Comments on: A Little Arithmetic

By: Tom Noble

Tom Noble — Thu, 19 Nov 2009 22:36:38 +0000

I’m afraid you lost me on line 1×1x1…, but thanx anyway.

By: Matthew Wampler-Doty

Matthew Wampler-Doty — Sun, 15 Nov 2009 08:57:26 +0000

I really liked this… you probably know that a closed form for Zeta(2n+1) is not known for integers n>1, but you can use the fact that the geometric series has an easy solution, and (valid) column tricks like the ones you are using to show that Sum_{n=1}^{infty} (zeta(2n+1) – 1) = 1/4 , remarkably.

By: Cos

Cos — Sat, 14 Nov 2009 01:08:41 +0000

Why should I believe Step One, when I could line them up slightly differently to get 2N = 0 ?

By: SteveJ

SteveJ — Fri, 13 Nov 2009 13:51:10 +0000

“this link is WAY too technical”

If you’re going to have your kids confront their middle school teachers with spurious identities between divergent series, why not also to have them ask, “but *why* is it that two holomorphic functions which are equal on an open set in the complex plane are equal on any connected domain containing that set, and hence the Riemann zeta function defined as an analytic continuation is unique? It’s not true of smooth real functions, so what’s special about complex functions?”

Maybe not.

By: MattF

MattF — Fri, 13 Nov 2009 13:07:09 +0000

Baez’s ‘5′ lecture is fun too– but he omits a neat elementary point. He notes that the golden ratio ‘phi’ can be written as a continued fraction:

phi = 1 + 1/(1 + 1/(1 + …))

but I think a mathematically talented middle-schooler would be intrigued by the fact this infinitely-long expression can be “summed” right away by noting that the continued fraction in its first denominator is none other than phi, all over again… So, therefore

phi = 1 + 1/phi

which leads to a quadratic equation.

By: David Pinto

David Pinto — Fri, 13 Nov 2009 12:40:07 +0000

Maybe Abbott and Costello were on to something: