series inversion

The method of series inversion allows us to derive the power seriesof 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 differentiableas 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 calculussuch as $\sin$, $\cos$, $\tan$ and $\exp$.

For the method to work smoothly, it is best to assume that we wantto invert $f$ at around the origin, and its value there is zero.There is no loss of generality, sinceif $f(b)=a$, we can apply series inversion to the function $g$ definedby $f(z)=a+g(z-b)$.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 $f^{\prime }(0)\ne 0$, otherwise $f$ will not evenbe 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 beginswith $z^n$. 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 determinesthe number of terms that we will be able to obtainin the power series expansion of $f^{-1}$.So in this case, we are going to seek an expansionof $f^{-1}(w)=\tan w$ 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 itselfwe do not get anything, but here it works because we canperform 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 inequation (2) simplified to $z=w+O(z^3)$.Why we should choose this simplication will be clear in a moment:

$z$$=w+{\displaystyle \frac{1}{3}}{\left(w+O(z^3)\right)}^3-{\displaystyle \frac{z^5}{5}}+O(z^7)$(3)

$=w+{\displaystyle \frac{1}{3}}\left(w^3+O(z^3)(w^2+\mathrm{\cdots })\right)-{\displaystyle \frac{z^5}{5}}+O(z^7)$(4)

$=w+{\displaystyle \frac{1}{3}}\left(w^3+O(w^3)\cdot O(w^2)\right)+O(w^5).$(5)

In equation (5) we used the fact that the expansionfor $z=f^{-1}(w)$ must begin with a $w$ term, i.e.$f^{-1}(0)=0$. Also note that we are guaranteed that the $w$ and $z$ termshave non-zero coefficients, because $f^{\prime }(0)\ne 0$. (Otherwisewe 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 identifiedexactly what the $w^3$ term must be.

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

$z$$=w+{\displaystyle \frac{1}{3}}{\left(w+{\displaystyle \frac{w^3}{3}}+O(w^5)\right)}^3-{\displaystyle \frac{1}{5}}{\left(w+{\displaystyle \frac{w^3}{3}}+O(w^5)\right)}^5+O(z^7)$(7)

$=w+{\displaystyle \frac{1}{3}}\left(w^3+\left({\displaystyle \genfrac{}{}{0pt}{}{3}{1}}\right){\displaystyle \frac{w^3}{3}}{w}^2+O(w^7)\right)-{\displaystyle \frac{1}{5}}\left(w^5+O(w^7)\right)+O(w^7)$(8)

$=w+{\displaystyle \frac{1}{3}}{w}^3+{\displaystyle \frac{2}{15}}{w}^5+O(w^7).$(9)

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

Summary

To summarize the procedure in general,we start with the expansion

$$w=f(z)=a_{1}z+a_2{z}^2+a_3{z}^3+\mathrm{\cdots },a_1\ne 0,$$(10)

and rearrange it to,

$$z=b_1\left(w-a_2{z}^2-a_3{z}^3-\mathrm{\cdots }-a_n{z}^n\right)+O(w^{n+1}),b_1=\frac{1}{a_1}.$$(11)

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

We probably should normalizethe functions so that $a_1=b_1=1$to make the computations easier.