Blog

Today is the day I understood Yoneda’s lemma. Of course I’ve proven it several times before; in a sense the proof is trivial, since there’s only one thing you can do at each stage. But I never felt that I really understood what the lemma was saying, or why it should be true, until something clicked today.

Here’s a nice solution of Putnam 1981 B5 that I haven’t seen anywhere else (so far). The main idea is to sum `bitwise,' rather than `termwise.'

Let $S_k$ denote the set of positive integers with the $k$^th bit set, counting from the right starting at $k=0$. Then we have

$$\sum_{n=1}^\infty \frac{B(n)}{n^2 + n} = \sum_{k=0}^\infty \sum_{n \in S_k} \frac{1}{n^2 + n}.$$

A bit of thinking shows that $n \in S_k$ if and only if $\lfloor 2^{-k} n\rfloor = 2m+1$ is odd. Thus the sum becomes

Update 2020-09-08: This solution has been added to Kiran Kedlaya’s solution page.

I've found an elegant geometry-only proof of Putnam 2003 B5, which, to the best of my knowledge, hasn't yet been discovered.

We begin with a diagram of the problem:

It’s well-known that the period of a simple harmonic oscillator (SHO) is independent of its oscillation amplitude. But is this the only oscillator for which this holds?

No. A simple counterexample is the SHO + ‘brick wall’ potential: