Megaminx II

To continue the story from last time, I followed this piece of advice and bought a new megaminx from Mefferts Puzzles. I can already say that this is a superior product, it turns much cleaner than my previous model, and I have managed to scramble and solve this model without having it fall apart on me.

In an ideal world, we would now say that they all lived happily ever after but this is a grumblog, so I’ll also record my negative experiences. Firstly, the mefferts site and online shop is not the most pleasant to deal with, but I did at least end up with the product I wanted. And secondly, during the first solve (which takes more time than is difficult, in terms of difficulty the megaminx is similar to the rubik’s cube), two of the coloured tiles came off. Still I believe that this is a solvable problem with a bit of glue.

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The American mobile phone system

It is clearly terrible. Actually terrible is too nice a word for it. The major problem that should be rectified is the billing of both caller and receiver for calls and messages. This goes against anything reasonable (naturally, as in any sensible phone or postal system, only the sender should pay), and since the companies don’t want to do anything about it, really should be legislated against. It is impossible to be tried twice for the same crime, so it should also be impossible for two billings to be made for the same service.

This is not the only problem with the system, but is the one which sticks out to me as by far the worst. I’m not really sure how best to go about fixing this.

A power series identity

Last year, I was giving a talk for PuMaGraSS, and I wanted to include the following power series identity:

The standard way that I have seen this proved is to go via the Jacobi Triple Product Formula which tends to have non-trivial proofs. I decided though that I wanted to find a more combinatorial way of proving this special case. This was successful and can be found at this site, though the proof would be more illuminating with pictures.

Essentially what one does is naively expands the product to get a sum over all paritions with distinct parts and then constructs an involution on a large subset of these partitions to cancel most of the terms, leaving only those that appear on the right hand side of the identity.

I had originally hoped for a combinatorial proof exploiting the fact that the product on the left hand side is the inverse of the generating function for the partition function. Such a proof does exist, is extremely elegant and can be found, amongst other gems in Aigner and Zagier’s book Proofs from the Book. This book is really an amazing read, consisting of a compilation of some of the most elegant pieces of mathematics known to man, and I would highly recommend that anyone with an interest in maths reads it.