Oxford Mathematics Admissions Test - Commutation computations

[@amritlohia8240]

Arguably a more natural approach - and the way I did this problem, as I had a certain amount of programming background - is to view the entire setup in terms of binary expansions. So g simply adds a 0 to the end of the binary expansion of a number, while f changes a 00 at the end to 01, or a 01 at the end to 10. This enables you to give slightly more rigorous arguments for why you've found all the possibilities in parts (ii) and (iv). The other thing to note is that in part (iii), you can just compute it directly, without using the commutator relations - we get f^k(x) = x+k, then applying g gives 2x+2k, f^j gives 2x+2k+j, g again gives 4x+4k+2j, and finally f^i gives 4x+4k+2j+i.

[@mscha]

For the final part, one way to do a sanity check on your answer, is to see if it matches known results. We know that for m=1, there are 4 results. Indeed, (1+1)² = 4, so that is promising.

[@mathoutloud]

Of course! I usually like to do sanity checks like that, it’s helped me find many mistakes in the past.

[@Fate-rh1ob]

I feel pretty confident that the answer to iv) is 4^m (although I have been confident a lot of times where it turned out not to be warranted). And in a funny way, it is a similar approach to that of my recent comment in your video kzbin.info/www/bejne/j2TGf6qnncSiqpI. Let m be a fixed nonnegative integer and suppose that h(m) is defined as the set of solutions (or 3-tuples (i,j,k)), such that i + 2j + 4k = 4m, with i,j,k being nonnegative integers. Then h(m+1) is the set of solutions to i + 2j + 4k = 4(m+1) = 4m + 4. Starting from any solution in h(m), I can get from 4m to 4m+4 in 4 ways: increasing i by 4, increasing i by 2 and j by 1, increasing j by 2 or increasing k by 1. This implies that |h(m+1)| = 4|h(m)|, where |A| is the number of elements in the set A. Since |h(0)| is 1, we get |h(m)| = 4^m. This also passes the sanity check for m=1. :) What do you think?

[@Fate-rh1ob]

Hah, and I already found that my confidence was indeed not warranted xD. For example, starting with the two solutions (i,j,k)=(4,0,0) and (i,j,k)=(0,2,0) in h(1), by increasing j in the first solution and increasing i by 4 in the second solution I end up with the same solution (i,j,k)=(4,2,0) in h(2). So I count stuff more than once... I now tried it in the same way that you did and 100% agree to your result!

[@mathoutloud]

Haha great to hear you found your mistake! Usually with my recordings I check the answer afterwards (if I have access to it) just to make sure I don’t leave everyone with something totally wrong. If I ended up with the wrong answer I would leave a text comment in the video itself or in the description, but some wrong things have slipped through in a few occasions without me knowing.