Question

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?

Answer 1

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

Answer 2

Stokes' Theorem

Answer 3

There are many, but here is one.

$d^2=0$

Answer 4

There's lots to choose from. Riemann-Roch and various other formulas from cohomology are pretty neat. But I think I'll go with

Σ_n=1^∞ n^-s = Π_{p prime} (1-p^-s)^-1.

Answer 5

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

Once suitably regularised of course :-)

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

$V - E + F = 2$

Euler's characteristic for connected planar graphs.

Answer 10

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

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

Answer 11

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.

Answer 12

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

Answer 13

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 2^nd, 3^rd, etc. exterior powers of the vector space.

Answer 14

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.

Answer 15

For a triangle with angles a, b, c $$\tan a + \tan b + \tan c = (\tan a) (\tan b) (\tan c)$$

Answer 16

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? :-)

Answer 17

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

Answer 18

I just cannot get this thing to make the 2 by 2 matrices of letters I want. Wait, fixed it myself. There is a thread in Meta about Latex/jsMath inconsistencies, one known problem is backslash being interpreted as an escape. So where I intended double backslash I just put three backslashes and that works for now. If it fails later I will switch to four or five backslashes.

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:

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

Answer 19

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!)

Answer 20

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.

Answer 21

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

Answer 22

$2^n>n $

Answer 23

$ D_A\star F = 0 $

Yang-Mills

Answer 24

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

j(p) - j(q) = p^-1 \prod_m>0,n>-1 (1-p^mqⁿ)^c(mn),

where j(q)-744 = \sum_{n >-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.

Answer 25

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

X_i+1 = 2 * X_i - X_i * A * X_i

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

Answer 26

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 (1870/1882), Linear Associative Algebra, § 15, p. 5.

Answer 27

$196884 = 196883 + 1$

Answer 28

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.

Answer 29

$\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 $

Answer 30

Bayes equations:

P(A|B) = P(A∩B)/P(B)

It is the basis of conditional probability.