The Geometry of the Basel Problem

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The Geometry of the Basel Problem
Sean B. Palmer
20/10/10 07:01
When Euler solved the Basel Problem, to calculate the sum of the
reciprocals of the squares 1 + 1/4 + 1/9 + 1/16 + 1/25 et seq., this
was the first time that pi appeared in a place which was not obviously
geometric. But the solution, pi squared over six, smells very strongly
of a geometric solution. The pi is circular, the exponentiation by two
is square, and could that division by six be something to do with a
triangle?

As far as I've been able to research, the most elementary proof of the
Basel Problem involves a lot of complex non-geometrical steps. There
is still no obvious basis for a circle in there, and though there is a
proof, and a generalisation that Euler discovered for the reciprocals
of not just squares but any even powers, it seems that the problem is
still not fully understood. The more general form of the Basel
Problem, for all powers as well as squares, is called the Riemann zeta
function.

The fact that the problem is not still understood is easily apparent
in the fact that a formula for the sum of the reciprocals of odd
powers is still unknown, and a hot topic in mathematics. The zeta
value of three has been discovered, but the odd powers above three are
get to be generalised.

If there is a geometric solution to the Basel Problem, it may be one
which can be extended to the Riemann zeta function in general. This
may in turn tell us something about the distribution of prime numbers,
since the Riemann zeta function is also the probability that a certain
amount of numbers chosen at random will be coprimes. A geometric
solution to the Basel Problem, then, is not likely to be a trivial or
elementary one. But it may provide insights that are not obvious from
non-geometric proofs.

I've been doing some work on this, and obviously I didn't get very far
since I'm not really a mathematician. But I did manage to come up with
a few things:

* Discovered an elegant expression of the Basel Problem as the series
of square reciprocals in geometric terms.
* Reasoned that though the Basel Problem series is an infinite one,
the convergence is very sharp because of the nature of the squares.
The 10th square reciprocal gives you a 100th of the original unit
square, and the 100th square reciprocal give you one 10,000th of the
original unit square. The general area is established quite quickly in
the sum.
* Discovered three elegant shapes and associated subshapes with pi/3
by pi/2 area, one conic, one rectangular, and one circular; with
various interesting relationships between the shapes.

What seems to be missing is a link to the original insight I had about
the geometric expression of squares with the one I had about the
geometric expression of pi. The important thing isn't the information
so much as the framing, because what we're looking for here is not a
proof of the Basel Problem, which has existed for centuries, but a
certain expression of it.

Even though I've come up with some elegant expressions of the pi
squared over six area, I still don't feel that I really understand the
geometric nature of this area. I'm not entirely sure that a geometric
solution would be best thought of in two dimensions, even. I can
imagine the reciprocal squares being stacked in three dimensions, and
finding an area based solution to the problem.

I also feel that intuitively there's going to be some very strange
stuff involving circles and squares, because pi squared is of course
the squaring of a circumference based on the relationship of a
diameter to side length with respect to the unit square. I'm not sure
exactly what sort of geometry would be required if you did this sort
of thing multiple times, though that might not be necessary.

--
Sean B. Palmer, http://inamidst.com/sbp/

Re: The Geometry of the Basel Problem
Sean B. Palmer
30/12/10 05:01
Some notes compiled on 11th November 2010:

http://en.wikipedia.org/wiki/Parity_of_zero
http://en.wikipedia.org/wiki/Aleph_number

http://www.mi.sanu.ac.rs/vismath/zen/zen5.htm
http://descmath.com/diag/short.html
http://ask.metafilter.com/25060/Whats-the-sum-of-all-integers
http://www.earlham.edu/~peters/courses/logsys/quirk.htm
http://www.evilmadscientist.com/article.php/SumTrick
http://www.arps.org/users/hs/kochn/qr/Transfinite.htm
http://en.wikipedia.org/wiki/Transfinite_number
http://blog.wolframalpha.com/2010/09/10/transfinite-cardinal-arithmetic-with-wolframalpha/
http://www.vordenker.de/gunther_web/achill1.htm
http://mathworld.wolfram.com/RiemannZetaFunction.html
http://www.wolframalpha.com/input/?i=%CE%B6(n)+%3D+pi**2
http://en.wikipedia.org/wiki/User:Sbp/Maths
http://en.wikipedia.org/wiki/Ramanujan_summation
http://mrob.com/pub/ries/index.html
http://en.wikipedia.org/wiki/Triangular_number
http://en.wikipedia.org/wiki/Indefinite_sum
http://en.wikipedia.org/wiki/Harmonic_number
http://www.physicsforums.com/showthread.php?t=267013
http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
http://rolfeschmidt.com/wordpress/?p=47

--
Sean B. Palmer, http://inamidst.com/sbp/

Re: The Geometry of the Basel Problem
Jason Davies
15/01/11 09:00
Dear Sean and friends,

You might be interested in this note titled "Summing inverse squares by euclidean geometry" by Johan Wästlund: http://www.math.chalmers.se/~wastlund/Cosmic.pdf

I discovered it via http://www.math.chalmers.se/~wastlund/ via http://news.ycombinator.com/item?id=2106927, the latter being a discussion about Knuth's latest Christmas Tree lecture, "Why Pi?", where he discusses some of Johan's work.

From Johan's page:

Another relation of pi to arithmetic is the famous identity by Euler summing the inverse squares of the positive integers to pi^2/6. One may ask (as one of the students at Knuth's lecture) whether there is an equally simple geometric explanation of this appearance of pi. The answer is yes, and the idea seems to go back to Yaglom and Yaglom 1953. This note is my contribution to making it better known.

So it seems there is indeed a proof using only classical Euclidean geometry!  Riemann zeta function, here we come! :-)