Jason Rasmussen (student) Thank you, for the response. I suppose my confusion comes into play when I am trying to figure out where the xn term comes from. I know that the Power Series notation is directly related to the Geometric Series of the form sigma [ brn ] where the limit is b/(1-r) for convergence at | r | <1. Therefore, the function f(x) needs to somehow take the form of b/(1-(x-a)), which may take some manipulation, and by setting r = (x-a) and b = Cn, we get the Geometric Series converted to the Power Series. Taking the nth order derivative of the Power Series gives Cn = fn(a)/n!. There must be a gap in my knowledge somewhere because I cannot seem to make f(x) = ex take the form of f(x) = b/(1-(x-a)). Maybe I should have labelled my question as "middle" because it may be more of a personal problem with algebra and logarithms. Or, am I to assume that all functions can be represented by sigma [fn(a) * (x-a)n / n!]? Hi again Jason, The usual way to approach Maclaurin series is to assume that you have a function f(x) that can be represented by a power series f(x) = C0 + C1(x - a) + C2(x - a)2+...+Cn(x - a)n+... for |x - a| < R and then show that Cn must satisfy Cn = fn(a)/n! where fn is the nth derivative of f. This would be a geometric series if each Cn had the same value, but that us usually not the case. The answer to your question "am I to assume that all functions can be represented by sigma [fn(a) * (x-a)n / n!]?" is no. Cheers Harley Go to Math Central