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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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Voting to close. People are at the repeating-other-people's-answers stage now. –  Qiaochu Yuan Aug 21 '10 at 18:23
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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
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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

up vote 29 down vote accepted

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

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I never liked writing it in the form e^(i pi) + 1 = 0, it's not a way I'd ever write anything. I think it looks a lot nicer when written e^(i pi) = -1. –  Sam Derbyshire Oct 28 '09 at 22:45
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More interesting and useful (and less mysterious) is: exp(i t) = cos(t) + i sin(t). –  Gerald Edgar Oct 29 '09 at 14:49
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@Gerald: I certainly agree. But I think this particular equation is so pretty I once had it henna tattoo-ed on my hand. @Sam: It's just a matter of eye, beholder etc. I don't like minus signs, but do like 0 on the RHS. –  Sonia Balagopalan Oct 29 '09 at 15:29
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I think people ascribe to this formula excessive mystery. The mystery disappears if one replaces the usual exponential with the matrix exponential on M_2(R) and then this is just the statement that the only trajectory of a particle whose velocity is always perpendicular to its displacement (and in the same proportion) is a circle. –  Qiaochu Yuan Dec 8 '09 at 18:04
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I think the people who ascribe to this formula excessive mystery might not find matrix exponentials particularly less mysterious. –  Cam McLeman Mar 4 '10 at 18:16

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:

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

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

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

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

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

as seen at Gaussianos

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

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It may be trivial, but I've always found

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

to be particularly beautiful.

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Once one decides that the "right" definition of $n!$ when $n$ isn't a natural number is $\Gamma(n+1)$, this formula becomes equivalent to one of my favorites: $(-{\frac{1}{2}})!=\sqrt{\pi}$. –  Andreas Blass Aug 21 '10 at 18:55

$(A-\lambda _1) (A-\lambda _2) \ldots = 0$, 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

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

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I like Riemann-Roch the most!!!

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

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With the stuff I've seen in the literature of sequence transformations, I've started to love the formulae for Aitken's Δ² process:

$S_n^{\prime}=S_{n+1}-\frac{(\Delta S_n)^2}{\Delta^2 S_n}$

and its generalization the Wynn ε algorithm:

$\varepsilon_{k+1}^{(n)}=\varepsilon_{k-1}^{(n+1)}+\frac1{\varepsilon_{k}^{(n+1)}-\varepsilon_{k}^{(n)}}$

for the latter one especially because it is nicely represented as a lozenge diagram:

Wynn epsilon

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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

d/dx (e^x) =e^x

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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.

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

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

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

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

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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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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

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

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$196884 = 196883 + 1$

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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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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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Bayes equations:

P(A|B) = P(A∩B)/P(B)

It is the basis of conditional probability.

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