Midterm Madness

Well, the midterm was this morning, and it didn't go too badly I think. The first question about the Fibonacci sequence was a breeze, almost suspiciously so, but unless I've really gotten bad at reading trick questions then that was fine. The second question, I realized 15 minutes after the test, I pulled some real bull algebra on, so I know I screwed that up, but at least the structure is there, so here's hoping for part marks.

The third question about sets was something more interesting! I stared at it for a good 15 minutes or so, having gotten as far as writing down that all sets have 2^n subsets and that each following set's subsets were those of the previous + subsets containing the element n. With two minutes left on the clock, I had a revelation (though it might have been wrong, I felt pretty good writing it down and remain confident it was a valid proof). By complete induction and partitioning S into S_1 and S_2, where S_1 is S less all the subsets containing {n}, and S_2 is S less all the subsets without {n}, you get two subsets of equal size, and better yet, you know S_1 is just the set of subsets of S:{1,..,n-1} and thus by the I.H. has 3 * 2^n-3 subsets. S_2 being the same size means it will have the same number of valid subsets, so the total will end up being 3*2^n-3 + 3*2^n-3 = 3*(2^n-3 + 2^n-3) = 3*(2^n-2)

Exactly what was needed. That was a fun proof.

I still feel pretty stupid about the algebra in question 2 though, because for some idiotic reason I wrote down 3^n+1 = 3^n + 3^n instead of 3^n + 3^n + 3^n or 3*3^n. Hopefully there'll be some mercy, because I think I did end up pseudo-legitimately showing that 3^n + 3^n > 2*(n+1)^3, so it follows 3^n + 3^n + 3^n could only be larger as well. Oh well, all in all it could've definitely gone a lot worse.

Now with that midterm out of the way and nothing due immediately after the weekend, time to enjoy the long weekend and go party before thanksgiving.