Recent content by Leumas13

Nevermind, Solved Use the substitution that sinu = 0.5i (e^iu -e^-iu) and the equivalent for cosine and shift the register of the summation.

So, I've been asked to show that f(x)=Ʃn=0∞ [ansin(nπx/a) + bncos(nπx/a) is equivalent to f(x)=Ʃn=-∞∞ c[SUB]n[SUB]einx/a I'm kinda stuck. i want to turn this into the exponential form but I don't see how to condense this into a single term. Thanks!