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Let $$\mathbb{C}[[x]] := \left\{ \sum_{n\geq 0} a_n x^n : a_n \in \mathbb{C} \right\}$$ be the set of formal power series of x and $$F(x) = \sum_{n\geq 0} a_n x^n, \; G(x) = \sum_{n\geq 0} b_n x^n \in \mathbb C[[x]] \text{ with } G(0) = b_0 = 0.$$



Prove that $$(F \circ G)(x) = F(G(x)) = \lim\limits_{k \rightarrow\infty} \sum\limits_{n=0}^k a_n G(x)^n$$ exists in $\mathbb{C}[[x]]$.


Prove that $$(F\circ G)'(x) = F'(G(x))G'(x).$$

To i)

I am not sure what "existing in $\mathbb{C}[[x]]$" really means. What exactly do I need to show here?

To ii)

I had few problems differentiating the terms. If $b_n$ and $a_n$ would be constant it would be easy and e.g. $G(x)'$ would be $(\sum_{n\geq 0} b x^n)' = \frac{b}{(x-1)^2}$, but it isn't. So how do I handle that?

Thanks in advance!

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See my answer here. – Bill Dubuque Jun 4 '11 at 15:10
Yes, the question immediately reminded me of that answer :-) – joriki Jun 4 '11 at 15:11
For formal power series I recommend Peter Henrici: Applied and computational complex analysis, Vol. 1, John Wiley (1974). – Christian Blatter Jun 4 '11 at 15:53
@Bill-Dubuque Thank you for this hint! Your explanation shows by Proposition 1.1.8 that the composition F(G(x)) convergences. But why does this imply that F(G(x)) is a well-defined formal power series (if F(x) and G(x) are well-defined)? – muffel Jun 4 '11 at 15:57
@muffell What do you mean by a "well-defined" formal power series? Did you read the last paragraph of the excerpt in my linked post? – Bill Dubuque Jun 4 '11 at 16:01
up vote 5 down vote accepted

I don't know if this is according to the site rules/practice, but since no one else bites, I will move my comments here in more editable and hopefully also edible form. I will remove my comments now as that seems to be the usual practice.

To prove part i) we start with the observation that the assumption $b_0=0$ implies that the for all positive integers the power series $G(x)^k$ is of the form $$ G(x)^k=b_1^kx^k+\sum_{n=k+1}^\infty c_{k,n}x^n $$ for some coefficients $c_{k,n}$. In the language of the notes (the link is given in Bill Dubuque's comment) we have $\deg G(x)^k\ge k$. Therefore the sum $$ \sum_{n=0}^\infty a_n(G_n(x))^n $$ converges to a formal power series $H(x)\in\mathbf{C}[[x]]$ with respect to the $I$-adic topology. Here $I$ is the ideal $I=x\mathbf{C}[[x]]$ (see also Prop. 1.1.8 in Dubuque's link). Part i) is now solved.

To do part ii) we need two Lemmas. I don't know, if they have been given in your textbook and/or lecture notes. The first Lemma is easy.

Lemma 1. If $F_1(x)$ and $F_2(x)$ are power series in $\mathbf{C}[[x]]$, and $F_3(x)=F_1(x)F_2(x)$ is their product, then their formal derivatives satisfy the usual 'derivative of the product' formula $$ F_3'(x)=F_1(x)F_2'(x)+F_1'(x)F_2(x). $$

If you have problems in proving this result (or finding a proof), please comment, and I will insert one here.

Corollary. If $F(x)\in \mathbf{C}[[x]]$ and $k$ is a positive integer, then $$ D(F(x)^k)=k F'(x) F(x)^{k-1}. $$

Proof. This follows from Lemma 1 as usual by induction on $k$.

Lemma 2. If the series $$ \sum_{n=0}^\infty F_n(x) $$ converges to a sum $F(x)$ in the ring $\mathbf{C}[[x]]$ w.r.t to the $I$-adic topology (i.e. in the sense of Dubuque's notes), then so does the series $$ \sum_{n=0}^\infty F_n'(x). $$ Furthermore, we have the identity $$ F'(x)=\sum_{n=0}^\infty F_n'(x). $$

Proof. If $x^\ell$ divides a summand $F_n(x)$, then clearly $x^{\ell-1}$ divides its derivative $F_n'(x)$. In other words, $\deg F_n'(x)=\deg F_n(x)-1$. As we assume that $\deg F_n(x)\to\infty$ as $n\to\infty$, this implies that $\lim_{n\to\infty}\deg F_n'(x)=\infty$, so the series $\sum_{n=0}F_n'(x)$ converges by Prop. 1.1.8. The claim of the Lemma follows from this, because the sequence of coefficients of any power $x^i$ in the sum eventually becomes a constant.

Again, if you want more details here, just ask!

Now we are in a position to finish off part ii). Let $H(x)=F(G(x))$ that we know to exist in the ring $\mathbf{C}[[x]]$ by part i). First $$ H'(x)=\sum_{n=0}^\infty D(a_n(G(x))^n $$ by Lemma 2. Here for each $n$ we have $D(a_n(G(x))^n=na_n(G(x))^{n-1}G'(x)$ by our Corollary. Therefore $$ H'(x)=\sum_{n=1}^\infty na_n(G(x))^{n-1}G'(x). $$ By applying part i) to the power series $F'(x)$ and $G(x)$ we see that the series on the right hand side is actually $F'(G(x))G'(x)$. This completes the proof of part ii).

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Lahtonen Great, thank you! – muffel Jun 13 '11 at 18:11

For question #1, I would ask you to make sure you understand why, if the constant term of $G$ is nonzero, the composition need not exist.

Then, as a satisfactory answer to #1, I would accept an argument showing very explicitly that you know how to calculate the $n$-th degree coefficient of $F\circ G$ in finitely many steps. In particular, I might want to see the remark that this coefficient depends only on the polynomials you get from $F$ and $G$ by lopping off the terms of degree greater than $n$.

For question #2, you might notice that it's sufficient to show the identity for $F(x)=x^m$, each $m\ge 0$. This would require some argument, of course. Then, proceeding along the lines suggested by Jyrki, verify that the identity does indeed hold for $x^m$.

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