Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
1  
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
1  
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|improve this answer

Your Answer

 
discard

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.