Sign up ×

Let $f(x)=\sum\limits_{n=0}^\infty a_nx^n$ a power series and $f(0)\ne0$. (w.l.o.g. $f(0)=1$) Suppose the power series has radius of convergence $r>0$.

A power series is continuous in her convergence interval.

So there is a $\delta\in]0,r[$ so that for $|x|<\delta$ it's $|a_1x+a_2x^2+\dots|<1$.

My Questions:

why is $|a_1x+a_2x^2+\dots|<1$?

why did they choose $...<1$? Why $1$ ?

Thanks for helping!!

share|cite|improve this question
It seems to be a step in a proof. Maybe you could give us what they are trying to prove. – Davide Giraudo May 15 '12 at 15:49
Most likely, they are going to prove $f(z)\ne 0$ using the reverse triangle inequality. The number 1 comes from f(0)=1. – user31373 May 15 '12 at 15:50

1 Answer 1

up vote 2 down vote accepted

If $a_0+a_1x+\dots$ has a positive radius of convergence $r$, then so does $f(x)= a_1+a_2x+\dots$, in fact, its radius of convergence is also $r$.

By continuity, if $s<r$, then there is a constant $K>0$ (that may very well depend on $s$) such that $|a_1+a_2x+\dots|<K$ as long as $|x|\le s$. This is simply the fact that continuous functions on closed bounded sets are bounded.

Now, $a_0+a_1x+\dots = a_0+xf(x)$. We know that $a_0=1$ since $f(0)=1$. Presumably you want to argue that if $|x|$ is sufficiently small, then $f(x)\ne0$. To do this, you need to argue that $xf(x)\ne -1$.

But if $|x|$ is sufficiently small (say, $|x|\le s$ with $s$ as above), then $|f(x)|<K$. Now, let $\delta>0$ be so small that $\delta<s$, and $\delta K<1$. Then, if $|x|<\delta$, we have $|xf(x)|=|x||f(x)|<\delta K<1$. In particular, this gives that $xf(x)\ne -1$, which (again) I assume is what you really want.

share|cite|improve this answer

Your Answer


