A Positive Reminder by J. A. Lindon

“A carpenter named Charlie Bratticks,
Who had a taste for mathematics,
One summer Tuesday, just for fun,
Made a wooden cube side minus one.

Though this to you may well seem wrong,
He made it minus one foot long,
Which meant (I hope your brains aren’t frothing)
Its length was one foot less than nothing,

Its width the same (you’re not asleep?)
And likewise minus one foot deep;
Giving, when multiplied (be solemn!),
Minus one cubic foot of volume.

With sweating brow this cube he sawed
Through areas of solid board;
For though each cut had minus length,
Minus times minus sapped his strength.

A second cube he made, but thus:
This time each one-foot length was plus:
Meaning of course that here one put
For volume, plus one cubic foot.

So now he had, just for his sins,
Two cubes as like as deviant twins:
And feeling one should know the worst,
He placed the second in the first.

One plus, one minus — there’s no doubt
The edges simply canceled out;
So did the volume, nothing gained;
Only the surfaces remained.

Well may you open wide your eyes,
For those were now of double size,
On something which, thanks to his skill,
Took up no room and measured nil.

From solid ebony he’d cut
These bulky cubic objects, but
All that remained was now a thin
Black sharply-angled sort of skin

Of twelve square feet — which though not small,
Weighed nothing, filled no space at all.
It stands there yet on Charlie’s floor;
He can’t think what to use it for!”

This poem by J. A. Lindon[1] was popularized by the inimitable Martin Gardner in “The Magic and Mystery of Numbers[2]”.

Thank you Alon Amit in Quora


[1] J. A. Lindon – Wikipedia

[2] https://books.google.com/books?i…


Analytic Continuation of the Riemann Zeta Function

The Riemann Zeta function is defined as the series

$$ \zeta(s)=\sum_{n=1}^{+\infty} \frac{1}{n^s} \ \ \ \ \ \ \ \ \ \ (1)$$

which converges for \(\Re (s)>1\).

It can be defined by analytic continuation to the whole complex plane except \(s=1\) through the Riemann functional equation

$$ \zeta(s)=2^s \pi^{s-1} \sin \left({\pi s \over 2} \right) \Gamma(1-s) \zeta(1-s) \ \ \ \ \ \ \ \ \ \ (2)$$

where \(\Gamma(s)\) is the gamma function.  

We want to derive the integral equation and obtain the values of the \(\zeta\) function for negative integers

$$\zeta(-n)=-{B_{n+1}\over n+1} ,\ \ \ \ \ \ \ \ \ \ (3)$$

and in particular the value

$$\zeta(-1)=\sum_{n=1}^{+\infty} n=-{1\over 12}.\ \ \ \ \ \ \ \ \ \ (4) $$

Using smoothed sums

Terry Tao gives a beautiful interpretation to the dry analytic continuation result using smoothed sums. It shows, for example, that

$$ \sum_{n=1}^{+\infty} \eta(n/N) = -{1\over 2} + C_{\eta,0} N + O(\frac{1}{N}) \ \ \ \ \ \ \ \ \ \ (5) $$

where \(\eta(x)\) is a smooth cutoff function with support in \([0,1]\) and \(C_{\eta,s}\) is the Mellin transform of the cutoff function

$$ C_{\eta,s}=\int_0^\infty x^s\eta(x)dx \ \ \ \ \ \ \ \ \ \ (6) $$

This expression, \((5)\), shows that the result of an analytic continuation of a divergent sum like this gives the unique non-divergent part of the series.


Poisson summation formula

In this note we will follow the presentation given by Ziman .

Consider \(f(x)\) an arbitrary function defined in the interval \([n,n+1]\). We can write

\begin{equation}f(x)=\sum_{s=-\infty}^{+\infty}e^{-i2\pi xs}g_s\label{fourier}\end{equation}

with \(g_s\) given by

\begin{equation} g_s=\int_n^{n+1} dx f(x) e^{i2\pi sx} \label{fourier-coefficient}\;.\end{equation}

We can evaluate Eq. \eqref{fourier} at the point \(n+\gamma\) with \(\gamma\in[0,1]\) to obtain

\begin{equation}f(n+\gamma)=\sum_{s=-\infty}^{+\infty} e^{i2\pi s\gamma} \int_n^{n+1} dx f(x) e^{i2\pi sx}\;.\label{fn}\end{equation}

Now we can sum over \(n\) to obtain the Poisson summation formula

\begin{equation}\sum_{n=-\infty}^{+\infty}f(n+\gamma)=\sum_{s=-\infty}^{+\infty} e^{i2\pi s\gamma} \int_{-\infty}^{+\infty} dx f(x) e^{i2\pi sx}\;.\label{pois}\end{equation}

Application: A very useful identity in Condensed Matter Physics

If \(f(x)=\delta(x)\) then
\begin{equation}\sum_{n=-\infty}^{+\infty}\delta(n+\gamma)=\sum_{s=-\infty}^{+\infty} e^{i2\pi s\gamma} \;.\label{conmat}\end{equation}

Application: Jacobi imaginary transformation

We can use the Poisson summation formula in Eq. \eqref{pois} just derived to obtain the formula used in the evaluation of Ewald sums.

\[\sum_{k=-\infty}^{+\infty} e^{-k^2t}=\sqrt{\pi \over t}\sum_{p=-\infty}^{+\infty} e^{-{\pi^2 p^2\over t}}\label{jacobi}\]