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

Jokes in the sense of Littlewood: examples?

2010-09-15T18:30:40Z

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 by Timothy Chow for Jokes in the sense of Littlewood: examples?

2010-09-15T20:35:37Z

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 by Colin Reid for Jokes in the sense of Littlewood: examples?

2010-09-15T21:32:12Z

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 by Vivek Shende for Jokes in the sense of Littlewood: examples?

2010-09-15T21:50:11Z

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 by Richard Borcherds for Jokes in the sense of Littlewood: examples?

2010-09-15T23:47:29Z

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 by Daniel Moskovich for Jokes in the sense of Littlewood: examples?

2010-09-15T23:54:06Z

The field with one element seems a good example.

Answer by Peter Arndt for Jokes in the sense of Littlewood: examples?

2010-09-15T23:54:39Z

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

Answer by Gerry Myerson for Jokes in the sense of Littlewood: examples?

2010-09-16T00:27:31Z

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 by dke for Jokes in the sense of Littlewood: examples?

2010-09-16T00:35:29Z

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 by Gerry Myerson for Jokes in the sense of Littlewood: examples?

2010-09-16T00:36:07Z

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 by dfranke for Jokes in the sense of Littlewood: examples?

2010-09-16T00:44:54Z

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

Answer by J. M. for Jokes in the sense of Littlewood: examples?

2010-09-16T01:04:59Z

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.

Answer by fedja for Jokes in the sense of Littlewood: examples?

2010-09-16T01:10:43Z

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 by Jamie Weigandt for Jokes in the sense of Littlewood: examples?

2010-09-16T01:20:09Z

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

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

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

Answer by Zev Chonoles for Jokes in the sense of Littlewood: examples?

2010-09-16T01:39:37Z

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 by Victor Protsak for Jokes in the sense of Littlewood: examples?

2010-09-16T05:36:32Z

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:

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

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

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

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

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

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

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.

Answer by gowers for Jokes in the sense of Littlewood: examples?

2010-09-16T09:26:02Z

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.

Answer by Yemon Choi for Jokes in the sense of Littlewood: examples?

2010-09-16T10:38:21Z

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).

Answer by Denis Serre for Jokes in the sense of Littlewood: examples?

2010-09-16T11:57:00Z

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

Answer by Keivan Karai for Jokes in the sense of Littlewood: examples?

2010-09-16T12:04:13Z

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 by Thierry Zell for Jokes in the sense of Littlewood: examples?

2010-09-16T12:31:59Z

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 by Joel David Hamkins for Jokes in the sense of Littlewood: examples?

2010-09-16T12:51:29Z

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 by Qfwfq for Jokes in the sense of Littlewood: examples?

2010-09-16T12:56:54Z

Lagrange's equation:

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

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

Answer by Someone for Jokes in the sense of Littlewood: examples?

2010-09-16T14:00:25Z

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 by Richard Stanley for Jokes in the sense of Littlewood: examples?

2010-09-16T16:10:20Z

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 by Dan Piponi for Jokes in the sense of Littlewood: examples?

2010-09-16T18:13:27Z

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 ${\partial^n\over\partial x^n}(fg) = (D+E)^n(fg) = \sum_{i=0}^n \binom{n}{i} D^i E^{n-i}(fg) = \sum_{i=0}^n \binom{n}{i} {\partial^i f\over\partial x^i}{\partial^{n-i} g\over\partial x^{n-i}}$.

Answer by Qiaochu Yuan for Jokes in the sense of Littlewood: examples?

2010-09-16T19:58:37Z

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.

Answer by Qiaochu Yuan for Jokes in the sense of Littlewood: examples?

2010-09-16T20:04:40Z

An expansion on Timothy Chow's example of Grandi's series $1 - 1 + 1 - 1 \pm ... = \frac{1}{2}$. 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 by Jesse Madnick for Jokes in the sense of Littlewood: examples?

2010-09-16T21:28:08Z

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 by Jesse Madnick for Jokes in the sense of Littlewood: examples?

2010-09-16T22:30:53Z

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

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

Answer by Haha tell me a Noether one for Jokes in the sense of Littlewood: examples?

2010-09-17T00:18:45Z

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 by ABayer for Jokes in the sense of Littlewood: examples?

2010-09-17T02:03:57Z

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 by Per Vognsen for Jokes in the sense of Littlewood: examples?

2010-09-17T03:17:53Z

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 $\pm n\pi$.

Answer by Richard Borcherds for Jokes in the sense of Littlewood: examples?

2010-09-17T03:20:44Z

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 by Peter Humphries for Jokes in the sense of Littlewood: examples?

2010-09-17T04:13:55Z

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 by Veit Elser for Jokes in the sense of Littlewood: examples?

2010-09-17T12:36:39Z

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}

Answer by Michael Hardy for Jokes in the sense of Littlewood: examples?

2010-09-17T12:49:41Z

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.

Answer by zeb for Jokes in the sense of Littlewood: examples?

2010-09-17T14:28:15Z

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 $A_m$.)

Answer by Darsh Ranjan for Jokes in the sense of Littlewood: examples?

2010-09-19T04:06:09Z

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 by Josh Guffin for Jokes in the sense of Littlewood: examples?

2010-09-19T14:11:38Z

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.

Answer by J. M. for Jokes in the sense of Littlewood: examples?

2010-09-19T18:50:03Z

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.

Answer by J. M. for Jokes in the sense of Littlewood: examples?

2010-09-19T18:59:49Z

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.

Answer by Christian Bjartli for Jokes in the sense of Littlewood: examples?

2010-09-19T19:45:20Z

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

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

Answer by ABayer for Jokes in the sense of Littlewood: examples?

2010-09-19T20:31:12Z

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. $n \cdot 2^{n-1}$.

Answer by Fedor Petrov for Jokes in the sense of Littlewood: examples?

2010-09-19T20:34:14Z

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?

Answer by Gerry Myerson for Jokes in the sense of Littlewood: examples?

2010-09-24T07:17:19Z

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 $\pi^2/4$.

Answer by Gunnar Magnusson for Jokes in the sense of Littlewood: examples?

2010-09-24T08:30:29Z

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 by Martin Brandenburg for Jokes in the sense of Littlewood: examples?

2010-09-24T09:52:38Z

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.

Answer by auniket for Jokes in the sense of Littlewood: examples?

2010-09-25T08:16:22Z

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.

Answer by Gerald Edgar for Jokes in the sense of Littlewood: examples?

2010-09-25T14:07:14Z

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 by Maurizio Monge for Jokes in the sense of Littlewood: examples?

2010-09-25T16:56:26Z

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.

Answer by Pablo Lessa for Jokes in the sense of Littlewood: examples?

2010-09-25T18:26:22Z

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.

Answer by I. J. Kennedy for Jokes in the sense of Littlewood: examples?

2010-09-28T16:43:53Z

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.

Answer by Jim Bryan for Jokes in the sense of Littlewood: examples?

2010-10-03T06:00:14Z

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).

Answer by Mahdiyar for Jokes in the sense of Littlewood: examples?

2010-10-03T07:28:51Z

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.

Answer by Harrison Brown for Jokes in the sense of Littlewood: examples?

2010-10-29T19:16:15Z

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.)

Answer by Allen Knutson for Jokes in the sense of Littlewood: examples?

2010-11-30T04:32:50Z

Given a function $f$ on the real line, let's compute the function $\Sigma f$, taking $n \mapsto f(1) +\ldots + f(n)$. Well, $\Sigma = 1/\Delta$, where $\Delta$ is the differencing operator $shift - 1$. And the shift operator is the exponential of the differentiation operator (this being, essentially, Taylor's theorem). Hence $$ \Sigma = \frac{1}{e^D - 1} = \frac{1}{D} \frac{D}{e^D-1} $$ 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: $\Sigma = \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 by Jesus Martinez Garcia for Jokes in the sense of Littlewood: examples?

2011-05-12T04:21:30Z

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

Answer by unknown (google) for Jokes in the sense of Littlewood: examples?

2011-05-12T07:12:13Z

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

Answer by Goldstern for Jokes in the sense of Littlewood: examples?

2011-05-12T13:59:23Z

(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)".