Im trying to solve this exercise from the book Analysis I of Amann and Escher (page 216, exercise 9).

Let $a=\sum a_k X^k\in\Bbb C[\![X]\!]$ with $a_0=1$.

(a) Show that there is some $b\in\Bbb C[\![X]\!]$ such that $ab=1$. Provide a recursive algorithm for calculating the coefficients $b_k$.

(b) Show that the radius of convergence of $\rho_b$ of $b$ is positive if $\rho_a$ of $a$ is positive.

The first part is easy, we have that $b_0=1$ and

$$b_n=-a_n-\sum_{k=1}^{n-1}a_kb_{n-k},\quad\forall n\ge 1$$

(with the convention that the empty sum is zero). But Im stuck in the second part. To context the exercise: this exercise comes prior to any definition of continuity, derivative or analyticity in the book, then, from this context, I dont know how to prove it or if it is provable.

My work so far: let $a:=\sum a_k X^k$ a formal power series with radius of convergence $\rho_a>0$, then for $a(x):=\sum_{k=0}^\infty a_k x^k$ for $x\in\Bbb B(0,\rho_a)$ we have

$$a_0=1\le\left|\sum_{k=0}^\infty a_kx^k\right|\le \sum_{k=0}^\infty |a_k|r^k=M<\infty,\quad\forall x\in\Bbb B(0,r),\text{ with }0<r<\rho_a$$

If we define $b:=\sum b_k X^k$ such that $ab=1$ then from above we have that

$$\frac1M\le\left|\sum_{k=0}^\infty b_kx^k\right|\le 1,\quad\forall x\in\Bbb B(0,r),\text{ with }0<r<\rho_a$$ but from the last expression I cannot conclude that $b$ is absolutely convergent for all $|x|<\rho_b$ for some $\rho_b$, hence I cannot conclude that $b$ have a positive radius of convergence.

Some help will be appreciated, thank you.

  • $\begingroup$ I found a complete solution here, sadly it imply some knowledge on continuity of series in it radius of convergence. $\endgroup$ – Masacroso Feb 1 '17 at 9:58

I found a solution that dosn't use continuity theorems, just basic properties of convergent sequences, so it fit well to the context where it is asked.

(a) If $ab=1$ it must be the case that $a_0=b_0=1$ and $$ c_n:=\sum_{k=0}^n a_k b_{n-k}=0,\quad\forall k\in\Bbb N_{\ge 1}\tag{1} $$ From (1) we have that $$ 0=\sum_{k=1}^{n-1}a_kb_{n-k}+b_na_0+a_nb_0\implies b_n=-a_n-\sum_{k=1}^{n-1}a_kb_{n-k}\tag{2} $$ From (2) is clear that the coefficients $b_n$ are well defined, hence $b$ exists for all $a$ such that $a_0=1$.

(b) Let $a:=\sum a_k X^k$ a formal power series with radius of convergence $\rho_a>0$, then for $\underline a(x):=\sum_{k=0}^\infty a_k x^k$ for $x\in\rho_a\Bbb B$ we have $$ \left|\sum_{k=0}^\infty a_kx^k\right|\le \sum_{k=0}^\infty |a_k|r^k=M<\infty,\quad\forall x\in r\Bbb B,\text{ with }0<r<\rho_a\tag3 $$ Then suppose that there is an $\epsilon\in(0,r)$ such that $$ \sum_{k=1}^\infty |a_k|\epsilon^k\le1\tag4 $$ From $(2)$ we can write the bound $$ |b_n|\le \sum_{k=1}^n|a_kb_{n-k}|\tag5 $$ Now from (4) and (5) we want to prove by induction that $|b_n|\le \epsilon^{-n}$. Observe that $b_0=1$ so the base case holds. Now assume that $|b_n|\le \epsilon^{-n}$ and from $(5)$ we find that $$ |b_{n+1}|\le \sum_{k=1}^{n+1}|a_k|\epsilon^{-n-1+k}\le \epsilon^{-n-1}\tag6 $$ where the second inequality is a consequence of $(4)$. Thus by Hadamard formula we find that $$ \rho_b=\frac1{\limsup\sqrt[n]{|b_n|}}\ge \epsilon>0\tag7 $$ So we only need to prove that such $\epsilon>0$ exists. Set $K:=\max_{k\in\Bbb N_{\ge0}}|a_k|$, then we have the bound $$ \sum_{k=1}^\infty|a_k| x^k\le K\sum_{k=1}^\infty x^k=\frac{Kx}{1-x},\quad\text{for }x\in(0,1)\tag8 $$ Then for any $\epsilon\in(0,\frac1{K+1} \wedge r)$ we find that $$ \sum_{k=1}^\infty|a_k|\epsilon^k\le K\frac{\epsilon}{1-\epsilon}\le 1\tag9 $$ as desired.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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