Jason Rasmussen (student) Thank you, for the response. I suppose my confusion comes into play when I am trying to figure out where the x^{n} term comes from. I know that the Power Series notation is directly related to the Geometric Series of the form sigma [ br^{n} ] where the limit is b/(1r) for convergence at  r  <1. Therefore, the function f(x) needs to somehow take the form of b/(1(xa)), which may take some manipulation, and by setting r = (xa) and b = C_{n}, we get the Geometric Series converted to the Power Series. Taking the nth order derivative of the Power Series gives C_{n} = f^{n}(a)/n!. There must be a gap in my knowledge somewhere because I cannot seem to make f(x) = e^{x} take the form of f(x) = b/(1(xa)). 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 [f^{n}(a) * (xa)^{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 for x  a < R and then show that C_{n} must satisfy where f^{n} is the nth derivative of f. This would be a geometric series if each C_{n} 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 [f^{n}(a) * (xa)^{n} / n!]?" is no. CheersHarley
