1. The problem statement, all variables and given/known data Given f = a0 + sum(ancos(nx) + bnsin(nx)) and f' = a0' + sum(an'cos(nx) + bn'sin(nx)) The sums are over all positive integers up to n. show that a0' = 0, an' = nbn, bn' = -nan Then prove a similar formula for the coefficients of f(k) using induction. 2. Relevant equations 3. The attempt at a solution OK I showed the first part (for the first derivative coefficients) by taking the derivative of f, that was easy enough. I have to confess that the teacher's question also said HINT: (Integrate by parts) but it was much simpler to just take the derivative of f. I don't know if that's part of the reason i'm having trouble with the induction part or not, but it seems inefficient to integrate backward when it's such a simple thing to show these coefficients with the derivative. I have not had a lot of experience with induction, and I haven't had to do much as an engineering (as opposed to math) major. So induction works like this: prove it for a single case, assume something is true for some integer k, then use that assumption to show it is also true for k+1. (This is to the best of my understanding) After considering the problem for some time, i don't think it's too much to show that for any k, a0(k) = 0, since the derivative of 0 is 0, and I've shown that the derivative of a0 = 0. It's the other coefficients that trip me up, because every time the derivative is taken, the coefficients will take on a value related to its opposite (in terms of functionally odd or even, i mean) coefficient. For example an' = nbn, notice that the derivative an' relies on the value of bn. I guess I'm just not sure how to start. I feel like I could just do the derivative a bunch of times and look for a pattern, but I feel like that defeats the purpose of proof by induction. EDIT: Ok, I wrote out f(k), then took the derivative and got that an(k+1) = nbn(k), but this isn't actually true, because, I did the bunch of derivatives and found that the pattern is actually an(k+1) = nkbn(k) How can I show that by induction? I can see that every time the function is differentiated it's going to cause another multiplication by n because of the chain rule, I'm just not quite sure how to show it mathematically. EDIT2: I integrated f(k+1) but it didn't seem to really add anything new as far as a solution goes... also the HINT was to integrate by parts but looking at the expression, I don't really see anything to integrate by parts even if i wanted to.