series inversion

The method of series inversion allows us to derive the power seriesMathworldPlanetmath of an inverse function f-1 given the power series of f.

Clearly, since we are representing functionMathworldPlanetmath f and f-1 by power series, the function f must necessarily be holomorphic; that is, it is differentiableMathworldPlanetmathPlanetmath as a complex-valued function on an open subset of the complex planeMathworldPlanetmath. (It follows that f-1 must also be holomorphic.) Holomorphic functions include the elementary functionsMathworldPlanetmath 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(0)0, otherwise f will not even be invertiblePlanetmathPlanetmath around the origin.

An example

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

First, we start with the well-known power series expansion for w=arctanz:

w=z-z33+z55+O(z7). (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)=tanw up to (but excluding) the w7 term.

A simple rearrangement of (1) gives

z=w+z33-z55+O(z7) (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 z3/3 on the right of equation (2), we put in equation (2) simplified to z=w+O(z3). Why we should choose this simplication will be clear in a moment:

z =w+13(w+O(z3))3-z55+O(z7) (3)
=w+13(w3+O(z3)(w2+))-z55+O(z7) (4)
=w+13(w3+O(w3)O(w2))+O(w5). (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(0)0. (Otherwise we would not be able to isolate z in equation (2).)

Now equation (5) simplifies to

z=w+w33+O(w5), (6)

which is already an achievement, because we have identified exactly what the w3 term must be.

We substitute (6) into the z3 and z5 terms of (2), and obtain:

z =w+13(w+w33+O(w5))3-15(w+w33+O(w5))5+O(z7) (7)
=w+13(w3+(31)w33w2+O(w7))-15(w5+O(w7))+O(w7) (8)
=w+13w3+215w5+O(w7). (9)

And this gives our desired expansion of z=tanw of degree <7.


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

w=f(z)=a1z+a2z2+a3z3+,a10, (10)

and rearrange it to,

z=b1(w-a2z2-a3z3--anzn)+O(wn+1),b1=1a1. (11)

So we know that z=b1w+O(w2), and we can substitute this into the term z2 of equation (11). At the end we will get an equation of the form z=b1w+b2w2+O(w3), and we can substitute this into the terms z2 and z3 of (11). And what ever results we will substitute back into the terms z2, z3, z4 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 a1=b1=1 to make the computations easier.


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