Jokes in the sense of Littlewood: examples? [closed]

Question

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

$1+2+3+4+\cdots=-1/12$,

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.

Answer 1

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,

$$f=\left(1+\int+\iint+\iiint+\cdots\right)1=1+\int_0^x1~dx+\int_0^x\int_0^x1~dx+\cdots$$

Thus,

$$f=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots=e^x,$$

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

Answer 2

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

Answer 3

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).

Answer 4

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."

Answer 5

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.

Answer 6

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)

Answer 7

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.

Answer 8

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

Answer 9

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.

Answer 10

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.

Answer 11

The field with one element seems a good example.

Answer 12

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.

Answer 13

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!

Answer 14

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.

Answer 15

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!

Answer 16

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.

Answer 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

$$D=\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\quad\text{and}\quad \gamma_i\gamma_j+\gamma_j\gamma_i=2\delta_{ij}.$$ (With a $2$ on the right hand side.)

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

$$\mathbb{R}\langle\gamma_1,\ldots,\gamma_n\rangle/(\gamma_i\gamma_j+\gamma_j\gamma_i - 2\delta_{ij}).$$

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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:

$$[\Delta,r^2]x_1^{\lambda}=((\lambda+2)(\lambda+1)+2(n-1)-\lambda(\lambda-1))x_1^{\lambda}=(4\lambda+2n)x_1^{\lambda}.$$

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

$$[\Delta,r^2]=4E+2n.$$

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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.

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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

$$\sum_{n\in\mathbb{Z}}\left(\frac{z}{w}\right)^n=\delta(z-w)$$

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

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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 $z_1,\ldots,z_N$.

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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.

Answer 18

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

Differentiate (remember the chain rule for partial derivatives):

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

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

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

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

Answer 19

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.

Answer 20

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.

Answer 21

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 $1/3$!

Answer 22

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.

Answer 23

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 $$ 1 + 1+ 1/2! + 1/3! + \dots = e $$

Answer 24

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.

Answer 25

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.

Answer 26

I do a double take every couple of months when I remember that

$(1 + 2 + 3 + \ldots + n)^2 = 1^3 + 2^3 + 3^3 + \ldots + n^3$

It just seems plain weird.

Answer 27

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.

Answer 28

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's TWF 202.

Answer 29

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 "$dt$".

Answer 30

Here's one of my own. Maybe I should publish it? A confused calculus student attempted to evaluate $$ \frac{d}{dx} \left( 1^n + 2^n + 3^n + \cdots + (x-1)^n \right) $$ and got $$ n1^{n-1} + n2^{n-1} + n3^{n-1} + \cdots + n(x-1)^{n-1} + \text{constant}. $$ That is in fact correct. The "constant" is a Bernoulli number.