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What's your favorite equation, formula, identity or inequality? [closed]
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?
soft-question big-list
2 revisions, 2 users
Kim Greene 100%
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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$\begingroup$ 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. $\endgroup$ – Sunni Mar 4 '10 at 1:20
$\begingroup$ Voting to close. People are at the repeating-other-people's-answers stage now. $\endgroup$ – Qiaochu Yuan Aug 21 '10 at 18:23
$\begingroup$ 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. Clark Aug 22 '10 at 9:02
$\begingroup$ Sadly my favourite $\sum \frac{1}{n^2 +a^2} = \frac{\pi}{a} cth(\pi a)$ wasn't listed $\endgroup$ – Ostap Chervak May 1 '11 at 16:57
$\begingroup$ $$Rf_*R\hbox{Hom}(F,f^!G)\approx R\hbox{Hom}(Rf_!F,G)$$ $\endgroup$ – Steven Landsburg Oct 22 '16 at 22:01
$e^{\pi i} + 1 = 0$
Sonia.B 55%
$\begingroup$ 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. $\endgroup$ – Sam Derbyshire Oct 28 '09 at 22:45
$\begingroup$ More interesting and useful (and less mysterious) is: exp(i t) = cos(t) + i sin(t). $\endgroup$ – Gerald Edgar Oct 29 '09 at 14:49
$\begingroup$ @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. $\endgroup$ – Sonia Balagopalan Oct 29 '09 at 15:29
$\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Dec 8 '09 at 18:04
$\begingroup$ I think the people who ascribe to this formula excessive mystery might not find matrix exponentials particularly less mysterious. $\endgroup$ – Cam McLeman Mar 4 '10 at 18:16
Stokes' Theorem
Sam Derbyshire
Trivial as this is, it has amazed me for decades:
$(1+2+3+...+n)^2=(1^3+2^3+3^3+...+n^3)$
Yaakov Baruch
$\begingroup$ I wouldn't call it trivial. There is also a nice combinatorial proof for it. $\endgroup$ – Victor Protsak Jun 3 '10 at 14:12
$\begingroup$ I am a bit late in seeing this post, but I completely agree with you, Yaakov. The equality is a bit related to the following pattern which I discovered as a child but continues to amaze me to this day: $1=1^3$; $3+5=2^3$; $7+9+11=3^3$; $13+15+17+19=4^3$;... Many mathematicians know that the sum of the first n odd numbers is n2, but I think very few are aware of this trivial yet incredible pattern. I actually wrote a little post about this on a math blog (the Everything Seminar): $\endgroup$ – Peter Luthy Nov 3 '10 at 5:34
$\begingroup$ oops, here is the link: cornellmath.wordpress.com/2008/02/15/… $\endgroup$ – Peter Luthy Nov 3 '10 at 5:34
$\begingroup$ Also see mathoverflow.net/questions/67117/… and in particular Vladimir Dotsenko's answer, where other similar relations are listed, including $2P_3^2=P_7+P_5$. $\endgroup$ – Yaakov Baruch Jun 10 '11 at 12:13
$\begingroup$ Actually, I asked a question about this formula (mathoverflow.net/questions/67117). It turns out that it is only the first one in an infinite list. Thus it is not so beautiful, at least not by itself. $\endgroup$ – Denis Serre Dec 17 '11 at 11:18
There are many, but here is one.
$d^2=0$
$\begingroup$ que?``````````` $\endgroup$ – BlueRaja Jul 15 '10 at 16:24
$\begingroup$ I'm sorry, but I don't get it. Could someone please give more information about what this equation is all about? $\endgroup$ – Joel Reyes Noche Oct 27 '11 at 13:35
$\begingroup$ en.wikipedia.org/wiki/Exterior_derivative $\endgroup$ – Robin Saunders Mar 3 '12 at 16:22
$$ \frac{24}{7\sqrt{7}} \int_{\pi/3}^{\pi/2} \log \left| \frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}\right| dt\\ = \sum_{n\geq 1} \left(\frac n7\right)\frac{1}{n^2}, $$ 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).
Richard Stanley
$\begingroup$ @Richard: Could you tell me the name of this conjecture. $\endgroup$ – crskhr Jun 20 '12 at 8:01
$\begingroup$ @Chandrasekhar: I don't know whether it has a name. $\endgroup$ – Richard Stanley Jun 20 '12 at 20:55
$\begingroup$ It may be formulated with integrals only, and quite similar: \begin{align*} 7\int_0^1 \log x\cdot \frac{1+x-x^2+x^3-x^4-x^5}{1-x^7}dx=\\ 12 \int_{\frac{\sqrt{21}-5}2}^1 \frac{\log |x|}{2+3x+2x^2}dx \end{align*} $\endgroup$ – Fedor Petrov Mar 9 '15 at 0:07
$\begingroup$ As mentioned in this question, a proof of this identity can be found in Section 5 of arxiv.org/abs/1005.0414 $\endgroup$ – j.c. May 7 at 19:19
$\begingroup$ For anyone else who was confused by the notation, the notation $(\frac{n}{7})$ is something called a Legendre symbol. It's not $n/7$ in parens. $\endgroup$ – Ben Crowell May 8 at 22:50
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.
Csar Lozano 50%
$\begingroup$ I honestly find the infinite product expression of this formula to be more intriguing - this now relates pi to just the primes. $\endgroup$ – Nick Salter Dec 18 '09 at 18:47
$\begingroup$ pi also appears in other values of $\zeta(2n)$, for positive integers $n$. $\endgroup$ – Regenbogen Mar 4 '10 at 18:05
$\begingroup$ @ Regenbogen: It appears in ALL values of $\zeta(2n)$. $\endgroup$ – Max Muller Jul 8 '10 at 15:32
$\begingroup$ This identity is the subject of one of my favourite proofs: For $0 < x < \pi/2$ we have $0 < \sin x < x < \tan x$, thus $1/\sin^2 x -1 < 1/x^2 < 1/\sin^2 x$. Split the interval $(0,\pi/2)$ into $2^n$ equal parts, and sum the inequality over the interior grid points $x_k=(k/2^n)(\pi/2)$, $k=1,\ldots,2^n-1$. This gives $S_n - (2^n-1) < (4^{n+1}/\pi^2) \sum_{k=1}^{2^n-1} k^{-2} < S_n$, where $S_n=\sum_k 1/\sin^2 x_k$ satisfies $S_1=2$ and $S_n = 2+4 S_{n-1}$ (since $1/\sin^2 x+1/\sin^2(\pi/2-x) = 4/\sin^2 2x$), so that $S_n=2(4^n-1)/3$. Multiply by $\pi^2/4^{n+1}$ and let $n\to\infty$. Voilà! $\endgroup$ – Hans Lundmark Jul 15 '10 at 14:08
$\begingroup$ I should also mention that variants of the proof above can be found in Aigner & Ziegler's "Proofs from the book" and in a paper by Josef Hofbauer (Amer. Math. Monthly, February 2002). $\endgroup$ – Hans Lundmark Jul 15 '10 at 14:10
There's lots to choose from. Riemann-Roch and various other formulas from cohomology are pretty neat. But I think I'll go with
$$\sum\limits_{n=1}^{\infty} n^{-s} = \prod\limits_{p \text{ prime}} \left( 1 - p^{-s}\right)^{-1}$$
George Lowther
1+2+3+4+5+... = -1/12
Once suitably regularised of course :-)
Dan Piponi
$\begingroup$ Ramanujan got rejected by many famous mathematicians (looking for a scholarship outside of India) because they did not understand this.. $\endgroup$ – BlueRaja Jul 15 '10 at 5:26
$\begingroup$ @BlueRaja What do you mean by that? Do you mean that his work, without this regularization, was insufficient for a scholarship? $\endgroup$ – Acccumulation May 7 at 21:35
$V - E + F = 2$
Euler's characteristic for connected planar graphs.
Federico Ramponi
$$\frac{1}{1-z} = (1+z)(1+z^2)(1+z^4)(1+z^8)...$$
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.
Michael Lugo
$\begingroup$ That's simple, but quite neat too. Haven't thought about that expansion before. Does it have any applications? $\endgroup$ – George Lowther Oct 29 '09 at 23:25
$\begingroup$ Well, it says binary expansions are unique, which is kind of a nice fact if you want to make computers that depend on that. As for applications within mathematics itself, I don't really know of any. $\endgroup$ – Michael Lugo Oct 30 '09 at 0:15
$\begingroup$ multiplying the RHS by $(1-z)$ starts an impressive chain reaction $\endgroup$ – Pietro Majer May 25 '11 at 6:33
$\begingroup$ @MichaelLugo Sorry, undergrad here. Could you please explain how this says that binary expansions are unique? $\endgroup$ – Ovi May 7 at 22:51
$\begingroup$ @Ovi The coefficient of $z^n$ in the Taylor series of $\frac1{1-z}$ counts how many subsets of the powers of $2$ sum up to $n$. Picking a term from each factor in Lugo's product picks out a subset of the powers of $2$ $\endgroup$ – man and laptop May 8 at 12:10
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.
$\begingroup$ So what is its combinatorial meaning, anyway? $\endgroup$ – Darsh Ranjan Oct 29 '09 at 4:17
$\begingroup$ If A is the adjacency matrix of a finite graph G, then the coefficient of t^k counts the number of non-negative integer linear combination of aperiodic closed walks on G (without a distinguished vertex) with a total of k vertices. This is an equivalent to an Euler product for the RHS which is again straightforward to prove algebraically and very interesting combinatorially. $\endgroup$ – Qiaochu Yuan Oct 29 '09 at 4:24
$\begingroup$ That's awesome! $\endgroup$ – Sonia Balagopalan Oct 29 '09 at 10:47
$\begingroup$ I should also note that as a special case of the Euler product one gets the cyclotomic identity: en.wikipedia.org/wiki/Cyclotomic_identity $\endgroup$ – Qiaochu Yuan Oct 29 '09 at 14:40
$\begingroup$ Funny. I got obsessed with that one for a while. I guess you know it's connected with my answer to this question about 1+2+3+...=-1/12 because you combine both when you compute regularised determinants of Laplacians. $\endgroup$ – Dan Piponi May 14 '10 at 21:05
For a triangle with angles a, b, c $$\tan a + \tan b + \tan c = (\tan a) (\tan b) (\tan c)$$
$\begingroup$ It is very interesting. So if inverse is also true we have interesting algebraic structure connected to set of triangles with equivalence relation given by similarity of triangles. Very interesting indeed! $\endgroup$ – kakaz Feb 9 '10 at 8:44
$196884 = 196883 + 1$
muad
$\begingroup$ Can you explain what this is about? $\endgroup$ – M. Winter Dec 26 '18 at 23:06
$\begingroup$ This apparently references Monstrous Moonshine conjectures, which started with an observation that the coefficient $c_1=196884$ of $q$ in the $q$-expansion of the modular invariant $j(\tau)$ is the sum $1+196883$ of the dimensions of two smallest irreducible representations of the largest sporadic finite simple group $M$, known as the Monster. $\endgroup$ – Victor Protsak May 8 at 14:22
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:
$$ \left( \begin{array}{cc} A & B \\\ 0 & I \end{array} \right) \left( \begin{array}{cc} A^t & C^t \\\ B^t & D^t \end{array} \right) = \left( \begin{array}{cc} I & 0 \\\ B^t & D^t \end{array} \right). $$
Will Jagy
$\begingroup$ Ooh... that's a cute proof. $\endgroup$ – Willie Wong Apr 25 '10 at 21:34
$\begingroup$ I can't remember the guy's name who found it, he was at UCSD and did lots of stuff with matrices. He was one of many people I told this little fact, nobody had ever heard of it, although it pops out if you go through the material on the Hodge star in Frank Warner's book, "Foundations of Differentiable Manifolds and Lie Groups." I think later editions have a different title. $\endgroup$ – Will Jagy Apr 27 '10 at 2:28
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 2nd, 3rd, etc. exterior powers of the vector space.
Darsh Ranjan
I always thought this one was really funny: $1 = 0!$
François G. Dorais
$\begingroup$ Yeah, so do I. It is also convention that zero product is $1$. $\endgroup$ – Sunni Mar 4 '10 at 1:15
$\begingroup$ True, but that's not what's going on here. One can define 0! by extending the rule n!=n*(n-1)! with n=1, or by computing that \Gamma(1) in its integral form is indeed 1. And it's only very mildly a convention that we choose \Gamma as the interpolating function for the factorial function. $\endgroup$ – Cam McLeman Mar 4 '10 at 18:23
$\begingroup$ Or, 0! is the number of bijective maps on a set with 0 elements, since the empty set is the only such map. $\endgroup$ – Jonas Meyer Mar 19 '10 at 3:52
$\begingroup$ Ha! Very funny! Reminded me of when one of my professors ended a proof with the statement to the effect that the Q is an element in set D. (get it?) $\endgroup$ – Burhan May 21 '10 at 18:25
$\begingroup$ At first I thought you said $0=1$ with an exclamation point for emphasis. Maybe it was your favorite since it is a prototype of a false statement ;) $\endgroup$ – David Corwin Jul 15 '10 at 10:39
$2^n>n $
$\begingroup$ What is so special about that? It's pretty easy to prove. $\endgroup$ – Kevin H. Lin Jan 10 '10 at 18:14
$\begingroup$ I didn't say it was hard to prove, but you've asked a fair question. It struck me as beautiful when I first learned as an undergrad of this way to see that there are infinitely many different infinite cardinals. $\endgroup$ – Jonas Meyer Jan 10 '10 at 18:40
$\begingroup$ Perhaps I should have written $|2^X|>|X|$ to better indicate what I had in mind. $\endgroup$ – Jonas Meyer Jan 10 '10 at 18:43
$\begingroup$ Or $2^\kappa>\kappa$. But I like it in the way that you stated it. $\endgroup$ – Carsten S Jan 15 '10 at 0:06
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.
Sammy Black
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.
Ian Morris
Gauss-Bonnet, even though I am not a geometer.
Carsten S
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.
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.
David Corwin
$\begingroup$ This is tantalizingly close to giving a variant of $e$ as a period. $\endgroup$ – Matt F. May 7 at 15:31
It may be trivial, but I've always found
$\sqrt{\pi}=\int_{-\infty}^{\infty}e^{-x^{2}}dx$
to be particularly beautiful.
$\begingroup$ 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: $(-{\frac{1}{2}})!=\sqrt{\pi}$. $\endgroup$ – Andreas Blass Aug 21 '10 at 18:55
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!)
$\begingroup$ This is a pretty fact, but I would say that -- despite the way it is written! -- it is not an "equation, formula, identity or inequality". Rather it is a Galois correspondence, or an equivalence of categories. (Brief justification: it's not asserting that any two particular objects are equal. The content is more functorial than that.) $\endgroup$ – Pete L. Clark Aug 22 '10 at 9:06
E[X+Y]=E[X]+E[Y] for any 2 random varibles X and Y
Jonathan Kariv
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? :-)
$\prod_{n=1}^{\infty} (1-x^n) = \sum_{k=-\infty}^{\infty} (-1)^k x^{k(3k-1)/2}$
Bo Peng
$ D_A\star F = 0 $
Yang-Mills
Alexander Noll
My favorite is the Koike-Norton-Zagier product identity for the j-function (which classifies complex elliptic curves):
j(p) - j(q) = p-1 \prodm>0,n>-1 (1-pmqn)c(mn),
where j(q)-744 = \sumn >-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.
S. Carnahan
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.
Jay Kangel
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