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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.

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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
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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

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: $$ \begin{array}{ccl} \left( \sum_{n} \frac{(-1)^n}{2n+1} \right)^2 &=& \sum_{m,n} \frac{(-1)^{m+n}}{(2m+1)(2n+1)} \\ &=& \sum_{m,k} \frac{(-1)^k}{(2m+1)(2m+2k+1)} \\ &=& \sum_{m} \frac{1}{(2m+1)^2} + \sum_{k \neq 0 } \sum_m \left( \frac{(-1)^k}{2 k (2m+1)} - \frac{(-1)^k}{2 k (2m+2k+1)} \right) \\ &=& \sum_{m} \frac{1}{(2m+1)^2} \\ \end{array}$$ Of course, one needs to justify rearranging the conditionally convergent series, but that spoils the joke.

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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)

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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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"Hmmm... And how is it obvious that evaluating ∫ without any function gives x?" Well, doesn't "Let "$\int$" denote $∫_0^x$." clearly mean: "Let "$\int f\;$" denote $∫_0^xf(x) dx$." If so, then $\int = \int _0 ^x dx = x$. –  Dick Palais Sep 16 '10 at 3:40
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I think it should read $1 = (1 - \int) f$. Then $f = (1 - \int)^{-1} 1$, which resolves the question about evaluating $\int$ without an input. –  Jonathan Wise Sep 16 '10 at 16:33
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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

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-Alvarez 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

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

(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)".

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

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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

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

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The journal is a "joke in the sense of Littlewood"? –  Gerry Myerson May 12 '11 at 11:41
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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

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.

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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.

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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

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.)

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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.

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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.

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This was already mentioned by Per Vognsen in his answer of 19 September, no? –  Gerry Myerson Oct 3 '10 at 12:22
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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 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).

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This is the same as the answer of AByer on 17 September, no? –  Gerry Myerson Oct 3 '10 at 12:24

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.

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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
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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

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.

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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

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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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.

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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

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

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

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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

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.

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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
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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
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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

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.

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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?

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

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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.

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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.

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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.

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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

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 $\pm n\pi$.

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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

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

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 $A_m$.)

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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

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