arrow_back We shall show that the power function x^a has a power series expansion about x₀=1. (We shouldn't expect that this function have a power series expansion centered on x₀=0, since for general real a, x^a isn't even defined for x=0.)

First, let's consider what the power series for x^a could possibly be. If it is in fact true that x^a has a power series expression, then by Proposition 7.22, x^a must be equal to its Taylor Series. So let's compute the Taylor Series centered at x=1 for this function.

Of course computing the Taylor Series in no way proves that x^a is equal to this series. All it establishes is that this is the only power series which x^a could possibly be equal to. But but now we have something something concrete to work with, and we can use other tricks to prove the two functions are equal .