 at 3:52
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?) – Burhan May 21 '10 at 18:25

I'm surprised that nobody said


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Perhaps because is a formula from physics. – Sunni Apr 25 '10 at 18:22
I love the "Einstein meets Pythagoras" version: $E=m(a^2+b^2)$ – Federico Poloni Aug 20 '10 at 20:46

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


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I wouldn't call it trivial. There is also a nice combinatorial proof for it. – Victor Protsak Jun 3 '10 at 14:12
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): – Peter Luthy Nov 3 '10 at 5:34

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