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?

2$\begingroup$ Voting to close. People are at the repeatingotherpeople'sanswers stage now. $\endgroup$– Qiaochu YuanCommented Aug 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. ClarkCommented Aug 22, 2010 at 9:02

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

1$\begingroup$ $$Rf_*R\hbox{Hom}(F,f^!G)\approx R\hbox{Hom}(Rf_!F,G)$$ $\endgroup$– Steven LandsburgCommented Oct 22, 2016 at 22:01

1$\begingroup$ Mine is Poisson formula which can take the form $\int_H f = \int_{H^{\perp}} \hat f$. $\endgroup$– WatsonCommented Dec 21, 2019 at 9:36
62 Answers
The EulerLagrange equations, $$\frac{\partial L}{\partial q_j} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}$$
The Pythagorean Theorem for RightCorner Tetrahedra[*]:
Euclidean: $A^2 + B^2 + C^2 = D^2$
Hyperbolic: $\cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} \;  \; \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} = \cos\frac{D}{2}$
Spherical: $\cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} \; + \; \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} = \cos\frac{D}{2}$
where $A$, $B$, $C$ are the areas of the "legfaces" and $D$ is the area of the "hypotenuseface".
For rightcorner simplices in higher Euclidean dimensions, we have that the sum of the squares of the content of legsimplices equals the square of the content of the hypotenusesimplex. (I don't happen to know the nonEuclidean counterparts of this generalization. Perhaps this makes for a good MO question!)
As generalizations of the Pythagorean Theorem for Triangles, I always found these (Euclidean) results to be more satisfying than the diagonalofabox/distance formulas: instead of dealing only with segments, we have that, as the dimension of the ambient space goes up, so does the dimension of the objects involved in the relations.
[*] Edges meeting at the "right corner" are mutually orthogonal.

$\begingroup$ Yes, this one deserves to be better known. I like the Euclidean one since it looks so bizarre at first: "Area squared, what the heck is that?!?". :) $\endgroup$ Commented Jul 15, 2010 at 20:40

1$\begingroup$ Even better: the higherdimensional Pythagorean ones are all a corollary of the CauchyBinet formula (en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula), which has a combinatorial proof. $\endgroup$ Commented Jul 29, 2010 at 18:04
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.
$\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 $
Cauchy integral formula $$ f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{wz} dw $$
My favorite equation is
$$\frac{16}{64} = \frac{1}{4}.$$
What makes this equation interesting is that canceling the $6$'s yields the correct answer. I realized this in, perhaps, third grade. This was the great rebellion of my youth. Sometime later I generalized this to finding solutions to
$$\frac{pa +b}{pb + c} = \frac{a}{c}.$$
where $p$ is an integer greater than $1$. We require that $a$, $b$, and $c$ are integers between $1$ and $p  1$, inclusive. Say a solution is trivial if $a = b = c$. Then $p$ is prime if and only if all solutions are trivial. On can also prove that if $p$ is an even integer greater than $2$ then $p  1$ is prime if and only if every nontrivial solution $(a,b,c)$ has $b = p  1$.
The key to these results is that if $(a, b, c)$ is a nontrivial solution then the greatest common divisor of $c$ and $p$ is greater than $1$ and the greatest common divisor of $b$ and $p  1$ is also greater than $1$.
Two other interesting facts are (i) if $(a, b, c)$ is a nontrivial solution then $2a \leq c < b$ and (2) the number of nontrivial solutions is odd if and only if $p$ is the square of an even integer. To prove the latter item it is useful to note that if $(a, b, c)$ is a nontrivial solution then so is $(b  c, b, b  a)$.
For what it is worth I call this demented division.
$(A\lambda _1) (A\lambda _2) \ldots = 0$, the CayleyHamilton theorem.

3$\begingroup$ Surely something like $f_A(A)=0$ trumps that visually. $\endgroup$ Commented Aug 22, 2010 at 1:45
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".

$\begingroup$ The name for this is "Schulz's iteration". $\endgroup$ Commented Aug 20, 2010 at 15:42
The Spectral theorem for normal operators on a Hilbert space:
$T = \int_{\sigma (T)} \lambda dP(\lambda)$
where $\sigma (T)$ is the spectrum of $T$ and $P$ is a regular projectionvalued measure supported on $\sigma (T)$.
Pick's theorem $A = I + \frac 1 2 B  1$, where $A$, $I$, and $B$ are the area, number of interior integer points, and number of boundary integer points, respectively, of a polygon with vertices on the integer lattice. Picks identity is fascinating because it computes a continuous quantity completely discretely. (Of course, this is not quite correct, since we have quite a discrete requirement about the vertices of the polygon.) Also, the "1" is not an accident, but the Euler characteristic of the polygon (and so there are various natural extensions of Pick's theorem).

$\begingroup$ That reminds me, I remember seeing a book with title something like "Computing the continuous discretely". I can't remember the authors; do you know them by any chance? :) $\endgroup$ Commented Aug 21, 2010 at 6:52

1$\begingroup$ My favorite way to think of $I+{\frac{1}{2}}B1$ is as the number of lattice points in the polygon, with the "reasonable" convention for points on the boundary: A point in the interior of an edge is half in the polygon, and a vertex with interior angle $\alpha$ is $\alpha/(2\pi)$ in the polygon. $\endgroup$ Commented Aug 21, 2010 at 18:52

$\begingroup$ @ Robin: I'm not responsible for the title... :) But there is certainly a connection here. $\endgroup$ Commented Aug 22, 2010 at 6:05
One that I just learned recently is $$ (1 + q + q^3 + q^6 + q^{10} + q^{15} + \cdots)^4 = \sum_{k=0}^\infty \sigma(2k+1)q^k $$ which states that the number of ways of writing an integer $k$ as a sum of exactly 4 triangular numbers (paying attention to ordering) is equal to the sum of divisors of $2k+1$.
If that isn't cool and surprising, I don't know what is.
$$Var[X+Y]=Var[X]+Var[Y]$$ for any two independent random variables $X$ and $Y$, which is the statistics equivalent of the Pythagorean Theorem.
Bayes equations: $${\mathbb P}(AB) = {\mathbb P}(A∩B)/{\mathbb P}(B).$$ It is the basis of conditional probability.
Well, of course my favorite is Stokes theorem (it used to be the background of my mobile in the old days where you still manually designed monochromatic backgrounds pixel by pixel), but that is already suggested. And so are many others. So I'll go for Kontsevich formula for the number $N_d$ of rational curves through $3d1$ generic points in the plane:
$N_d + \sum_{\stackrel{d_A, d_B \geq 1}{d_A + d_B = d}} \binom{3d  4}{3 d_A  1} N_{d_A} N_{d_B} d_A^3 d_B = \sum_{\stackrel{d_A, d_B \geq 1}{d_A + d_B = d}} \binom{3d  4}{3 d_A  2} N_{d_A} N_{d_B} d_A^2 d_B^2$
Although I admit this looks ugly until you see the proof. Then it becomes so neat!
I think this fits the original question's request for something nicelooking: $\binom{2n}{n}=(4)^n\binom{1/2}{n}$


$\begingroup$ It's a simple verification. If you really think you need to cite it, you can probably find a reference in some combinatorics book somewhere, it's one of those things that comes up now and again... $\endgroup$ Commented Dec 23, 2014 at 18:45

$\begingroup$ I have not been introduced to negative binomials formally. This seems fairly advanced. Does it have algebraic, arithmetic, combinatorial or geometric meaning? Do you have a reference that could say something substantial about this interesting formulation? $\endgroup$– TurboCommented Dec 23, 2014 at 20:36
I have a soft spot for Heine's formula from the theory of orthogonal polynomials (since the proof is such a pretty calculation):
If $\mu$ is a measure with finite moments $\beta_k=\int x^k d\mu(x)$, then
$$\det(\beta_{i+j})_{i,j=0,\ldots,k1} = \frac{1}{k!} \int \cdots \int \Delta(x_1,\ldots,x_k)^2 d\mu(x_1) \cdots d\mu(x_k)$$
where $\Delta$ is the Vandermonde determinant.

$\begingroup$ Is there a multivariate version of this? i.e. where the x_i are vectors in R^n. $\endgroup$ Commented Jul 29, 2010 at 18:54

Lately, I really like the GreenleesMay duality: $RHom_A(R\Gamma_{\mathfrak{a}}M,N) \cong RHom_A(M,L\Lambda_{\mathfrak{a}}N)$ which holds for any pair of complexes over a noetherian ring.

1$\begingroup$ There are really nice formulas in derived categories! $\endgroup$ Commented Jul 12, 2016 at 21:33
The Gauss Formula from Riemannian geometry:
$\overline{\nabla}_XY = \nabla_XY + \text{II}(X,Y)$
It may just be a decomposition into tangential and normal parts, but I find it very aesthetically pleasing. (It's also not completely immediate that the tangential part of the ambient connection should actually be the intrinsic connection.)
How about $\displaystyle \sigma_7(n)=\sigma_3(n)+120\sum_{k=1}^{n1} \sigma_3(k) \sigma_3(nk)$? This is an utterly shocking result, and the only known proof uses complex analysis.

$\begingroup$ This result drops out easily from a couple of identities for Ramanujan's tau function that I published in my 1993 thesis: \tau(n) = 60 \sum_{k=0}^n (n3k)(2n3k)s_3(k)s_3(nk) and \tau(n)=n^2s_7(n) – 540 \sum_{k=1}^{n1} k(nk)s_3(k)s_3(nk), where s_3(0)=1/240 (for s read \sigma). What proof are you referring to? Thanks. $\endgroup$ Commented Aug 11, 2010 at 18:04

$\begingroup$ I just found the result you gave on wikipedia and it is derived using Eisenstein series, which is how I obtained my identities for the tau function, so I believe my proof of this result is pretty similar to what you had in mind. $\endgroup$ Commented Aug 11, 2010 at 18:19

$\begingroup$ It follows from the fact that $E_4^2 = E_8$, which follows from the dimension formulas for spaces of modular forms of a given weight. $\endgroup$ Commented Aug 11, 2010 at 23:04

2$\begingroup$ No, see Skoruppa, NilsPeter, A quick combinatorial proof of Eisenstein series identities., J. Number Theory 43 (1993), no. 1, 6873 for a nonanalytic proof. $\endgroup$ Commented Aug 21, 2010 at 7:47
The braid relation is probably my favorite equation, algebraically capturing the Reidemeister III move as $x y x = y x y$. Although to a younger person, I still find that suggesting that 5 is not prime is reliably charming revelation: $5 = (2 + i)(2  i)$.
There are many beautiful equations above, so I'll be a bit different and add something nonsensical. Namely
$$\langle f\rangle = \frac{\int_\ast f(\phi)e^{\frac{\mathrm{i}}{\hbar}\int_M\mathcal{L}(\phi)}\mathcal{D}\phi}{\int_\ast e^{\frac{\mathrm{i}}{\hbar}\int_M\mathcal{L}(\phi)}\mathcal{D}\phi}.$$
Just insert your favourite spacetime manifold $M$ and the classical Lagrangian $\mathcal{L}$ of your choice, and you get to learn the expectation value of any physical observable $f$... as soon as you figure out what the hell $\ast$ and $\mathcal{D}\phi$ are, that is.

2$\begingroup$ And after you reconcile the fact that the denominator is probably $\infty$. $\endgroup$ Commented Aug 20, 2010 at 15:26

1$\begingroup$ Ah yes, but after THAT it's great :) $\endgroup$– gsprCommented Aug 21, 2010 at 9:42
The isogeny theorem: $$\mathrm{Hom}_K(A,A') = \mathrm{Hom}_{G_K}(T_\ell(A),T_\ell(A')).$$
Ky Fan's inequality seems rather beautiful. The most beautiful proof can be found here
Jovanović, Milan V.; Pogány, Tibor K.; Sándor, József, Notes on certain inequalities by Hölder, Lewent and Ky Fan, J. Math. Inequal. 1, No. 1, 5355 (2007). ZBL1147.26011, MR2347705.
$\sum_{i=1}^m \sum_{j=1}^n a_{ij} = \sum_{j=1}^n \sum_{i=1}^m a_{ij}$
$$e=\lim_{n\to\infty}\sqrt[p_n]{\prod_{k=1}^np_n}$$ as seen at Gaussianos
$(\prod_{k=1}^np_n=p_n$# which is the primorial of the nth prime number $p_n)$
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.
I'm surprised that nobody said
$e=mc^2$

20$\begingroup$ Perhaps because is a formula from physics. $\endgroup$– SunniCommented Apr 25, 2010 at 18:22

19$\begingroup$ I love the "Einstein meets Pythagoras" version: $E=m(a^2+b^2)$ $\endgroup$ Commented Aug 20, 2010 at 20:46
I learned Quantum Mechanics and Linear Algebra in tandem, so Schrodinger's linear timeindependent equation has always had a special place in my heart. It shows that eigenvalues and eigenvectors are fundamental to our description of atomic physics. Also treating observables as operators was a great conceptual revolution.
$H\psi=E\psi$

$\begingroup$ I first saw this equation in the second semester of physical chemistry over a decade ago.That course got me considering a change of career from biochemistry to mathematics. $\endgroup$ Commented Jun 2, 2010 at 19:16