# Trouble with representing power series as polynomials.

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

Proof:

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?

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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

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 $2 \le j \le m$.

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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.

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

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

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