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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

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There are numerous significant formula and laws from our mathematical legacy. It should be noticed while we are now enjoying them freely, there are still quite a lot of basic problems that we are not able to solve using the techniques we know. – Sunni Mar 4 '10 at 1:20
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

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

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Surely something like $f_A(A)=0$ trumps that visually. – Cam McLeman Aug 22 '10 at 1:45

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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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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The name for this is "Schulz's iteration". – J. M. Aug 20 '10 at 15:42

$\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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Bayes equations:

P(A|B) = P(A∩B)/P(B)

It is the basis of conditional probability.

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

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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).

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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? :-) – Robin Chapman Aug 21 '10 at 6:52
My favorite way to think of $I+{\frac{1}{2}}B-1$ 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. – Andreas Blass Aug 21 '10 at 18:52
@ Robin: I'm not responsible for the title... :) But there is certainly a connection here. – matthias beck Aug 22 '10 at 6:05

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!

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

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Is there good reference? – Turbo Dec 23 '14 at 11:37
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... – Harry Altman Dec 23 '14 at 18:45
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? – Turbo Dec 23 '14 at 20:36

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

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

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

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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.

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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.

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There are really nice formulas in derived categories! – Leo Alonso Jul 12 at 21:33

Ky Fan's inequality seems rather beautiful. The most beatiful proof can be found here

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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 $\sigma (T)$.

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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.

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Is there a multivariate version of this? i.e. where the x_i are vectors in R^n. – Abdelmalek Abdesselam Jul 29 '10 at 18:54
None that I know of... – Hans Lundmark Aug 1 '10 at 9:36

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.

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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 (n-3k)(2n-3k)s_3(k)s_3(n-k) and \tau(n)=n^2s_7(n) – 540 \sum_{k=1}^{n-1} k(n-k)s_3(k)s_3(n-k), where s_3(0)=1/240 (for s read \sigma). What proof are you referring to? Thanks. – Derek Jennings Aug 11 '10 at 18:04
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. – Derek Jennings Aug 11 '10 at 18:19
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. – David Corwin Aug 11 '10 at 23:04
No, see Skoruppa, Nils-Peter, A quick combinatorial proof of Eisenstein series identities., J. Number Theory 43 (1993), no. 1, 68--73 for a non-analytic proof. – Robin Chapman Aug 21 '10 at 7:47

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

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

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

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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.

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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?!?". :) – Hans Lundmark Jul 15 '10 at 20:40
Even better: the higher-dimensional Pythagorean ones are all a corollary of the Cauchy-Binet formula (, which has a combinatorial proof. – Qiaochu Yuan Jul 29 '10 at 18:04

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.

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

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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.)

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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.

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And after you reconcile the fact that the denominator is probably $\infty$. – Nate Eldredge Aug 20 '10 at 15:26
Ah yes, but after THAT it's great :-) – gspr Aug 21 '10 at 9:42


as seen at Gaussianos

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

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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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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.


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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. – The Mathemagician Jun 2 '10 at 19:16

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