# 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 $\mathrm{sin}$, $\mathrm{cos}$, $\mathrm{tan}$ and $\mathrm{exp}$.

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 ${f}^{\prime}(0)\ne 0$, otherwise $f$ will not even
be invertible^{} around the origin.

## An example

We explain the method by an example, for $f(z)=\mathrm{arctan}z$. 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 $w=\mathrm{arctan}z$:

$$w=z-\frac{{z}^{3}}{3}+\frac{{z}^{5}}{5}+O({z}^{7}).$$ | (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 ${f}^{-1}(w)=\mathrm{tan}w$ 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:

$z$ | $=w+{\displaystyle \frac{1}{3}}{\left(w+O({z}^{3})\right)}^{3}-{\displaystyle \frac{{z}^{5}}{5}}+O({z}^{7})$ | (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 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 ${f}^{\prime}(0)\ne 0$. (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:

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

- 1 Lars V. Ahlfors. Complex Analysis. McGraw-Hill, 1979.

Title | series inversion |
---|---|

Canonical name | SeriesInversion |

Date of creation | 2013-03-22 15:39:09 |

Last modified on | 2013-03-22 15:39:09 |

Owner | stevecheng (10074) |

Last modified by | stevecheng (10074) |

Numerical id | 8 |

Author | stevecheng (10074) |

Entry type | Derivation |

Classification | msc 30B10 |