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May 10, 2010

This Week’s Finds in Mathematical Physics (Week 297)

Posted by John Baez

In week297 of This Week’s Finds, see some knot sculptures by Karel Vreeburg:

Read about special relativity in finance. Learn about lazulinos. Admire some peculiar infinite sums. Ponder a marvelous property of the number 12. Then: learn about the role of Dirichlet forms in electrical circuit theory!

Posted at May 10, 2010 4:38 AM UTC

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Re: This Week’s Finds in Mathematical Physics (Week 297)

You mentioned the

Barratt-Priddy-Quillen therem.

First, there’s a missing ‘o’. Second, is this the same as the Barratt-Priddy-Quillen-Segal theorem? Third, what precisely does it say?

Is there a place where you can find stored the group cohomology of S nS_n?

Posted by: David Corfield on May 10, 2010 9:27 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Thanks for the corrections and questions!

The Barratt-Priddy-Quillen theorem is essentially the same as the Barratt-Priddy-Quillen-Segal theorem. I’m not sure what the official statement of this theorem is, since (as the name suggests) it was built up a bit at a time by these four authors. In the interests of clarity I changed my remark to:

After all, stable homotopy groups of spheres are related to the cohomology of symmetric groups, since the group completion of the classifying space of the groupoid of finite sets is Ω S \Omega^\infty S^\infty — see “week199” if you don’t know what I’m talking about here. But I’m confused about the numbers 12 versus 24 here, and also the role of /\mathbb{Q}/\mathbb{Z} coefficients.

Does someone know a place where you can look up cohomology groups of the symmetric groups?

I can expand on the above a bit:

The group cohomology of the symmetric group S nS_n is the same as the cohomology of its classifying space BS nB S_n. The disjoint union of these spaces BS nB S_n becomes a topological monoid thanks to disjoint union of finite sets, which gives maps S n×S mS n+mS_n \times S_m \to S_{n+m} and thus BS n×BS mBS n+mB S_n \times B S_m \to B S_{n+m}. We can turn this topological monoid into a topological group by throwing in inverses — that’s ‘group completion’. And the result is Ω S \Omega^\infty S^\infty, the space whose homotopy groups are the stable homotopy groups of spheres.

Posted by: John Baez on May 10, 2010 5:09 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

…100ms is quite stale if you are seeing the markets plunge the way they were…

LOL. A collegue of mine sitting next to me today has recently installed trading software on a supercomputer cluster for a similar reason (it’s not that the “arcane mathematical models” need that computing power, but that the communication has to be as fast as possible).

But seriously, I would expect that the time consumed by the handshake of the used network protocols is far more important to the latency than the limited signal velocity.

(If you are looking for yet another career change we could be renting micro-wormwholes connecting the worlds major stock markets.)

Posted by: Tim van Beek on May 10, 2010 10:11 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Next time I want to bring inductors and capacitors into the game

Don’t forget memristors!

Posted by: Sjoerd Visscher on May 10, 2010 10:53 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

I’ll only be talking about linear circuit elements; memristors are inherently nonlinear. Or more precisely: a linear memristor is the same as a resistor. See week294 for an explanation of why.

Posted by: John Baez on May 10, 2010 6:52 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

For a while, about a trillion dollars had evaporated!

One wonders if the the dollar size of the world economy is canonical adjoint to the expected time to complete a stock transaction; the more precise we expect timing to be, the more prone is the market to HUGE variations. Hmmm… I therefore also want to ask whether the economy is closer to being thermodynamics or quantum mechanics? And is it renormalizable? And do we really want an economy that works like that?

Posted by: some guy on the street on May 10, 2010 3:56 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

I’m curious. Did nobody notice that one of the items in week297 was a joke — or did some of you notice but were just too blasé to say anything?

Actually I have a feeling that readership of the nn-Café went down after the recent crash. That could be part of it too.

Posted by: John Baez on May 15, 2010 12:54 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Actually I have a feeling that readership of the n-Café went down after the recent crash.

Is there any way to check that? Do we have statistics from the web server on readership? It would be sad if we lost a lot of people because we were down for a week. Maybe they’ll come back over time.

Posted by: Mike Shulman on May 15, 2010 2:05 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

I got three hundred and fifty visitors to my site the day you published it, but not one left a comment.

Posted by: Mike Stay on May 15, 2010 8:13 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Did nobody notice that one of the items in week297 was a joke…?

Are you going to tell us which one? There are many statements that are entertaining, several statements that obviously are jokes (I take it that you do not mean those), but I did not notice anything that is wrong (in such an obvious way that I would notice it) or nonsense

Posted by: Tim van Beek on May 15, 2010 9:39 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

You must be getting too subtle for us. I did notice that your references are numbered 1), 1), 2), 3),…, but that’s not very funny.

Posted by: David Corfield on May 15, 2010 9:46 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

I thought Mike Stay’s lazulino joke was delightful, but it’s always hard to know how to respond to whimsical semi-hoaxes like this without spoiling the fun for later arrivals.

Posted by: Greg Egan on May 15, 2010 10:17 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

I liked it and the short story by Borges, too, but the internet and the media are so full of much more absurd pseudoscientific content - that’s not intended to be a hoax - that I feel totally overfed.

(I still like the note on piffles that you all know, I presume).

Posted by: Tim van Beek on May 15, 2010 10:57 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Thanks for noting the mistake in numbering, David. Given the other joke, that could have been intentional — but it wasn’t: I often screw up the numbering, because I keep adding new items all over the place as I gradually compose an issue of This Week’s Finds. I’ve fixed this mistake.

Thanks for not giving away the joke, Greg. I found the mathematical story by Borges at the core of this joke to have the slippery texture of a nightmare, as ‘blue tigers’ give way to blue discs, and then… oh, the horror!

Adding to the dreamlike quality was the fact that I thought I’d read all of Borges’ stories… but I’d never seen this one.

It turns out that Borges did not invent the idea of a blue tiger.

Posted by: John Baez on May 15, 2010 3:47 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Maybe you will like - if you do not know it already - some of the works of Edgar Allen Poe like the raven, and of his follower Lovecraft.

This is less playful and more nightmare-like.

Posted by: Tim van Beek on May 15, 2010 4:11 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

I’ve read most of Poe and Lovecraft. If you asked me how much classic German fantasy and science fiction I’d read, then you’d have me at a disadvantage. But I used to read quite a lot in English. As I age, it seems harder to work up a thrill that way. The Borges story did it, but that’s rare. By contrast, it seems easier and easier for me to explore realms of mathematics that inspire awe and something akin to terror: Kronecker’s ‘Jugendtraum’, Grothendieck’s ‘standard conjectures’, the Langlands program, Witten’s ideas on the Monster group and black holes…

Even worse, all these are related! Something big is going on — something so big it’s downright scary.

In fact that’s why I’ve been saving this line from Lovecraft for a quote at the end of a suitable This Week’s Finds:

The sciences… have hitherto harmed us little; but some day the piecing together of dissociated knowledge will open up such terrifying vistas of reality, and of our frightful position therein, that we shall either go mad from the revelation or flee from the deadly light into the peace and safety of a new dark age.

I don’t believe it, but it describes a feeling I get sometimes…

Posted by: John Baez on May 15, 2010 5:47 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

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 “Nachtstücke”. 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, haveneed\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.

Posted by: Tim van Beek on May 16, 2010 9:15 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Unusual…a Borges story with a happy ending. Though not as happy as, say, ‘The Secret Miracle’…

Posted by: Tigers,... on May 16, 2010 7:22 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Thanks!

Posted by: Mike Stay on May 15, 2010 10:50 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

“it seems harder to work up a thrill that way …” - reviews are sometimes shorter and funnier, like here.

Posted by: Thomas on May 15, 2010 9:34 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Ah yes, A Perfect Vacuum. I read and loved everything I could get my hands on by Stanislaw Lem back when I was in college. If you haven’t read everything by him already, I recommend The Futurological Congress.

Posted by: John Baez on May 16, 2010 3:38 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

“Snail on the Slope” by the Strugatskii brothers is fascinating too. Here seems to be a translation online, but I have not looked closer to that. The two parallel stories in it are a long reflection on the troubled relation between organized civilisation and nature. I like especially the new word ‘knowledge-sickness’ in analogy to ‘home-sickness’ (“Wissensweh” in the German translation) from which the main characters suffer, but without which they wouldn’t like to live. In contrast to them, the well-adapted people in the science community resp. forest tribes live physically and mentally in a kind of dulling symbiosis with their surrounding. Some people claim that Cameron took many ideas of his “Avatar”-movie from this and other Strugatskii novels. This story’s main character is modeled after a well known mathematician.

Posted by: Thomas on May 16, 2010 11:06 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Some people claim that Cameron took many ideas of his “Avatar”-movie from this and other Strugatskii novels.

Maybe, if anyone needs an inspiration to come up with a story like that of “Avatar”…I was reminded of Winnetou I, but am sure that most US-Americans will never have heard of that, ironically.

But “Definitely Maybe” may have inspired this scientific work: Card game restriction in LHC can only be successful!

Posted by: Tim van Beek on May 17, 2010 9:08 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

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.

Posted by: Thomas on May 17, 2010 12:00 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

In week297 I mentioned an observation from Robert Baillie. Take this series:

π8=1+1/31/51/7+1/9+1/11 \frac{\pi}{\sqrt{8}} = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - \cdots

Square each term, add them up… and you get the square of what we had:

π8=1+1/3 2+1/5 2+1/7 2+1/9 2+1/11 2+ \frac{\pi}{8} = 1 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9^2 + 1/11^2 + \cdots

In email, Harun Šiljak mentions another “coincidence”, relating sums to integrals:

0 1x xdx= n=1 n n \int_0^1 x^{-x} d x = \sum_{n=1}^\infty n^{-n}

He says this is a famous identity due to Johann Bernoulli. I’d never seen it. Anyone know how it’s proved? Also this:

0 1x xdx= n=1 (n) n \int_0^1 x^x d x = -\sum_{n=1}^\infty (-n)^{-n}

Posted by: John Baez on May 17, 2010 12:15 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

0 1x xdx= n=1 n n\int_{0}^{1} x^{-x} d x = \sum_{n=1}^\infty n^{-n}

One of my favorite identities of all time!

It’s not that hard; if memory serves, just write

x x= n=0 (1) n(xlnx) nn!x^{-x} = \sum_{n=0}^\infty (-1)^n \frac{(x \ln x)^n}{n!}

and expand the integral term-by-term, using integration by parts and induction at each step. I should try this myself just to make sure, but I think it just pops right out like magic.

Posted by: Todd Trimble on May 17, 2010 2:25 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Those series for π/8\pi/\sqrt{8} and π 2/8\pi^2/8 are cute, but I think this is just one example of a pretty generic phenomenon. If (a n)(a_n) is any sequence of real numbers which converges to 00 but which is not absolutely summable, then by changing signs appropriately you can make na n\sum_n a_n converge conditionally to any sum you choose; this is a minor variation of Riemann’s famous rearrangement theorem. In particular, if you start with any (a n) 2 1(a_n) \in \ell^2 \setminus \ell^1, by sprinkling in some minus signs you can always arrange to have ( na n) 2= na n 2\big(\sum_n a_n\big)^2 = \sum_n a_n^2.

Posted by: Mark Meckes on May 17, 2010 3:05 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Good point! But just to keep things from getting too dull, here’s a question: does anyone know any other examples where the ‘sprinkling of minus signs’ can be done in such a systematic way? Is there any nice theory of this?

(In case it’s not obvious, the pattern of signs here:

π/8=1+1/31/51/7+1/9+1/11 \pi/\sqrt{8} = 1+ 1/3 − 1/5 − 1/7 + 1/9 + 1/11− \cdots

is supposed to go ++++++++--++--++-- \cdots.)

Posted by: John Baez on May 17, 2010 8:16 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Interesting questions! I don’t have answers to them, but here are some related thoughts. Suppose the sequence (a n)(a_n) is the Fourier coefficients of some periodic function ff. For simplicity for the moment I’ll assume (a n)(a_n) gives the coefficients for the exponential Fourier and is indexed by all integers: f(x)= na ne i2πnx f(x) = \sum_{n\in \mathbb{Z}} a_n e^{i2\pi n x} in L 2[0,1]L^2[0,1]. Then n|a n| 2=f 2 2 \sum_{n\in \mathbb{Z}} \vert a_n \vert ^2 = \Vert f \Vert_2^2 and na n=f(0) \sum_{n\in \mathbb{Z}} a_n = f(0) assuming (!) that ff is well-behaved enough. So after inverting the Fourier transform, the phenomenon we’re seeing here is that f(0)=f 2f(0) = \Vert f \Vert_2. Assuming (!) that ff is continuous, there is certainly some x 0x_0 such that f(x 0)=f 2f(x_0) = \Vert f \Vert_2, and it’s just a matter of translation to assume x 0=0x_0=0, which amounts to multiplying the (exponential) Fourier coefficients by constants of modulus 11.

Does anyone happen to know which function has the Fourier series with the coefficients 1,1/3,1/5,1/7,1/9,1/11,1,1/3,-1/5,-1/7,1/9,1/11,\ldots?

(!) One drawback of thinking about this phenomenon in this way is that one can’t necessarily easily check the required niceness of ff from the Fourier coefficients.

Posted by: Mark Meckes on May 18, 2010 4:14 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

I was being sloppy there. If ff is continuous then certainly |f(x 0)|=f 2\vert f(x_0) \vert = \Vert f \Vert_2 for some x 0x_0.

Posted by: Mark Meckes on May 18, 2010 8:05 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Mathematica (when suitably cajoled) gave me:

f(t)=ie it2(tanh 1(e 14i(π2t))+tanh 1(e 14i(3π2t))+tanh 1(e 14i(2t+π))+tanh 1(e 14i(2t+3π)))2f(t)=-\frac{i e^{-\frac{i t}{2}} \left(\tanh ^{-1}\left(e^{\frac{1}{4} i (\pi -2 t)}\right)+\tanh ^{-1}\left(e^{\frac{1}{4} i (3 \pi -2 t)}\right)+\tanh ^{-1}\left(e^{\frac{1}{4} i (2 t+\pi )}\right)+\tanh ^{-1}\left(e^{\frac{1}{4} i (2 t+3 \pi )}\right)\right)}{\sqrt{2}}

having Fourier coefficients a n=1,13,15,17...a_n = 1, \frac{1}{3}, -\frac{1}{5}, -\frac{1}{7} ..., where:

a n=12π π πf(t)e intdta_n = \frac{1}{2\pi} \int_{-\pi}^{\pi}\! f(t) e^{-i n t}\, d t

Posted by: Greg Egan on May 19, 2010 6:36 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

And I should add that with ff as defined above, f(0)f(0) is indeed 2×π82 \times \frac{\pi}{\sqrt{8}} (with the factor of 22 from the extension of the Fourier series to negative nn).

Posted by: Greg Egan on May 19, 2010 7:15 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

The function ff I gave above is discontinuous at ±π2\pm \frac{\pi}{2}. In fact I think f(t)=0f(t)=0 for all |t|>π2|t| \gt \frac{\pi}{2}, while lim t±π2 f(t)=(1i)π2\lim_{t\to \pm \frac{\pi}{2}^{\mp}}f(t)=(1\mp i)\frac{\pi}{2}.

Also, |f(t)|=0|f(t)|=0 for all |t|>π2|t| \gt \frac{\pi}{2}, while |f(t)|=π2|f(t)|=\frac{\pi}{\sqrt 2} for all |t|<π2|t| \lt \frac{\pi}{2}. So alas, I can’t make Mark’s technique work with this function.

Posted by: Greg Egan on May 19, 2010 9:34 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Ah, here’s an example that works, very closely related to the original one but with a less grungy function yielding the Fourier series.

Suppose we define:

f(t)=e it2(tan 1(e it2)+tan 1(e it2))f(t)=e^{\frac{i t}{2}} \left(\tan ^{-1}\left(e^{-\frac{i t}{2}}\right)+\tan ^{-1}\left(e^{\frac{i t}{2}}\right)\right)

Then ff has Fourier coefficients:

a n=12π π πf(t)e intdt=(1) n2n+1a_n=\frac{1}{2 \pi}\int_{-\pi}^\pi f(t) e^{-i n t} d t=\frac{(-1)^n}{2n+1}

Note that this is not quite our original series; the signs merely alternate, and of course this extends to negative nn. We have:

a 1=a 0=1a_{-1}=a_0=1

a 2=a 1=13a_{-2}=a_1=-\frac{1}{3}

a 3=a 2=15a_{-3}=a_2=\frac{1}{5}

a 4=a 3=17a_{-4}=a_3=-\frac{1}{7}

That is, the series is symmetrical under n1nn\to -1-n

The sums are:

S= na n=2(113+1517+...)=f(0)=π2S = \sum_{n\in \mathbb{Z}}a_n = 2 (1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...) = f(0) = \frac{\pi}{2}

na n 2=2(1+13 2+15 2+17 2+...)=π 24=S 2\sum_{n\in \mathbb{Z}}a_n^2 = 2 (1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...) = \frac{\pi^2}{4} = S^2

And if I’ve understood Mark correctly, the Fourier-esque explanation for this is that f 2=|f(0)|\Vert f\Vert_2=\vert f(0)\vert.

Posted by: Greg Egan on May 19, 2010 1:21 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Nice example. We can make it look a little more like the original one by writing it as (1+11313+15+15+) 2=1 2+1 2+13 2+13 2+15 2+15 2+. \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.

Posted by: Mark Meckes on May 19, 2010 2:40 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

I failed to notice that the ff I gave in the last comment simplifies greatly, down to just:

f(t)=π2e it2f(t)=\frac{\pi}{2} e^{\frac{i t}{2}}

I also got the numbering of the a na_n mirror-reversed by sign, using the definition of the Fourier series I specified, but that doesn’t affect the actual sums.

One way to generate more examples would be to simply pick any ff that has a constant modulus, and has f(0)f(0)\in \mathbb{R} and f(0)>0f(0)\gt 0, and hence f(0)=f 2f(0)=\Vert f \Vert_2. Then the sum of its Fourier coefficients (over all the integers), squared, would equal the sum of the squared magnitudes of those coefficients (again over all the integers).

Posted by: Greg Egan on May 19, 2010 2:50 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

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

Good point! Of course there’s the issue of whether ff is regular enough that f(0)f(0) is the sum of the Fourier coefficients. This is a sticky subject in general but one simple sufficient condition is that ff be piecewise continuous on [π,π][-\pi, \pi] and differentiable at 00.

Posted by: Mark Meckes on May 19, 2010 3:01 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Here’s another example. Pick a positive irrational number α\alpha.

f(t)=πe iαtsin(πα)f(t)=\frac{\pi e^{i \alpha t}}{sin(\pi \alpha)}

a n=(1) nαna_n=\frac{(-1)^n}{\alpha-n}

na n=...1α+3+1α+21α+1+1α1α1+1α21α3+...=πsin(πα)\sum_{n\in\mathbb{Z}}a_n=...-\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}+...=\frac{\pi}{sin(\pi \alpha)}

na n 2=...+1(α+3) 2+1(α+2) 2+1(α+1) 2+1α 2+1(α1) 2+1(α2) 2+1(α3) 2+...=π 2sin(πα) 2\sum_{n\in\mathbb{Z}}a_n^2=...+\frac{1}{(\alpha+3)^2}+\frac{1}{(\alpha+2)^2}+\frac{1}{(\alpha+1)^2}+\frac{1}{\alpha^2}+\frac{1}{(\alpha-1)^2}+\frac{1}{(\alpha-2)^2}+\frac{1}{(\alpha-3)^2}+...=\frac{\pi^2}{sin(\pi \alpha)^2}

Posted by: Greg Egan on May 19, 2010 3:27 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

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

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

1α+3+1α+21α+1+1α1α1+1α21α3+=πsin(πα)\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 α\alpha, with poles only at the integers. Both have simple poles at the integers, and for both the residue at each pole is 11. 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 αα+1\alpha \to \alpha + 1, and it goes to zero as the imaginary part of α\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 α\alpha that goes to zero as as the imaginary part of α\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

+(1α+3) n+(1α+2) n+(1α+1) n+(1α) n+(1α1) n+(1α2) n+(1α3) m+=π nsin(πα) n \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 n1n \ge 1. But now instead of simple poles we have poles of order nn.

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

Posted by: John Baez on May 21, 2010 12:10 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

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

...1(α+3) 3+1(α+2) 31(α+1) 3+1α 31(α1) 3+1(α2) 31(α3) 3+...=π 3(cos(2πα)+3)4sin 3(πα)...-\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, cos(2πα)+34\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!

Posted by: Greg Egan on May 21, 2010 2:28 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Egan wrote:

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

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

1sin 3z1(zz 3/6) 31z 3z 5/2=1z 311z 2/21z 3(1+z 2/2)=1z 3+2z\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 33 at zero we’re also getting a nonzero residue there, which screws things up. The case n=2n = 2 is interesting; I seem to get

1sin 2z1z 2+3 \frac{1}{\sin^2 z} \approx \frac{1}{z^2} + 3

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

Posted by: John Baez on May 21, 2010 1:43 PM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Yes, you need to match the whole principal part, not just the residues; for a pole of order nn, the principal part lives in an nn-dimensional vector space, so nn equations to check! If n=1n=1, that is just residues, but otherwise… !

Posted by: some guy on the street on May 22, 2010 2:04 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

JB wrote:

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

1sin 2z1z 2+3\frac{1}{\sin^2 z}\approx\frac{1}{z^2}+3

for small z, but somehow that’s okay.

I get:

1sin 2z1z 2+13\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 nn of all the constant terms in the Taylor expansions of the 1(αn) 2\frac{1}{(\alpha-n)^2} terms at α=0\alpha=0. Specifically:

1(αn) 21n 22αn 3+...\frac{1}{(\alpha-n)^2}\approx\frac{1}{n^2}-\frac{2\alpha}{n^3}+...

n,n01n 2=π 23\sum_{n\in\mathbb{Z}, n\neq 0}\frac{1}{n^2}=\frac{\pi^2}{3}

π 2sin 2πα1α 2+π 23\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.

Posted by: Greg Egan on May 22, 2010 1:29 AM | Permalink | Reply to this

Re: This Week’s Finds in Mathematical Physics (Week 297)

Mark and Greg: nice! You’re turning an isolated curiosity into the basis of a whole theory.

Mike passed on some posts from the “math-fun” mailing list conversation on this topic. Here are a couple of high points:

From: Robert Baillie
Date: Wed, Feb 22, 2006 at 8:38 AM
To: math-fun

HAKMEM 239 has this interesting question:

“119 (Schroeppel): Can someone square some series for Pi to give the series

Pi^2/6 = 1 + 1/2^2 + 1/3^2 + … ?”

Would you settle for squaring a series for Pi/Sqrt(8) to get a series for Pi^2/8?

Pi/Sqrt(8) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 –++ …

Pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + 1/9+2 + 1/11^2 + …

Also, notice that you can square the first series by adding the squares of the individual terms! I first noticed this by comparing exercises 6a and 6b on page 398 in Wilfred Kaplan’s “Advanced Calculus”.

The series for Pi/Sqrt(8) comes from the Fourier series for
f(x) = 0 for -Pi ≤ x < 0, f(x) = 1 for 0 ≤ x ≤ Pi.
The series is
f[x_] := (1/2) + (2/Pi) Sum[ Sin[(2n-1)x]/(2n-1), {n, 1, Infinity}]
f[Pi/4] gives the series for Pi/Sqrt[8].

The series for Pi^2/8 comes from the Fourier series for
g(x)= Abs[x], -Pi ≤ x≤ Pi.
g[x_] := (Pi/2) - (4/Pi) Sum[ Cos[(2n-1)x]/((2n-1)^2), {n, 1, Infinity}]
g[0] gives the series for Pi^2/8.


[Note that f is a square wave, while g is a sawtooth wave. - jb]


From: Eugene Salamin
Date: Wed, Feb 22, 2006 at 9:56 AM
To: math-fun

Yes. Fourier series exchanges pointwise multiplication with
convolution. The square wave convolved with itself gives the triangle wave, so the square of the Fourier coefficient of the former equals the Fourier coefficient of the latter.

Gene


From: Robert Baillie
Date: Wed, Feb 22, 2006 at 4:12 PM
To: math-fun

Here’s another series with a similar weird property:

a1 + a2 + a3 + … = a1^2 + a2^2 + a3^2 + …

Specifically:
Sum[ Sin[n]/n , {n, 1, Infinity}] = Sum[ (Sin[n]/n)^2 , {n, 1, Infinity}]
= (Pi - 1)/2

The Fourier series for this function:
f[x_] := -(Pi + x)/2 /; -Pi < x < -1;
f[x_] := x(Pi - 1)/2 /; -1 < x < 1;
f[x_] := (Pi - x)/2 /; 1 < x < Pi;
is: Sum[ Sin[n]/(n^2) * Sin[n x] , {n, 1, Infinity}].
At x = 1, you get the sum of (Sin[n]/n)^2

The function
g[x_] := (Pi-x)/2
has the Fourier series
Sum[ Sin[n x]/n , {n, 1, Infinity}].
At x = 1, you get the sum of Sin[n]/n.

This was problem 6241 in the Monthly about 25 years ago.

Posted by: John Baez on May 20, 2010 1:29 PM | Permalink | Reply to this

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