The n-Category Café

But I used to read quite a lot in English. As I age, it seems harder to work up a thrill that way.

Yes, I feel the same, after I went to college and learned about math and physics, I almost only read books about that, and almost no fiction any more (although Tammo tom Diek used to tell his students that they should do that from time to time, and not try to think about mathematics all day).

If you asked me how much classic German fantasy and science fiction I’d read, then you’d have me at a disadvantage.

Great, that is a chance that I’ll get only once in a lifetime, and I will jump at it even though it will be completly off topic for this blog:

As a teenager I used to spend as much time as I could afford in the public library in the small town near the village where I grew up - that was before there was an internet or even more than three TV channels. So my knowledge of literature consists of the books that the library had in these days.

They had mostly German stuff, and much from the era of romanticism. So I happen to know that e.g. Poe borrowed much for his story about the house of Usher from the German romanticist E.T.A. Hoffmann, from the story “Das Majorat” from 1819 from the collection “Nachtst¼cke”. This is explained in an article by Arno Schmidt. Schmidt himself promoted many forgotten German authors, and wrote a monumental book “Zettels Traum” where Poe plays a major role, and which is as ingenious and hard to read as Joyce’s “Finnegans Wake”.

BTW, one feat of Schmidt was the massive use of puns and punctuation to create some interesting effects, especially in the (much easier to read than Zettels Traum) “Kaff auch Mare Crisium” - which interlaces two stories, one of them about the rest of humanity living on the moon after the earth was made uninhabitable (that’s what the “Mare Crisium” in the title alludes to, “Kaff” is from a northern German dialect and means “chaff”, that alludes to humanity that is like chaff blown away by the wind of fate :-).

One trivial example of this would be:

“The (dis)advantage!; of using differential forms is that one does not, $\frac{have}{need}$, coordinates…”

It seems to me that he was - in this respect - at least 40 years ahead of his time…

Oh, and there is a German author who continued the Cthulhu myth of Lovecraft, Wolfgang Hohlbein (that is not mentioned by wikipedia). If you know any kids who are curious about the Nibelungen sage, I would recommend his book “Hagen von Tronje”, that is really interesting: The great hero Siegfrid is a warmonger and the villain Hagen is the hero that tries to stop Siegfrid from destroying the kingdom of Burgund by his foolishness.

The concept of science as antientropic activity and therefore rooted in an ethics of responsibility, embodied in the novel by “Vecherovsky” and opposed to the “jellyfishes” of the science community (embodied by “Malianov”: “a line of fire and brimstone that could never be crossed was drawn between Vecherovsky and me” … “Maybe he’ll just sit and pour over the work and try to discover the point of intersection of the theory of … and the qualitative analysis of … , and probably that will be a very strange point of intersection. And I will stay home, meet my mother-in-law and Bobchick at the plane tomorrow, and we’ll all go out and buy the bookshelves together.”), leads back to the Grothendieck school - as most readers of this blog surely know, “Vecherovsky” is modeled after one of it’s members. Grothendieck’s remarks on ‘responsibility’ look as if he would have extended it towards a responsibility of expressing mathematical ideas as clearly and explicit as possible, which somehow conflicts his statements that the published volumes of SGA are lacking expositions of some basic ideas or mental imagery behind the special constructions which he had discussed in the seminars. It would be interesting to know more about that.

I was being sloppy there. If $f$ is continuous then certainly $\vert f(x_0) \vert = \Vert f \Vert_2$ for some $x_0$.

Nice example. We can make it look a little more like the original one by writing it as $$\left(1+1-\frac{1}{3}-\frac{1}{3}+\frac{1}{5}+\frac{1}{5} + \cdots \right)^2 =1^2 + 1^2 + \frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{5^2} + \cdots.$$

One way to generate more examples would be to simply pick any $f$ that has a constant modulus, …

Good point! Of course there’s the issue of whether $f$ is regular enough that $f(0)$ is the sum of the Fourier coefficients. This is a sticky subject in general but one simple sufficient condition is that $f$ be piecewise continuous on $[-\pi, \pi]$ and differentiable at $0$.

Cool! We don’t really need $\alpha$ to be irrational. All we need is that $\alpha$ isn’t an integer — right?

I wonder why I don’t remembering seeing this formula before:

$$\cdots -\frac{1}{\alpha+3}+\frac{1}{\alpha+2}-\frac{1}{\alpha+1}+ \frac{1}{\alpha}-\frac{1}{\alpha-1}+ \frac{1}{\alpha-2}-\frac{1}{\alpha-3}+ \cdots =\frac{\pi}{sin(\pi \alpha)}$$

It’s very cute. Now that I’ve seen it, I can sketch an ‘effortless proof’ based on the principle that ‘two analytic functions with simple poles at the same places, with the same residues, can’t be all that different’.

Here it is:

Both sides are analytic functions of $\alpha$, with poles only at the integers. Both have simple poles at the integers, and for both the residue at each pole is $1$. So, the difference of the two sides is an entire function (an analytic function with no poles). But the difference is also invariant under the translation $\alpha \to \alpha + 1$, and it goes to zero as the imaginary part of $\alpha$ goes to $\pm \infty$. So the difference is bounded — this needs a bit of work). But Liouville’s theorem says a bounded entire function is constant — so a bounded entire function of $\alpha$ that goes to zero as as the imaginary part of $\alpha$ goes to $\pm \infty$ must be zero! So, the difference of the two sides is zero.

The same type of argument should show that

$$\cdots + \left(\frac{-1}{\alpha+3}\right)^n+ \left(\frac{1}{\alpha+2}\right)^n+ \left(\frac{-1}{\alpha+1}\right)^n +\left(\frac{1}{\alpha}\right)^n +\left(\frac{-1}{\alpha-1}\right)^n +\left(\frac{1}{\alpha-2}\right)^n+ \left(\frac{-1}{\alpha-3}\right)^m+ \cdots =\frac{\pi^n}{sin(\pi \alpha)^n}$$

for any natural number $n \ge 1$. But now instead of simple poles we have poles of order $n$.

So — if I haven’t screwed up — this is a very nice example, where taking the $n$th power of the summands has the effect of giving the $n$th power of the sum for all $n = 1,2,3, \dots$.

Though I agree with you that $\frac{\pi^n}{\sin(\pi\alpha)^n}$ has poles of order $n$ at every $m\in \mathbb{Z}$, with coefficients $(-1)^{n m}$, I can’t get agreement on the sums for $n\gt 2$. For example, I get:

$$...-\frac{1}{(\alpha +3)^3}+\frac{1}{(\alpha +2)^3}-\frac{1}{(\alpha +1)^3}+\frac{1}{\alpha ^3}-\frac{1}{(\alpha -1)^3}+\frac{1}{(\alpha -2)^3}-\frac{1}{(\alpha -3)^3}+...= \frac{\pi ^3 (\cos (2 \pi \alpha )+3)}{4 \sin ^3(\pi \alpha )}$$

So there’s an entire function, $\frac{\cos (2 \pi \alpha )+3}{4 }$, equal to 1 on the integers, multiplying the RHS.

Those Fourier series arguments from the mailing list are very nice!

Egan wrote:

Though I agree with you that $\frac{\pi^n}{\sin(\pi\alpha)^n}$ has poles of order $n$ at every $m\in \mathbb{Z}$, with coefficients $(-1)^{n m}$, I can’t get agreement on the sums for $n\gt 2$.

Whoops! Pride goeth before a fall. I think the problem is that, taking $n = 3$ for example, and looking near $z = 0$:

$$\frac{1}{\sin^3 z} \approx \frac{1}{(z - z^3/6)^3} \approx \frac{1}{z^3 - z^5/2} = \frac{1}{z^3} \frac{1}{1 - z^2/2} \approx \frac{1}{z^3} (1 + z^2/2) = \frac{1}{z^3} + \frac{2}{z}$$

so in addition to the pole of order $3$ at zero we’re also getting a nonzero residue there, which screws things up. The case $n = 2$ is interesting; I seem to get

$$\frac{1}{\sin^2 z} \approx \frac{1}{z^2} + 3$$

for small $z$, but somehow that’s okay. And for $n = 1$ my argument was really correct, I think.

JB wrote:

The case $n=2$ is interesting; I seem to get

$$\frac{1}{\sin^2 z}\approx\frac{1}{z^2}+3$$

for small z, but somehow that’s okay.

I get:

$$\frac{1}{\sin^2 z}\approx\frac{1}{z^2}+\frac{1}{3}$$

I think what makes this OK is that it matches the sum over $n$ of all the constant terms in the Taylor expansions of the $\frac{1}{(\alpha-n)^2}$ terms at $\alpha=0$. Specifically:

$$\frac{1}{(\alpha-n)^2}\approx\frac{1}{n^2}-\frac{2\alpha}{n^3}+...$$

$$\sum_{n\in\mathbb{Z}, n\neq 0}\frac{1}{n^2}=\frac{\pi^2}{3}$$

$$\frac{\pi^2}{\sin^2 \pi\alpha}\approx\frac{1}{\alpha^2}+\frac{\pi^2}{3}$$

It’s not a huge amount of work to show that all the terms of the Taylor series at 0 agree … but your argument about the difference being a bounded entire function looks simpler.$