Now that we know how to compose power series, we can invert them. But against expectations I’m talking about multiplicative inverses instead of compositional ones.
More specifically, say we have a power series expansion
within the radius , and such that . Then there is some radius within which the reciprocal has a power series expansion
In particular, we have .
In the proof we may assume that — we can just divide the series through by — and so . We can set
within the radius . Since we know that , continuity tells us that there’s so that implies .
Now we set
And then we can find a power series expansion of .
It’s interesting to note that you might expect a reciprocal formula to follow from the multiplication formula. Set the product of and an undetermined to the power series , and get an infinite sequence of algebraic conditions determining in terms of the . Showing that these can all be solved is possible, but it’s easier to come around the side like this.
Now that we can take powers of functions defined by power series and define them by power series in the same radii.. well, we’re all set to compose functions defined by power series!
Let’s say we have two power series expansions about :
within the radius , and
within the radius .
Now let’s take a with and . Then we have a power series expansion for the composite:
The coefficients are defined as follows: first, define to be the coefficient of in the expansion of , then we set
To show this, first note that the hypothesis on assures that , so we can write
If we are allowed to exchange the order of summation, then formally the result follows. To justify this (at least as well as we’ve been justifying such rearrangements recently) we need to show that
converges. But remember that each of the coefficients is itself a finite sum, so we find
On the other hand, in parallel with our computation last time we find that
So we find
which must then converge.