What's your favorite equation, formula, identity or inequality? [closed]

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 $$\int_M\mathrm{d}\omega=\oint_{\partial M}\omega.$$

Answer 3

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

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

Answer 4

There are many, but here is one.

$d^2=0$

Answer 5

$$ \frac{24}{7\sqrt{7}} \int_{\pi/3}^{\pi/2} \log \left| \frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}\right| dt\\ = \sum_{n\geq 1} \left(\frac n7\right)\frac{1}{n^2}, $$ 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).

Answer 6

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 7

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

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

Answer 8

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

Answer 9

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

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.

Answer 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.

Answer 11

$V - E + F = 2$

Euler's characteristic for connected planar graphs.

Answer 12

$196884 = 196883 + 1$

Answer 13

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

Answer 14

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:

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

Answer 15

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

Answer 16

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 17

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.

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

Answer 18

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

Answer 19

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.

Answer 20

It may be trivial, but I've always found

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

to be particularly beautiful.

Answer 21

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 22

$2^n>n $

Answer 23

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

Answer 24

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 25

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 26

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 27

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

Answer 28

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 29

$ D_A\star F = 0 $

Yang-Mills

Answer 30

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