# series inversion

The method of series inversion allows us to derive the power series of an inverse function $f^{-1}$ given the power series of $f$.

Clearly, since we are representing function $f$ and $f^{-1}$ by power series, the function $f$ must necessarily be holomorphic; that is, it is differentiable as a complex-valued function on an open subset of the complex plane. (It follows that $f^{-1}$ must also be holomorphic.) Holomorphic functions include the elementary functions studied in calculus such as $\sin$, $\cos$, $\tan$ and $\exp$.

For the method to work smoothly, it is best to assume that we want to invert $f$ at around the origin, and its value there is zero. There is no loss of generality, since if $f(b)=a$, we can apply series inversion to the function $g$ defined by $f(z)=a+g(z-b)$. Then $f^{-1}(w)=g^{-1}(w-a)+b$; we obtain the power series for $f^{-1}$ centred at $a$.

Also, it must be true that $f^{\prime}(0)\neq 0$, otherwise $f$ will not even be invertible around the origin.

## An example

We explain the method by an example, for $f(z)=\arctan z$. In the following, we will consistently use the notation $O(z^{n})$ to denote a holomorphic function $h(z)$ whose power series expansion begins with $z^{n}$. And similarly when $z$ is replaced with the variable $w$.

First, we start with the well-known power series expansion for $w=\arctan z$:

 $w=z-\frac{z^{3}}{3}+\frac{z^{5}}{5}+O(z^{7})\,.$ (1)

The number of explicit terms in the power series expansion determines the number of terms that we will be able to obtain in the power series expansion of $f^{-1}$. So in this case, we are going to seek an expansion of $f^{-1}(w)=\tan w$ up to (but excluding) the $w^{7}$ term.

A simple rearrangement of (1) gives

 $z=w+\frac{z^{3}}{3}-\frac{z^{5}}{5}+O(z^{7})$ (2)

Now we substitute equation (2) into itself. Of course, usually when we substitute an equation into itself we do not get anything, but here it works because we can perform simplications using the $O$ notation. So for instance, in the following, in second term $z^{3}/3$ on the right of equation (2), we put in equation (2) simplified to $z=w+O(z^{3})$. Why we should choose this simplication will be clear in a moment:

 $\displaystyle z$ $\displaystyle=w+\frac{1}{3}\Bigl{(}w+O(z^{3})\Bigr{)}^{3}-\frac{z^{5}}{5}+O(z^% {7})$ (3) $\displaystyle=w+\frac{1}{3}\Bigl{(}w^{3}+O(z^{3})(w^{2}+\cdots)\Bigr{)}-\frac{% z^{5}}{5}+O(z^{7})$ (4) $\displaystyle=w+\frac{1}{3}\Bigl{(}w^{3}+O(w^{3})\cdot O(w^{2})\Bigr{)}+O(w^{5% })\,.$ (5)

In equation (5) we used the fact that the expansion for $z=f^{-1}(w)$ must begin with a $w$ term, i.e. $f^{-1}(0)=0$. Also note that we are guaranteed that the $w$ and $z$ terms have non-zero coefficients, because $f^{\prime}(0)\neq 0$. (Otherwise we would not be able to isolate $z$ in equation (2).)

Now equation (5) simplifies to

 $z=w+\frac{w^{3}}{3}+O(w^{5})\,,$ (6)

which is already an achievement, because we have identified exactly what the $w^{3}$ term must be.

We substitute (6) into the $z^{3}$ and $z^{5}$ terms of (2), and obtain:

 $\displaystyle z$ $\displaystyle=w+\frac{1}{3}\left(w+\frac{w^{3}}{3}+O(w^{5})\right)^{3}-\frac{1% }{5}\left(w+\frac{w^{3}}{3}+O(w^{5})\right)^{5}+O(z^{7})$ (7) $\displaystyle=w+\frac{1}{3}\left(w^{3}+\binom{3}{1}\frac{w^{3}}{3}w^{2}+O(w^{7% })\right)-\frac{1}{5}\left(w^{5}+O(w^{7})\right)+O(w^{7})$ (8) $\displaystyle=w+\frac{1}{3}w^{3}+\frac{2}{15}w^{5}+O(w^{7})\,.$ (9)

And this gives our desired expansion of $z=\tan w$ of degree $<7$.

## Summary

 $w=f(z)=a_{1}z+a_{2}z^{2}+a_{3}z^{3}+\cdots\,,\quad a_{1}\neq 0\,,$ (10)

and rearrange it to,

 $z=b_{1}\Bigl{(}w-a_{2}z^{2}-a_{3}z^{3}-\cdots-a_{n}z^{n}\Bigr{)}+O(w^{n+1})\,,% \quad b_{1}=\frac{1}{a_{1}}\,.$ (11)

So we know that $z=b_{1}w+O(w^{2})$, and we can substitute this into the term $z^{2}$ of equation (11). At the end we will get an equation of the form $z=b_{1}w+b_{2}w^{2}+O(w^{3})$, and we can substitute this into the terms $z^{2}$ and $z^{3}$ of (11). And what ever results we will substitute back into the terms $z^{2}$, $z^{3}$, $z^{4}$ of equation (11). We can repeat this process until we have all the terms of $z=f^{-1}(w)$ that we need.

We probably should normalize the functions so that $a_{1}=b_{1}=1$ to make the computations easier.

## References

Title series inversion SeriesInversion 2013-03-22 15:39:09 2013-03-22 15:39:09 stevecheng (10074) stevecheng (10074) 8 stevecheng (10074) Derivation msc 30B10