Inverses of Power Series

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

$\displaystyle p(x)=\sum\limits_{n=0}^\infty p_nz^n$

within the radius $r$ , and such that $p(0)=p_0\neq0$ . Then there is some radius $\delta$ within which the reciprocal has a power series expansion

$\displaystyle\frac{1}{p(x)}=\sum\limits_{n=0}^\infty q_nz^n$

In particular, we have $q_0=\frac{1}{p_0}$ .

In the proof we may assume that $p_0=1$ — we can just divide the series through by $p_0$ — and so $p(0)=1$ . We can set

$\displaystyle P(z)=1+\sum\limits_{n=1}^\infty\left|p_nz^n\right|$

within the radius $h$ . Since we know that $P(0)=1$ , continuity tells us that there’s $\delta$ so that $|z|<\delta$ implies $|P(z)-1|<1$ .

Now we set

$\displaystyle f(z)=\frac{1}{1-z}=\sum\limits_{n=0}^\infty z^n$
$\displaystyle g(z)=1-p(z)=\sum\limits_{n=0}^\infty -p_nz^n$

And then we can find a power series expansion of $f\left(g(z)\right)=\frac{1}{p(z)}$ .

It’s interesting to note that you might expect a reciprocal formula to follow from the multiplication formula. Set the product of $p(z)$ and an undetermined $q(z)$ to the power series $1+0z+0z^2+...$ , and get an infinite sequence of algebraic conditions determining $q_n$ in terms of the $p_i$ . Showing that these can all be solved is possible, but it’s easier to come around the side like this.

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September 24, 2008 - Posted by John Armstrong | Analysis, Calculus, Power Series

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