First, let me make it clear that I do not mean jokes of the "abelian grape" variety. I take my cue from the following passage in A Mathematician's Miscellany by J.E. Littlewood (Methuen 1953, p. 79):

I remembered the Euler formula $\sum n^{-s}=\prod (1-p^{-s})^{-1}$; it was introduced to us at school, as a joke (rightly enough, and in excellent taste).

Without trying to define Littlewood's concept of joke, I venture to guess that another in the same category is the formula


which Ramanujan sent to Hardy in 1913, admitting "If I tell you this you will at once point out to me the lunatic asylum as my goal."

Moving beyond zeta function jokes, I would suggest that the empty set in ZF set theory is another joke in excellent taste. Not only does ZF take the empty set seriously, it uses it to build the whole universe of sets.

Is there an interesting concept lurking here -- a class of mathematical ideas that look like jokes to the outsider, but which turn out to be important? If so, let me know which ones appeal to you.

closed as no longer relevant by Yemon Choi, Dan Petersen, Andy Putman, Bill Johnson, Mark Sapir Jan 4 '12 at 13:29

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    Don't think it's exactly the same concept but Scott Aaronson has a post on something similar here: – Harry Altman Sep 15 '10 at 18:40
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    In regard to the empty set being a joke, Frank Harary and Ronald Read wrote a 1974 paper entitled "Is the null graph a pointless concept?". – Richard Stanley Sep 15 '10 at 22:29
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    It's amazing to see how many such jokes involve geometric series. – Thierry Zell Sep 16 '10 at 12:16
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    I just noticed that there are two puns in "abelian grape variety," as variety can associate with abelian or with grape. Too bad this is the kind of joke you don't mean. – Gerry Myerson Sep 26 '10 at 6:34
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    The two most recent answers have been more-or-less duplicates of previous answers. Time to close? – Gerry Myerson Oct 3 '10 at 12:25

59 Answers 59

Let "$\int$" denote $\int_0^x$. We want to find the solution to

$$\int f = f-1.$$

We simply "factor out" $f$, getting $1=\left(1-\int\right)f$. Thus, $f=(1-\int)^{-1}1$.

Using the geometric series,




as expected. (This was told to me by Steve Miller)

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    I was going to give this example! It is from a classic book by W. W. Sawyer, Prelude to mathematics. What makes this example interesting is that Sawyer described precisely the phenomenon we are discussing here: to a mathematics student, it appears to be a joke, but in fact, this is a standard technique of solving integral equations. Just to appreciate how a good a joke it is, the passage in Sawyer's book I quoted by memory I read over 20 years ago! – Victor Protsak Sep 16 '10 at 3:07
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    @Victor. I read some of those W.W. Sawyer books too, and they may have planted the seed of this question in my mind. Alas, it was 50+ years ago, and I no longer remember what I read in them. – John Stillwell Sep 16 '10 at 19:20

The geometric series expansion of projective space: $\frac{\mathbf{C}^{n+1} - \mathrm{pt}}{\mathbf{C} - \mathrm{pt}} = \mathbf{C}^n + \cdots + \mathbf{C}^1 + \mathrm{pt}$

  • This is great! There's no actual "good" reason for this (e.g. a perspective in which CW complexes correspond to formal polynomials / power series), is there? – Aaron Mazel-Gee Sep 16 '10 at 5:50
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    Aaron, I find John Baez's perspective here to be quite interesting. – Mike Skirvin Sep 16 '10 at 13:55
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    @Aaron: This equation holds in the Grothendieck ring of varieties (sometimes called ``ring of motives of varieties'' or similar). Whenever you have a Zariski locally trivial fibration $X \to Y$ with every fiber isomorphic to $F$, then $[X] = [F] * [Y]$ in the Grothendieck ring. – Arend Bayer Sep 17 '10 at 14:57
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    This equation is fantastic! – Martin Brandenburg Sep 24 '10 at 9:13
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    @Vivek: that doesn't mean the equation doesn't have content, e.g. it is meaningful to know that in the Grothendieck ring the terms don't all collapse to zero. (At least, I assume that they don't.) – Qiaochu Yuan Sep 28 '10 at 16:51

The Cayley-Hamilton Theorem:

If $A$ is a square matrix with characteristic polynomial $p(\lambda) = \det(A-\lambda I)$, then $p(A) = 0$.

Because you know, you "just plug in."

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    There are some funny proofs of this, too. Here's one that works over any field: First notice that the theorem holds for diagonalizable matrices. Then, adjoin $n^2$ indeterminants to our field and take the algebraic closure. But the $n \times n$ matrix whose entries are those indeterminants is now diagonalizable! Thus, we've proved the Cayley-Hamilton theorem as a polynomial identity over our original field. – Gene S. Kopp Sep 25 '10 at 17:19
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    That's the first proof of Cayley-Hamilton that I've actually liked.... – Brian Lawrence Aug 1 '11 at 9:09
  • This works for selfadjoint operator right? (is it call functional calclus or something?) – J. GE Oct 2 '12 at 19:37
  • @GeneS.Kopp: doesn't your proof apply verbatim to the polynomial $\Pi(x-\lambda_i)$, where $\lambda_i$ are distinct eigenvalues without multiplicity? – Michael Oct 18 '13 at 22:39
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    @GeneS.Kopp: why is that matrix diagonalizable again? – Michael Jan 5 '16 at 17:29

If $1-ab$ is invertible for $a$, $b$ in a (noncommutative) ring then so is $1-ba$.

Proof: $$(1-ba)^{-1} = 1+ba +baba+\cdots = 1+b(1+ab+abab+\cdots)a = 1+b(1-ab)^{-1}a,$$
The meaningless infinite series give the right answer (which is hard to guess).

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    There was a "rational" explanation for this sometime ago on MO; maybe someone with better MO-searching fu than I can find it. – Mariano Suárez-Álvarez Sep 15 '10 at 23:57
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    If you think of products of a's and b's as regular expressions, with $(1-x)^{-1}$ playing the role of $x^{\ast}$, the result is fairly obvious. This type of reasoning is, to some extent, captured by Kleene algebras. – Dan Piponi Oct 4 '10 at 17:31

In the same vein as the "Freshman's dream" $$(a + b)^p = a^p + b^p,$$ which is true in characteristic $p$, there is also the "Sophomore's dream", which is the identity $$\int_{0}^{1}{x^{-x} \: dx} = \sum_{n = 1}^{\infty}{n^{-n}}.$$ Surprisingly enough, this identity is actually correct.

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    Nice! This joke has a long pedigree, going back to a formula of Johann Bernoulli (1697): $\int^{1}_0 x^x dx=\sum^{\infty}_{n=0}(-1)^n n^{-n}$. But it makes a much better joke with the sign change. – John Stillwell Sep 17 '10 at 14:48

Tim Gowers mentioned infinities that may sound like jokes, especially to outsiders. Here is one specific example: you are standing in a room; at every tick of the clock, someone throws in a pair of numbered ping-pong balls: 1 & 2, then 3 & 4, etc... and you only have enough time to throw out one of them before the next tick. If you throw out the one with the largest number, then after $\omega$ ticks of the clock, you are in the room with all the odd-numbered balls, whereas if you always threw out the ball with the smallest number, you would be rid of them all!

And what if the balls are not numbered? A good way to get non-mathematicians thinking about infinity.

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    A similar example is The Gnome and the Pearl of Wisdom: A Fable by Richard Willmott (animations at This is about a sequence of numbered boxes and numbered marbles. The boxes start out empty, then in step $ t $, the gnome puts the stone number $ t $ to box $ 0 $, then resolves the conflict of two stones being in the same box by repeatedly moving the stone with the higher number (in the first variation; lower number in the second variation) to the next box. The question is where are the stones after $\omega$ steps. – Zsbán Ambrus Sep 16 '10 at 13:53
  • An interesting practical application. Very Clever! – Newb Oct 31 '13 at 3:48

Nobody mentioned the third isomorphism theorem yet? If $B$ and $C$ are normal subgroups of $A$ and $C \le B$ then $\frac{A/C}{B/C} \cong \frac{A}{B}$.

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    +1 for the user name. – Thierry Zell Sep 17 '10 at 13:11
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    I don't get this one. The proof for integers is very close to the proof for groups (if you are thinking about integers as cardinalities of sets) – Steven Gubkin Sep 11 '12 at 20:45

The fundamental axioms of mathematics are inconsistent if and only if we can prove that they are consistent.

(Because, you know, it follows from "logic." See Second Incompleteness theorem)

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    As Torkel Franzen pointed out in his wonderful book Godel's Theorem: An Incomplete Guide to Its Use and Abuse, if you harbored serious doubts about the consistency of your axioms, why would you be seek a consistency proof in that same setting? – Thierry Zell Sep 16 '10 at 16:33
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    But would a naive person be upset to find a proof of consistency in a supposedly rock-solid system? Probably not---but the joke here is that, nevertheless, they should be upset, as it reveals inconsistency. – Joel David Hamkins Sep 16 '10 at 21:55

Mazur's proof that knots do not have inverses under addition of knots:
If $A+B=0$, then $$A = A + (B+A)+(B+A)+\cdots=(A+B)+(A+B)+\cdots=0.$$
This is like the traditional joke proof that $1=0$ with $A=1$, $B=-1$; the difference is that the proof with knots is valid because the infinite sums of knots are meaningful: make the knots smaller and smaller.

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    That is the Eilenberg swindle (or Eilenberg-Mazur swindle) I briefly mentioned in my answer. A great "Littlewood joke". – Victor Protsak Sep 17 '10 at 4:37

The chain rule, in the form $${dy\over dx}={dy\over du}{du\over dx}$$ is a joke - you just cancel the $du$, top and bottom.

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    This is not a joke, you CAN cancel those du's, if understood properly. Namely, if we have graph of f(x), and a tangent vector v at point (c, f(c)), then dy = dy(v) = projection of v to y-coordinate, dx = dx(v) = projection of v to x-coordinate. Their quotient dy(v)/dx(v) is equal to f'(c). – mathreader Sep 16 '10 at 9:20
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    mathreader: indeed you can do that, but it's still a joke in the sense this question asks. – Zsbán Ambrus Sep 16 '10 at 13:39
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    Well, if you have two differentials which are dependent in the sense that $dx \wedge dy = 0$ then there is of course some function such that $dy$ = $f dx$. What could $f$ be called but $dy / dx$? Since $\mathbb{R}$ is one-dimensional, the chain rule joke always works. – Matt Noonan Sep 16 '10 at 14:28
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    @mathreader: Be very careful. What happens if u'(x) = 0? – Theo Johnson-Freyd Sep 16 '10 at 21:55
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    In fact, this is the very first thing that came to mind when I read the question. – David Corwin Dec 21 '12 at 11:27

The field with one element seems a good example.

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    e.g. $A_n = \mathbb{P}GL_n(\mathbb{F}_1)$ – Peter Arndt Sep 16 '10 at 0:00
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    I've met this one in a presentation where I was told that the Gauss binomial coefficient $ \genfrac{[}{]}{0pt}{}{n}{k}_q $ is the number of $ k $ dimensional subspaces in an $ n $ dimensional projective space over the field $ GF_q $. – Zsbán Ambrus Sep 16 '10 at 14:13

Another example from intro calculus: I once put a question of the form "$y=f(x)^{g(x)}$, find $y'$" on an exam. One student reasoned, if the exponent were a constant, the answer would be $g(x)f(x)^{g(x)-1}f'(x)$, but that's not right; if the base were a constant, the answer would be $g'(x)f(x)^{g(x)}\log f(x)$, but that's not right either; so I'll put them together to get $g(x)f(x)^{g(x)-1}f'(x)+g'(x)f(x)^{g(x)}\log f(x)$. This joke was on me, since that turns out to be correct.

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    In graduate school I graded one of the questions on a multiple section exam in beginning calculus. The correct answer was 4, and many many students got that, but few by any correct route. My conclusion, as the derivative is a limit process, was that the set consisting of the single number 4 is dense in the real line. – Will Jagy Sep 16 '10 at 0:52
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    Of course, this is just the multivariate chain rule $\frac{d}{dx}f(u(x),v(x)) = \frac{\partial f}{\partial u}\frac{du}{dx}+\frac{\partial f}{\partial v}\frac{dv}{dx}$. Perhaps your student was motivated by the product rule $(fg)'=f'g+fg'$, which works the same way. – Ricky Liu Sep 16 '10 at 17:33
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    It has naturaly physical sense: if $x$ arises several times in our expression, we may consider small peturbations (from the definition of derivative) of $x$ being independent and then sum up. – Fedor Petrov Sep 19 '10 at 20:09
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    I noticed this myself a long time ago (for the special case y=x^x), but always thought it was a cute coincidence -- I never realized that it was the multivariate chain rule! – Harrison Brown Oct 29 '10 at 19:02

A typical proof of Kolmogorov's zero-one law has as its punchline 'therefore A is independent of A'. Perhaps not a joke in the sense of Littlewood, but amusing nonetheless.

I want to evaluate $f(x+t)$. This is a function of two variables, but let's consider it a function $F(t)$ whose value is a function of $x$, i. e., $F(t)(x) = f(x+t)$. Note that $F(0) = f$, and in general $F$ satisfies the differential equation $$F'(t) = D_x(F(t))$$ (both sides being the function $x\mapsto f'(x+t)$). But $D_x$ is just a linear operator, so this is just a homogeneous linear ODE with constant coefficients. The solution is thus $$f(x+t) = F(t)(x) = (e^{tD_x}F(0))(x) = (e^{tD_x}f)(x) = \sum_{n=0}^\infty \frac{((tD_x)^nf)(x)}{n!} = \sum_{n=0}^\infty \frac{t^n f^{(n)}(x)}{n!}.$$ Voilà, Taylor series!

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    I love this joke! It appears as Exercise 14.6.3 in Arnol'd's book "Ordinary Differential Equations". – Tom Church Sep 19 '10 at 5:10
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    a trivial remark: this joke is apparently taken seriously by Lie theorists (their exp map) – Aknazar Kazhymurat May 21 at 16:14

In characteristic $p$, the so-called biologists' rule

$$(a+b)^p = a^p + b^p$$ (which got its name by mathematics students that worked as teaching assistants for "mathematics for biologists") is correct.

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    I've heard this called the "freshman's dream". – Michael Lugo Sep 16 '10 at 15:31
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    Yes, in fact Thomas Hungerford's book Algebra calls it that way in Exercise 11 in page 121. And according to it, the terminology is due to V.O. McBrien. – Adrian Barquero-Sanchez Sep 19 '10 at 15:41
  • or "freshman's delight" – Sean Tilson Sep 25 '10 at 3:19
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    In German: Klein Fritzchens binomische Formel. – Martin Brandenburg Mar 23 '12 at 17:17

An expansion on Timothy Chow's example of Grandi's series $1 - 1 + 1 - 1 \pm \cdots = \frac{1}{2}$. It is possible to interpret the left hand side as computing the Euler characteristic of infinite real projective space $\mathbb{R}P^{\infty}$, which is a $K(\mathbb{Z}/2\mathbb{Z}, 1)$ and therefore rightfully has orbifold Euler characteristic $\frac{1}{2}$! I think I learned this example from somewhere on Wikipedia.

  • I think you learned it from TWF, right? – Unknown Oct 3 '10 at 9:25

A good joke about infinity is the following. A hotel has rooms $1,2,\dots$. Every room is full when a new guest arrives. The clerk moves the occupant of room $n$ to $n+1$ to make room for the new guest in room 1. An hour later another guest arrives and the clerk repeats the process. 30 minutes later a third guest arrives and the process is repeated. Then 15 minutes, 7.5 minutes, etc., until two hours after the first new guest infinitely many guests have arrived and been accommodated. The clerk is very pleased with himself for dealing with these infinitely many guests, when he notices to his horror that all the rooms are empty! All the guests have mysteriously disappeared!

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    Is this Achilbert and the Tortoise, or what?... – darij grinberg Sep 16 '10 at 16:37
  • Right, this is isomorphic to the take about The Gnome and the Pearl of Wisdom which I just mentioned above. – Zsbán Ambrus Sep 18 '10 at 10:17

We owe Paul Dirac two excellent mathematical jokes. I have amended them with a few lesser known variations.

A. Square root of the Laplacian: we want $\Delta$ to be $D^2$ for some first order differential operator (for example, because it is easier to solve first order partial differential equations than second order PDEs). Writing it out,

$$\sum_{k=1}^n \frac{\partial^2}{\partial x_k^2}=\left(\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\right)\left(\sum_{j=1}^n \gamma_j \frac{\partial}{\partial x_j}\right) = \sum_{i,j}\gamma_i\gamma_j \frac{\partial^2}{\partial x_i x_j}, $$

and equating the coefficients, we get that this is indeed true if

$$D=\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\quad\text{and}\quad \gamma_i\gamma_j+\gamma_j\gamma_i=\delta_{ij}.$$

It remains to come up with the right $\gamma_i$'s. Dirac realized how to accomplish it with $4\times 4$ matrices when $n=4$; but a neat follow-up joke is to simply define them to be the elements $\gamma_1,\ldots,\gamma_n$ of

$$\mathbb{R}\langle\gamma_1,\ldots,\gamma_n\rangle/(\gamma_i\gamma_j+\gamma_j\gamma_i - \delta_{ij}).$$

Using symmetry considerations, it is easy to conclude that the commutator of the $n$-dimensional Laplace operator $\Delta$ and the multiplication by $r^2=x_1^2+\cdots+x_n^2$ is equal to $aE+b$, where $$E=x_1\frac{\partial}{\partial x_1}+\cdots+x_n\frac{\partial}{\partial x_n}$$ is the Euler vector field. A boring way to confirm this and to determine the coefficients $a$ and $b$ is to expand $[\Delta,r^2]$ and simplify using the commutation relations between $x$'s and $\partial$'s. A more exciting way is to act on $x_1^\lambda$, where $\lambda$ is a formal variable:


Since $x_1^{\lambda}$ is an eigenvector of the Euler operator $E$ with eigenvalue $\lambda$, we conclude that


B. Dirac delta function: if we can write

$$g(x)=\int g(y)\delta(x-y)dy$$

then instead of solving an inhomogeneous linear differential equation $Lf=g$ for each $g$, we can solve the equations $Lf=\delta(x-y)$ for each real $y$, where a linear differential operator $L$ acts on the variable $x,$ and combine the answers with different $y$ weighted by $g(y)$. Clearly, there are fewer real numbers than functions, and if $L$ has constant coefficients, using translation invariance the set of right hand sides is further reduced to just one, $\delta(x)$. In this form, the joke goes back to Laplace and Poisson.

What happens if instead of the ordinary geometric series we consider a doubly infinite one? Since

$$z(\cdots + z^{-n-1} + z^{-n} + \cdots + 1 + \cdots + z^n + \cdots)= \cdots + z^{-n} + z^{-n+1} + \cdots + z + \cdots + z^{n+1} + \cdots,$$

the expression in the parenthesis is annihilated by the multiplication by $z-1$, hence it is equal to $\delta(z-1)$. Homogenizing, we get


This identity plays an important role in conformal field theory and the theory of vertex operator algebras.

Pushing infinite geometric series in a different direction,

$$\cdots + z^{-n-1} + z^{-n} + \cdots + 1=-\frac{z}{1-z} \quad\text{and} \quad 1 + z + \cdots + z^n + \cdots = \frac{1}{1-z},$$

which add up to $1$. This time, the sum of doubly infinite geometric series is zero! Thus the point $0\in\mathbb{Z}$ is the sum of all lattice points on the non-negative half-line and all points on the positive half-line:

$$0=[\ldots,-2,-1,0] + [0,1,2,\ldots] $$

A vast generalization is given by Brion's formula for the generating function for the lattice points in a convex lattice polytope $\Delta\subset\mathbb{R}^N$ with vertices $v\in{\mathbb{Z}}^N$ and closed inner vertex cones $C_v\subset\mathbb{R}^N$:

$$\sum_{P\in \Delta\cap{\mathbb{Z}}^N} z^P = \sum_v\left(\sum_{Q\in C_v\cap{\mathbb{Z}}^N} z^Q\right),$$

where the inner sums in the right hand side need to be interpreted as rational functions in $z_1,\ldots,z_N$.

Another great joke based on infinite series is the Eilenberg swindle, but I am too exhausted by fighting the math preview to do it justice.

  • In a similar spirit as the first joke: how does one "get" a "root" (=usually a complicated thingie) of a polynomial $f(X)\in A[X]$? Well, consider the element $\overline{X}\in A[X]/(f)$. – M.G. Jul 22 '17 at 12:40

Multiplication is repeated addition: $$ x^2 = \underbrace{x+\cdots+x}_{x\text{ times}} $$

Differentiate (remember the chain rule for partial derivatives):

$$\frac{d}{dx} \big(x^2\big)\qquad\qquad$$

$$ \qquad= {\underbrace{1+\cdots+1}_{x\text{ times}}}$$

$$\qquad{}+ \underbrace{x+\cdots+x}_{1\text{ times}}$$

$$ \qquad= x + x = 2x .$$

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    I took the liberty of changing "+x+\cdots+x" to "{}+x+\cdots+x". That puts a larger amount of space between the first "+" and the "x", thus: $+x+\cdots+x$ versus ${}+x+\cdots+x$. When you understand why that is just how the software ought to work, then you will understand something about typesetting and about the fact that Donald Knuth (inventor of TeX, among other things) knew what he was doing. ${}\qquad{}$ – Michael Hardy May 7 '15 at 20:01
  • Ha Ha, great!!! – user86003 Jun 8 '16 at 14:48

The classical Stokes formula $\int_{\partial\Omega}\omega=\int_\Omega d\omega$ is certainly a Littlewood type joke. That is especially true if you learn it after you've spent a few months covering vector calculus, learned rotor, divergence, path and surface integrals of 2 kinds, etc., which is the standard route to follow.

There are other divergent series that fit the bill, such as $1-1+1-1+ \cdots = 1/2$. Here's one from formal language theory: Suppose we define a language $L$ recursively by the rule $L = 1 | aL$, meaning that the empty string $1$ is in $L$, and the letter $a$ followed by any element in $L$ is also in $L$. Jokingly, we note that $|$ is akin to addition and concatenation is akin to multiplication, so we can solve for $L$: $1 = L - aL = L (1-a)$, so $$L = {1\over 1-a} = 1|a|aa|aaa|aaaa \ldots,$$ which is the right answer.

The umbral calculus could also be considered an elaborate joke.

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    Nice example. I like the fact that the very name "umbral calculus" admits its shady nature. – John Stillwell Sep 15 '10 at 20:51
  • There is a nice piece of algebra with structure types at . Actually, John Baez's TWF is full of "jokes" (as in, concrete mathematics from a ridiculously abstract viewpoint that actually makes hard things trivial). – darij grinberg Sep 15 '10 at 23:03
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    Dear Mariano, I think that "umbral calculus" was coined by Sylvester. Looking on Mathworld confirms this, but (according to Mathworld) the shadows being alluded to are the combinatorial identities obtained, which "shadow" more obvious polynomial or Taylor series idenities. – Emerton Sep 16 '10 at 0:10
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    @Mariano: The term "umbra" (shadow) is attributed to Sylvester, and its first occurrence seems to be the following, from 1851 (in his Collected Mathematical Papers vol. 1, p.242): "Each quantity is now represented by two letters; the letters themselves, taken separately, being symbols neither of quantity nor of operation, but mere umbrae or ideal elements of quantitative symbols." – John Stillwell Sep 16 '10 at 0:40
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    Dear John, This is more appealing and romantic than the Mathworld explanation, and presumably more accurate, since you are quoting Sylvester directly. I wonder if the Mathworld explanation is just based on speculation rather than primary sources? – Emerton Sep 16 '10 at 1:52

In a probability oral exam, a student is asked to compute the probability that a random number chosen from the interval $[0,1]$ is larger than $2/3$. The students answers $1/3$. The teacher asks him to explain his argument, and he says: well, there are three possibilities: the number is either less than, or bigger than, or equal to $2/3$, so, the probability is $1/3$!

I know you say "moving beyond zeta function jokes", but I'd say the following two zeta-regularizations deserve to be alongside your Ramanujan example: $$\infty!= \sqrt{2\pi}\qquad\qquad\mbox{and }\qquad\qquad \prod_{\mbox{$p$ prime}}p =4\pi^2.$$ One can also entertain beginning calculus students with $\frac{1}{2}!=\frac{1}{2}\sqrt{\pi}$ as a way of introducing the Gamma function.

How many finite sets are there?

Well, there is one set of cardinality 0; one set of cardinality one; one set of cardinality 2, but since its automorphism group has order 2, we only count it with multiplicity 1/2; there is one set of cardinality 3, counted with multiplicity 1/3!; ... So the number of sets is $$ 1 + 1+ 1/2! + 1/3! + \dots = e $$

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    So $e$ is the Euler number of the space, or category, or something, of finite sets. – Tom Goodwillie Sep 17 '10 at 12:59
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    It's the Euler number of the groupoid of finite sets. – David Roberts Sep 19 '10 at 22:12

This is a souped-up version of the freshman's dream: as Jon Borwein pointed out to me: if $a_n=(-1)^n/(2n+1)$, then $$\left(\sum_{n=-\infty}^{\infty}a_n\right)^2=\sum_{n=-\infty}^{\infty}a_n^2$$ as they are both $\pi^2/4$.

Moreover, this can be proved by telescoping sums: $$ \begin{align} \left( \sum_{n} \frac{(-1)^n}{2n+1} \right)^2 &= \sum_{m,n} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \\ &= \sum_{m,k} \frac{(-1)^k}{(2m+1)(2m+2k+1)} \\ &= \sum_{m} \frac{1}{(2m+1)^2} + \sum_{k \neq 0 } \sum_m \left( \frac{(-1)^k}{2 k (2m+1)} - \frac{(-1)^k}{2 k (2m+2k+1)} \right) \\ &= \sum_{m} \frac{1}{(2m+1)^2} \end{align}$$ Of course, one needs to justify rearranging the conditionally convergent series, but that spoils the joke.

Let $C(x) = \sum_{n \ge 0} \frac{1}{n+1} {2n \choose n} x^n$ be the generating function for the Catalan numbers. Then $C(x) = \frac{1 - \sqrt{1 - 4x}}{2}$. In particular, $C(1) = \frac{1 - \sqrt{-3}}{2}$ is a sixth root of unity - except of course that $C(1)$ does not converge.

Nevertheless, there is a remarkable sense in which $C^7 = C$; see Blass' Seven Trees in One and the generalization in Fiore and Leinster's Objects of Categories as Complex Numbers. As with many results of this type, I learned about this from John Baez.

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    I think the second author ought to be "Leinster," though I would love to take credit since I like that paper very much. :) – Eric Finster Sep 24 '10 at 16:00
  • That Fiore-Leinster paper is hilarious! I must read it someday, when I find the time. – Vectornaut Oct 9 '12 at 3:43

Lagrange's equation:

$$\frac{d}{dt} \left( \frac{\partial}{\partial \dot{q}} \mathcal{L} \right) =\frac{\partial}{\partial q} \mathcal{L},$$

as $\dot{q}=\frac{dq}{dt}$, you can simplify "$dt$".

Given a function $f$ on the real line, let's compute the function $\sum f$, taking $n \mapsto f(1) +\cdots + f(n)$. Well, $\sum = 1/\Delta$, where $\Delta$ is the differencing operator $\text{shift} - 1$. And the shift operator is the exponential of the differentiation operator (this being, essentially, Taylor's theorem). Hence $$ \sum = \frac{1}{e^D - 1} = \frac{1}{D} \frac{D}{e^D-1} $$ Using L'Hopital's rule on the latter as $D\to 0$, whatever THAT means, we see the limit is $1$. So expand in a power series: $$ \frac{1}{D} \frac{D}{e^D-1} = \frac{1}{D} (1 + \text{power series in $D$}) $$ The first term is $1/D$, which is of course $\int$.

No surprise: $\sum = \int + $ correction terms. What the above suggests is that those correction terms come from the Taylor expansion of $\frac{D}{e^D - 1}$. This leads to the Euler summation formula (and eventually, to Hirzebruch-Riemann-Roch).

I learned this from "Concrete Mathematics", where I recall this joke being attributed to Laguerre. Part of why it is a joke is that the Euler summation formula has an error term, that can't be neglected for most functions, e.g. $\ln(x)$ which one wants to sum up to compute $\ln(n!)$. It can be neglected for polynomials times exponentials.

  • Victor Protsak hinted at this joke in his comment of 17 September on Per Vognsen's contribution. – Gerry Myerson Nov 30 '10 at 5:35

This isn't a particularly interesting example, but the existence of different sizes of infinity fits your criterion of being something that makes outsiders laugh (as I know from experience) and that is also very important to mathematicians.

The familiar argument that says that if you want an explicit example of $a^b=c$ with a and b irrational and c rational, then one of $a=b=\sqrt{2}$ or $a=\sqrt{2}^{\sqrt{2}}$ and $b=\sqrt{2}$ will work is certainly an argument that makes people laugh. Though the result itself is not very important, the phenomenon it illustrates is quite important.

Added two minutes later: I've just had a look at Scott Aaronson's post and seen that Erik, one of the earlier commenters, chose precisely the same two examples. It was a coincidence -- honest.

This was once presented to me as a kind of proof, though I think it works better as a kind of joke:

To compute ${\partial^n\over\partial x^n}(fg)$, split ${\partial\over\partial x}$ into the sum of a piece $D$ that just acts on $f$ (acting as the identitiy on $g$) and a piece $E$ that just acts on $g$ (acting as the identity on $f$) and write ${\partial^n\over\partial x^n}(fg) = (D+E)^n(fg) = \sum_{i=0}^n \binom{n}{i} D^i E^{n-i}(fg) = \sum_{i=0}^n \binom{n}{i} {\partial^i f\over\partial x^i}{\partial^{n-i} g\over\partial x^{n-i}}$.

  • Some things about this kind of derivative are clearer if you first think of it as $\displaystyle\frac{\partial^n}{\partial x_1\,\cdots\,\partial x_n}(fg)$ and then after finding the derivative let the $n$ variables $x_1,\ldots,x_n$ coalesce into indistinguishability. What happens is that before they coalesce, every coefficient in the expansion is $1$, but after they coalesce the coefficients simply count the number of terms in a class of terms that become indistinguishable from each other. ${}\qquad{}$ – Michael Hardy May 7 '15 at 20:04

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