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

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.

• Don't think it's exactly the same concept but Scott Aaronson has a post on something similar here: scottaaronson.com/blog/?p=392 – Harry Altman Sep 15 '10 at 18:40
• 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
• It's amazing to see how many such jokes involve geometric series. – Thierry Zell Sep 16 '10 at 12:16
• 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
• 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

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

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.

Another geometric series joke:

Let $S_\epsilon$ be the $\epsilon$-shift operator and let $D = (1 - S_\epsilon)/\epsilon$ be the formal derivative operator. Then $D^{-1} = \epsilon \sum_{n=0}^\infty {S_\epsilon}^n$, so $(D^{-1}f)(x) = \epsilon(f(x) + f(x-\epsilon) + f(x-2\epsilon) + \cdots)$, the integral of $f$ from $-\infty$ to $x$. Thus differentiation and integration are inverses.

Playing a prank with the binomial formula on $D$ yields fractional derivatives, another famous joke.

John kicked off the thread with zeta functions; my favorite zeta function joke has to be Euler's solution to the Basel problem where he factored $sin(x)$ as a power series according to its roots $\pm n\pi$.

• An even more impressive version of the shift operator-derivative "geometric series joke" is the Euler-Maclaurin summation formula. – Victor Protsak Sep 17 '10 at 4:39

The number of finite sets is $e$.

(Since of course we should count in the stack sense: up to isomorphism, dividing by the number of automorphisms).

• This is the same as the answer of AByer on 17 September, no? – Gerry Myerson Oct 3 '10 at 12:24
• Oops! I missed his answer when I read through them the first time. I think he actually heard this joke from me in last summer though. :-). – Jim Bryan Oct 3 '10 at 13:41

The chain rule "joke" reminded me of a similar notation joke: Radon-Nikodym derivatives.

If $\mu$, $\nu$, $\lambda$ are $\sigma$-finite measures with $\nu \ll \mu \ll \lambda$, and $f \geq 0$ is measurable, then:

$$\int f\ d\nu = \int f \left[\frac{d\nu}{d\mu}\right]\ d\mu$$

and

$$\left[\frac{d\nu}{d\lambda}\right] = \left[\frac{d\nu}{d\mu}\right]\left[\frac{d\mu}{d\lambda}\right]$$

• When taken seriously, this joke could lead one to discover Yamagami's modular algebra and Tomita-Takesaki theory. – Dmitri Pavlov Mar 3 '13 at 11:43

Hausdorff dimension. Try showing a Sierpinski triangle to a non-mathematician and explaining that it is a 1.585-dimensional object.

• And then try the same thing with the sierpinksi tetraeder (dimension = 2) – Johannes Hahn Sep 16 '10 at 10:43
• I just did this to an 11-year-old – David Corwin Dec 23 '12 at 0:10
• @JohannesHahn I think this makes sense. Imagine your tetrahedron sitting on the table. Pick up the top one of the four mini-tetrahedra and put it onto the table so that its base fits into the triangular hole between the other three. Repeat this process recursively and your Sierpinski tetrahedron will be reduced to a triangle. – Oscar Cunningham Apr 3 '16 at 19:25

$$\dfrac{16}{64} = \dfrac{1\not{6} }{\not{6} 4} = \dfrac{1}{4}$$

$$25^{1/2} = \not25^{1/\not2} = 5^1 = 5$$

$$\sqrt{64} = \sqrt[\not 6]{\not 64} = \sqrt{4}$$

• For related developments in this important area, see B. Ruekberg, "Simplified Mathematics,". J. Irreproducible Results 35 (1), 1990. – Nate Eldredge Sep 21 '10 at 18:59
• Don't forget that by crossing off the $9$'s we get $\frac {95}{19}=5$. – Eric Naslund Feb 16 '11 at 21:32
• $\frac{sin x}{n} = 6$, just cancel n's, you get "six" – mathreader Dec 31 '18 at 11:14

In Cardano's formula for the roots of the cubic there are negative numbers under the square roots if and only if all roots of the polynomial are real. That, and the fact that Cardano and his contemporaries didn't even believe in negative numbers.

The yoga of motives, in the parts which are conjectural but inspire the right guesses, is a huge joke!

In german elementary schools you learn "In Summen kürzen nur die Dummen.", i.e. it is dumb to cancel sums in quotients. But later in algebra or category theory you may learn

$\sum_{i \in I} A_i / \sum_ {i \in I} B_i = \sum_{i \in I} A_i / B_i$,

which is correct.

• You also frequently divide by zero when dealing with abelian groups ;-) – Johannes Hahn Sep 24 '10 at 12:22
• And does it work? I don't mean the identity; do students remember not to cancel in quotients? If yes, I might start using it... – Thierry Zell Sep 24 '10 at 12:52
• Martin, thanks for the great slogan, which I'd like to translate as "Only the dumb cancel in sums." Probably couldn't use it in the US, alas, it might hurt the students' self-esteem ... – John Stillwell Sep 25 '10 at 15:54
• The whole thing is because sums in abelian categories are more or less the same as products. If such a rule would hold for tensor division (whatever this could mean), that would be interesting... – darij grinberg Oct 3 '10 at 14:24
• It is a general fact that coproducts commute with cokernels, this holds in every Ab-category. More generally, colimits commute with colimits. Thus it has nothing to do with abelian categories, where also only finite sums are finite products. – Martin Brandenburg Oct 3 '10 at 20:41

Gerry's answer reminded me of the "Lucky Larry" series, which is a regular column in The AMATYC Review.

One particularly nice one is

$$\lim_{x\to\infty}\frac{\ln\ln x}{\ln x}$$

where you can cancel the $\ln x$ and wind up with $\ln 1=0$, which turns out correct.

I've always thought of compactness arguments for passing from a finite result to an infinite one as sorts of jokes. The general idea, after all, is that to get the infinite analogue of the finite result, you just "make it bigger and bigger!" (Of course it doesn't work in all situations, but when it does it's often forehead-slappingly simple.)

• Nonstandard analysis is one of these jokes. – Zsbán Ambrus Jan 31 '11 at 21:50

Many infinite, periodic expression involving numbers is a joke, according to non-mathematicians. Three examples: $$0,999999999\cdots=1,$$ $$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}=\frac{1+\sqrt5}{2},$$ $$\sqrt{-a^2+\sqrt{-a^2+\sqrt{-a^2+\cdots}}}=\frac12(1+\sqrt{1-4a^2}).$$

Two ways to calculate $1+2+3+\dots$

1st way. Let $$1-1+1-1+\dots=a.$$ Then also $$0+1-1+1-1+\dots=a.$$ Sum up (summing respective terms), get $$2a=1+0+0+\dots=1,$$ $a=1/2$.

Let now $$1-2+3-4+\dots=b.$$ Again, write down $$0+1-2+3-\dots=b.$$ Sum up and get $$2b=1-1+1-1+\dots=1/2,$$ $b=1/4$. Now let $$1+ 2+3+ 4+5+\dots=с.$$ Then $$0+4\cdot 1+0+4\cdot 2+0+4\cdot 3+0+\dots=4с.$$ Substract and get $$-3с=1-2+3-4+5+\dots=1/4,$$ hence

$$1+2+3+\dots=-1/12.$$

2nd way. Let $$1+2+3+\dots=S.$$ Then $$2+4+6+\dots=2\cdot(1+2+3+\dots)=2S.$$ Hence $$1+3+5+\dots=(1+2+3+4+5+\dots)- (0+2+0+4+0+\dots)=-S.$$

Now sum up $$1+2+3+4+\dots=S$$ and $$0+1+3+5+\dots=-S.$$ We get $$0=1+3+6+9+\dots=1+3\cdot(1+2+3+\dots)=1+3S,$$

$S=-1/3$. So

$$1+2+3+\dots=-1/3.$$

The question is: why the first way gives correct answer $\zeta(-1)=-1/12$, while the second leads to incorrect answer?

The Eilenberg swindle was briefly mentioned by Victor Protsak, and something which always struck me as similar (and seemed like black magic to me when I first saw it) is the Pelczynski decomposition method for proving that complemented subspaces of $\ell^p$ are isomorphic to $\ell^p$. I don't have a link to hand but might try to write something out later (or someone else is welcome to edit this).

I can't say I share Littlewood's sense of humor, but here is a formula that made me grin the first time I saw it ($A$ and $B$ are non-commuting square matrices):

\begin{equation}\frac{d}{dt}e^{A+B t}=\int_0^1 e^{s(A+B t)}\ B\ e^{(1-s)(A+B t)} \ ds. \end{equation}

• I don't get it. – Martin Brandenburg Sep 24 '10 at 9:44
• Well, it's true -- do you need a proof? I thought it was funny because integrals are in a sense the opposite of derivatives and yet here the result of a derivative is an integral. – Veit Elser Sep 24 '10 at 12:06
• The Cauchy integral formula would be even funnier then. – Matt Young Sep 24 '10 at 17:18
• @MartinBrandenburg: I know one proof and one "proof" of this identity, and they're both pretty amusing. The proof (from "Exponential Operators and Parameter Differentiation in Quantum Physics," by R.M. Wilcox) is done by showing that both sides of the identity satisfy the same first-order linear differential equation, with the same initial condition. The "proof," which I've always associated with Richard Feynman, is done by writing exp(A + Bt) as (1 + (A + Bt)/N)^N, for N extremely large, and applying the product rule. – Vectornaut Oct 9 '12 at 3:32

I'm not sure if this counts, but here's one of my own devising:

How do we find second derivatives of inverse functions?

Well, the second derivative of $y = f(x)$ is defined as $(f')' = \frac{d\frac{dy}{dx}}{dx}$, and by the quotient rule we can write this as $\frac{dxd^2y-dyd^2x}{dx^3} = \frac{d^2y}{dx^2}-\frac{dyd^2x}{dx^3}.$ (Since $x$ usually varies linearly, we normally substitute $d^2x = 0$.)

So by symmetry, the second derivative of the inverse function is $(f^{-1})'' = \frac{d^2x}{dy^2}-\frac{dxd^2y}{dy^3} = -\frac{dx^3}{dy^3}(\frac{d^2y}{dx^2}-\frac{dyd^2x}{dx^3}) = -f''/(f')^3.$

We can also directly derive the chain rule for second derivatives: Let $x = f(v)$, $v = g(u)$, and we get $(f\circ g)'' = \frac{d^2x}{du^2} - \frac{dxd^2u}{du^3} = (\frac{d^2x}{dv^2}-\frac{dxd^2v}{dv^3})\frac{dv^2}{du^2}+\frac{dx}{dv}(\frac{d^2v}{du^2}-\frac{dvd^2u}{du^3}) = (f''\circ g)(g')^2 + (f'\circ g)g''.$

Here's another one:

A friend of mine (Sam Elder) was trying to calculate the number of combinations on a simplex lock with $n$ buttons. After some work he had gotten a recurrence which I'm going to write as $2A_m = 1+\sum_{k=0}^m \binom{m}{k}A_{k}$, which he showed to me. My thought process went like this:

Hmm, this looks like the recurrence for the Bernoulli numbers. How did we prove the recurrence for the Bernoulli numbers again? One way is to use the well-known fact that $\sum_{n=0}^{\infty} n^m = -\frac{B_{m+1}}{m+1}$, so $-\frac{B_{m+1}}{m+1} = \sum_{n=0}^{\infty} n^m = \sum_{n=0}^{\infty}(n+1)^m = -\sum_{k=0}^m\binom{m}{k}\frac{B_{k+1}}{k+1}$.

Working backwards, what I need to do is find a sequence of functions $f_m(n)$ satisfying $2f_m(n+1) = \sum_{k=0}^m\binom{m}{k}f_{k}(n)$. This leads naturally to the choice $f_m(n) = \frac{n^m}{2^n}$. So the number of simplex combinations on m buttons is

$A_m = \sum_{n=1}^{\infty} \frac{n^m}{2^n}.$

Sam did not find this formula incredibly helpful.

(Continuing the analogy with the Bernoulli numbers, we can also derive the formula $\sum_{n=1}^x \frac{n^m}{2^n} = A^m-\frac{(x+A)^m}{2^x},$ where we interpret $A^m$ as $A_m$.)

• zeb, I think the formula you've given for A_m is very interesting. It says that A_m is the mth moment of a geometric distribution and I would love if anyone could give a conceptual explanation why this is the same thing as what it actually counts (which I can't quite recall at the moment). – Qiaochu Yuan Sep 19 '10 at 17:36
• The sequence $A_m$ is on the Online Encyclopedia of Integer Sequences as A000629. Apparently I'm not the first one to come up with that formula; maybe the references on OEIS have a more conceptual proof. – zeb Sep 20 '10 at 15:45

When solving the linear recursive equation $a_{n+2} = a_{n+1} + 2 a_n$, you solve the quadratic equation $x^2 = x + 2$, which has the solutions $x= 2$ and $x = -1$, and then you get a basis of solutions $2^n$ and $(-1)^n$.

What if you start with $a_{n+2} = 4a_{n+1} - 4a_n$, so that the quadratic equation $x^2 = 4x - 4$ has only one solution $x = 2$? Easy, you take the solution $2^n$, and its derivative with respect to 2, i.e. $n \cdot 2^{n-1}$.

You can't prove something exists just by computing the probability of its existence - right? The first application of the "probabilistic arguments" in Combinatorics I have encountered was this; took me a long time to get it.

For consecutive Farey fractions $\frac{a}{b}, \frac{c}{d}$ the mediant is obtained via a "simple's man addition": $$\frac{p}{q} = \frac{a+c}{b+d}$$ which since $\frac{a}{b},\frac{c}{d}$ are consecutive if and only if $det\begin{pmatrix}a & c\\\\ b & d\end{pmatrix} = 1$ also turns out to be the rule of invariance of the determinant when you add a column to another column.

• For consecutive fractions in a Farey sequence $\frac{a+c}{b+d}$ is not the sum of $\frac ab$ and $\frac cd$ but rather the next Farey fraction to appear between these two. – Andreas Blass Sep 25 '10 at 20:22
• Of course... for some reason i've been thinking that the mediant also had to be equal to the sum. Thanks for the correction! – Maurizio Monge Sep 25 '10 at 23:49

(An example from set theory, more specifically: forcing):

The backslash $\backslash$ is used for set difference, the forward slash $/$ is often used for taking a quotient. In my experience, students often have a hard time distinguishing between the two. (Between the two symbols, not the two concepts.)

But there is one case where the difference does not matter: if $B$ is a Boolean algebra, and $I$ an ideal, then the sets $B\setminus I$ (= $I$-positive elements) and $B/I$ (quotient Boolean algebra) are equivalent as forcing notions: $B\setminus I = B/I$.

For example, random forcing can be seen as "Borel sets modulo measure zero sets", or equivalently as "Lebesgue-positive Borel sets (without any identification)".

Another joke in the spirit of the chain rule; you solve separable differential equations by "multiplying by g(y)dx" $$g(y) dx \left(\frac{dy}{dx} = \frac{f(x)}{g(y)}\right) \Rightarrow g(y)dy = f(x) dx$$

Then, there's nothing to be done but integrating to get rid of the dx and dy.

I also like to point out to students who ask about cancellation in the chain rule that you can cancel there just like you can cancel the sixes and nines respectively in

$$\frac {16}{64} = \frac 1 4 \qquad \text{and} \qquad \frac{19}{95}=\frac 1 5;$$ that is, carefully, and when it makes sense to do so.

• Canceling these digits is a mistake that coincidentally yields the right answer, but multiplying by $g(y)\,dx$ is a "joke" only because the rules haven't been properly codified---like the way Dirac's delta function and its derivatives were a "joke". – Michael Hardy Sep 19 '10 at 16:31

Yet another divergence joke:

$$\sum_{j=0}^\infty (-1)^j j!=e E_1(1)\approx 0.596347362\dots$$

where $E_n(z)$ is an exponential integral.

This is intimately related to the formal hypergeometric series ${}_2 F_0$ being the asymptotic expansion of a certain convergent improper integral.

This also serves as a warning of sorts to users of convergence acceleration methods like Wynn ε or Levin t ; attempts to sum a divergent series might give "correct" yet still unexpected answers.

Here's more divergence craziness:

$$\int_0^\infty \sin\;u\mathrm{d}u=1$$

and

$$\int_0^\infty \ln\;u\;\sin\;u\mathrm{d}u=-\gamma$$

($\gamma$ is of course the Euler-Mascheroni constant)

which only makes sense when interpreted as

$$\lim_{\varepsilon\to 0} \int_0^\infty \exp(-\varepsilon u)\ln\;u\;\sin\;u\mathrm{d}u$$

and similarly for the first one.

Results obtained from numerical quadrature methods specially designed for infinite oscillatory integrals (e.g. the Ooura-Mori double exponential quadrature and the Longman scheme) agree with these closed forms.

I've always seen the following identity for the determinant of block matrices as a 'joke'

$\det\left( \begin{array}{cc} A & B \newline C & D \end{array} \right) = \det(AD-BC)$

• It isn't true for arbitrary square blocks of equal size, but if for example the four matrices commute, then the identity holds. – Darsh Ranjan Sep 19 '10 at 23:36

I suppose this is a silly example, but as far back as grade school I found it amusing (and maybe a little profound) that to invert a fraction $\frac{a}{b}$, you literally invert it! As I learned about more and more mathematical objects through the years, I kept waiting for this kind of thing to happen again, but it never really did.

• I had a 6th grade teacher who demonstrated inversion by calling a small kid out of the front row and literally turning him upside-down. With the possible exception of some John Conway's performances, it's the most dramatic event I've seen in a mathematics classroom. – John Stillwell Sep 28 '10 at 21:48
• If $y=f(x)$, and you invert the function to get $x=f^{-1}(y)$, then you get the derivative of the inverse function by inverting the derivative of the original function: $dx/dy=1/(dy/dx)$. – Gerry Myerson Sep 28 '10 at 23:01

I recall that the following simple "proof" of $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ is attributed to Euler:

Begin with the fact that for a polynomial $a_0 + a_1 x + \cdots + a_N x^N$ the sum of the inverses of the roots is given by $\sum_{n=1}^N \frac{1}{x_n} = -\frac{a_1}{a_0}$. (If you only remember the formula for the sum of the roots just make a change of variable $y=1/x$). Now consider the "polynomial" $$\frac{\sin\sqrt{x}}{\sqrt{x}} = 1 - \frac{x}{3!} + \cdots$$ whose roots are $x_n = (n\pi)^2$ for $n\in N$. By applying the aforementioned fact the desired result is immediate.

• This was already mentioned by Per Vognsen in his answer of 19 September, no? – Gerry Myerson Oct 3 '10 at 12:22
• Oh yes; you're right. I did skim through all responses before submitting mine but somehow it didn't capture my attention since it didn't have a big $\pi^2/6$ right at the center! Thanks for the comment and sorry for duplicate answer. – Mahdiyar Oct 4 '10 at 3:48

The journal of unpublishable mathematics, which seems to be down at the moment is one of my favourites

• The journal is a "joke in the sense of Littlewood"? – Gerry Myerson May 12 '11 at 11:41
• I do not know, I never had the chance to meet him to ask ;) – Jesus Martinez Garcia May 23 '11 at 14:29
• Right now their is nothing in the journal. I guess unpublishable mathematics does not get published. – Spice the Bird Sep 23 '12 at 20:48
• That is the joke – Jesus Martinez Garcia Oct 10 '12 at 17:34

If $\frac{8}{0}= \infty$, then $\frac{n}{0}= n^{'}$, where $n^{'}$ is $n$ rotated by 90 degrees on the right.

• I think a better version of this is $\lim_{\omega\to\infty}3=8$, but I'm not sure this a "joke in the sense of Littlewood." – Gerry Myerson May 12 '11 at 11:39