Average of sin^100(x).
I found an interesting math problem on The Reference Frame blog, concerning solving the average of the 100th power of sin(x). Mr. Motl, the blog author seemed to think that there is no fast 5 minute way to solve this problem, only to discover several exact attacks on the comments to complement his approximation.
All it takes is some fundamental concepts in complex analysis. To be fair, it took me more than 5 minutes to work it thru and correct some errors along the way. I’d say something like 10-15 minutes. Have to say to my defense though, that I did the work under being influenced by sleep deprivation (unfortunately I suffer from a sleep disorder).
Here’s a fast way to work out an exact result. First recognizing that sin(x)=(exp(i*x)-exp(-i*x))/(2*i) gets you (1/2^100)*sum(k=0..100)(100!/(k!*(100-k)!)*exp(i*x*(100-2*k)) after applying the 100th power and some binomial magic. Expanding the sum you can group the terms to form cosines in the form of (exp(i*x)+exp(-i*x))/2, whose average is ofcourse zero. The only term that contributes to the average is the exp(0)/2 term (since the binomial expansion is symmetric in this way) and you end up with 100!/(2^100*50!*50!) as the answer.