Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am a math student trying to wrap my head around complex analysis through self-study. I am using Complex Analysis by Serge Lang, but I find myself struggling with some of his power series manipulations and his representations of power series as polynomials. For instance, given the following theorem/proof (taken from Theorem 6.1, p.76):

Theorem Let $f(T) = a_1 T + higher$ $terms$ be a formal power series with $a_1 \not= 0$. Then there exists a unique power series $g(T)$ such that $f(g(T)) = T$. This power series also satisfies $g(f(T)) = T$.


For convenience of notation we write $f(T)$ in the form

$$f(T) = a_1T - \sum_{2}^{\infty} a_nT^n$$

We seek a power series

$$g(T) = \sum_{1}^{\infty} b_nT^n$$

such that

$$f(g(T)) = T$$

The solution to this problem is given by solving the equation in terms of the coefficients of the power series

$$a_1g(T) - a_2g(T)^2 - ... = T$$

These equations are of the form

$$a_1b_n - P_n(a_2, ..., a_n, b_1,..., b_{n - 1}) = 0 \quad \text{and} \quad a_1b_1 = 1 \quad \text{for} \quad n = 1$$

where $P_n$ is a polynomial with positive integer coefficients (generalized binomial coefficients)


I can follow this all right up until the polynomial representation is used. What I would like to do is follow this type of argument with greater ease. What readings and/or exercises should I do accomplish this?

share|cite|improve this question
It is more or less pattern recognition: notice that every coefficient of $T^n$ will have a contribution from only finitely many of the terms $a_1g(T), -a_1g(T)^2,\cdots$, and thus will be composed of sums and products of powers of the $a_i$ and $b_i$ variables. – anon Aug 14 '12 at 6:29
Try an example: take $f(T)=T+T^2+T^3+...$ and compute $f(g(T))$ up to $T^4$ or $T^5$. You will have to remember how the multinomial theorem allows you to expand $(b_1T+...)^n$ to the desired order. – Georges Elencwajg Aug 14 '12 at 6:41
@anon I can see that it would be equivalent to $a_nP_n(b_1,...,b_{n-1})$ but I don't see how each $a_n$ then gets to be on its own in the polynomial. – providence Aug 14 '12 at 6:42
@GeorgesElencwajg I have not yet learnt the multinomial theorem. Perhaps I should read in this area before continuing with this text? What sort of text would contain the multinomial theorem and related information? – providence Aug 14 '12 at 6:43
The form $P_n(a_2,\cdots,a_n,b_1,\cdots,b_{n-1})$ can surely be made more narrow and specific, but what does it matter when all we need to recognize is that the coefficients of $T^n$ will be polynomials in the coefficients of the $a_i,b_i$? There's no need to overthink this. – anon Aug 14 '12 at 6:53
up vote 1 down vote accepted

It's probably clearest if you look at the first few terms explicitly.

$$\eqalign{f(g(T)) &= a_1 g(T) - a_2 g(T)^2 - a_3 g(T)^3 - a_4 g(T)^4 + \ldots\cr &= a_1 (b_1 T + b_2 T^2 + b_3 T^3 + b_4 T^4 + \ldots) - a_2 (b_1 T + b_2 T^2 + b_3 T_3 + \ldots)^2 \cr& \ \ \ \ - a_3 (b_1 T + b_2 T^2 + \ldots)^3 - a_4 (b_1 T + \ldots)^4 + \ldots\cr &= a_1 b_1 T + (a_1 b_2 - a_2 b_1^2) T^2 + (a_1 b_3 - 2 a_2 b_1 b_2 - a_3 b_1^3) T^3\cr& \ \ \ \ + (a_1 b_4 - a_2 (b_2^2 + 2 b_1 b_3) - 3 a_3 b_1^2 b_2 - a_4 b_1^4) T^4 + \ldots\cr}$$ Do you see the pattern? For the coefficient of $T^m$, $a_1 g(T)$ contributes $a_1 b_m $ and $-a_j g(T)^j$ contributes $-a_j$ times a sum of products of $j$ of the $b_k$ with indices adding up to $m$, for $2 \le j \le m$.

share|cite|improve this answer
Thank you for your help. This answer in addition to anon's comments above allowed me to resolve the issue. – providence Aug 14 '12 at 6:57

There's an intuitive idea here that the "size" of a formal power series is inversely related to the degree of its leading term. $T$ is 'small'. $T^2$ is even 'smaller' than $T$.

This is even more well-behaved than the usual notion of size, because errors can't accumulate: no matter how many things of the same 'size' as $T^2$ we add together, we cannot affect the coefficient on $T$.

The leading term of $f(T)$ and $g(T)$ have degree 1: they are both "small". So their powers are very small. If we write down the power series

$$ f(g(T)) = \sum_{n = 1}^{+\infty} a_n g(T)^n $$

then the terms get successively smaller. If we are interested in the coefficient on $T^m$, then we can completely ignore powers of $g(T)$ with exponents greater than $m$, because all of those are "smaller" than $T^m$. In other words

$$ f(g(T)) = \sum_{n = 1}^{m} a_n g_m(T)^n + \left\langle \text{ Terms of degree > m } \right\rangle $$

where I've defined

$$ g_m(T) = \sum_{n=1}^m b_n T^n $$

so that we also have

$$ g(T) = g_m(T) + \left\langle \text{ Terms of degree > m } \right\rangle $$

So, if we are interested in the $m$-th coefficient, we can reduce the problem to one entirely about polynomials.

If you are familiar with modular arithmetic, all of the above can be interpreted as working modulo $T^{m+1}$. i.e.

$$g(T) \cong g_m(T) \pmod{T^{m+1}}$$

$$ f(g(T)) \cong \sum_{n = 1}^{m} a_n g(T)^n \cong \sum_{n = 1}^{m} a_n g_m(T)^n \pmod{T^{m+1}} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.