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

The Euler-Lagrange equations, $$\frac{\partial L}{\partial q_j} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_j}$$

Answer 2

The Pythagorean Theorem for Right-Corner 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 "leg-faces" and $D$ is the area of the "hypotenuse-face".

For right-corner simplices in higher Euclidean dimensions, we have that the sum of the squares of the content of leg-simplices equals the square of the content of the hypotenuse-simplex. (I don't happen to know the non-Euclidean 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 diagonal-of-a-box/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.

Answer 3

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 4

$\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 5

Cauchy integral formula $$ f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z} dw $$

Answer 6

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.

Answer 7

$(A-\lambda _1) (A-\lambda _2) \ldots = 0$, the Cayley-Hamilton theorem.

Answer 8

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 9

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 projection-valued measure supported on $\sigma (T)$.

Answer 10

$\sin^2 A + \cos^2 A = 1$

Answer 11

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).

Answer 12

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.

Answer 13

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

Answer 14

Bayes equations: $${\mathbb P}(A|B) = {\mathbb P}(A∩B)/{\mathbb P}(B).$$ It is the basis of conditional probability.

Answer 15

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 $3d-1$ 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!

Answer 16

I think this fits the original question's request for something nice-looking: $\binom{2n}{n}=(-4)^n\binom{-1/2}{n}$

Answer 17

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

$$\det(\beta_{i+j})_{i,j=0,\ldots,k-1} = \frac{1}{k!} \int \cdots \int \Delta(x_1,\ldots,x_k)^2 d\mu(x_1) \cdots d\mu(x_k)$$

where $\Delta$ is the Vandermonde determinant.

Answer 18

I'm a fan of $\Omega SU \simeq BU$.

Answer 19

Lately, I really like the Greenlees-May 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.

Answer 20

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.)

Answer 21

How about $\displaystyle \sigma_7(n)=\sigma_3(n)+120\sum_{k=1}^{n-1} \sigma_3(k) \sigma_3(n-k)$? This is an utterly shocking result, and the only known proof uses complex analysis.

Answer 22

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: $5 = (2 + i)(2 - i)$.

Answer 23

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.

Answer 24

The isogeny theorem: $$\mathrm{Hom}_K(A,A') = \mathrm{Hom}_{G_K}(T_\ell(A),T_\ell(A')).$$

Answer 25

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, 53-55 (2007). ZBL1147.26011, MR2347705.

Answer 26

$\sum_{i=1}^m \sum_{j=1}^n a_{ij} = \sum_{j=1}^n \sum_{i=1}^m a_{ij}$

Answer 27

$$e=\lim_{n\to\infty}\sqrt[p_n]{\prod_{k=1}^np_n}$$ as seen at Gaussianos

$(\prod_{k=1}^np_n=p_n$# which is the primorial of the nth prime number $p_n)$

Answer 28

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.

Answer 29

I'm surprised that nobody said

$e=mc^2$

Answer 30

I learned Quantum Mechanics and Linear Algebra in tandem, so Schrodinger's linear time-independent 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.

$H\psi=E\psi$