coefficient formula
If a function happens to be the sum of a power series, can you reach in and read off the coefficients one at a time? Yes — and there is a clean formula to do it. The trick is that each coefficient is exactly tied to one derivative at the center, so differentiating the right number of times and evaluating extracts the coefficient you want.
Statement: if f(x) = sum_{n=0}^infinity c_n (x - a)^n on some interval around a, then differentiating term by term n times and setting x = a kills every term except the n-th, giving f^{(n)}(a) = n! c_n. Solving for the coefficient, c_n = f^{(n)}(a)/n!. This is precisely why the power series of f must be its Taylor series: the coefficients are forced, not free.
The formula works because term-by-term differentiation is valid inside the radius of convergence, and at x = a all the positive powers of (x - a) vanish, isolating a single term. The same logic shows the constant term is c_0 = f(a). In complex analysis there is a parallel integral formula (Cauchy's) for the coefficients, which sometimes recovers them even when differentiating is awkward.
For f(x) = e^x with center a = 0, every derivative is e^x, so f^{(n)}(0) = 1 and c_n = 1/n!. This recovers e^x = sum x^n/n! and confirms the coefficients are uniquely the Taylor coefficients.
Reading off exponential coefficients via c_n = f^{(n)}(0)/n!.