5
$\begingroup$

I am trying to prove that for $z \in \mathbb C, |z|\le 1$:

$$|e^z -1| \le 2 |z|$$

But I'm stuck and I need help. I showed that for all $z$:

$|e^z -1| \le |z|e^{|z|}$

but it does not seem useful. Also, $|e^z -1| \le \sum_{n \ge 1}{|z|^n \over n!}$ does not seem useful.

Could someone show me how one would prove this inequality?

$\endgroup$
0
6
$\begingroup$

Let $|z| \le 1$ and $N \in \Bbb N$. Then

$$ \left|\sum_{n = 1}^N \frac{z^n}{n!}\right| \le \sum_{n = 1}^N \frac{|z|^n}{n!} \le \sum_{n = 1}^N \frac{|z|^n}{2^{n-1}} \le |z| \sum_{n = 1}^\infty \left(\frac{1}{2}\right)^{n-1} = 2|z|.$$

Letting $N\to \infty$ results in

$$|e^z - 1| \le 2|z|.$$

$\endgroup$
1
  • $\begingroup$ Thank you for your help. I understand it now. $\endgroup$ – Anna Apr 20 '15 at 8:06
4
$\begingroup$

$\frac {e^z - 1}{z}= 1 + z/2 + z^2/3! + \cdots .$ For $|z|\le 1,$ the absolute value of this is $\le 1 + 1/2 + 1/3! + \cdots = e - 1 <2.$

$\endgroup$
1
  • $\begingroup$ But $2 |z| \le 2$... $\endgroup$ – Anna Apr 20 '15 at 8:05
2
$\begingroup$

We have $|z|\le 1$. $$ \sum_{n\ge 1} \frac{|z|^n}{n!} \overset{\text{?}}\le 2|z| $$ $$ \sum_{n\ge1} \frac{|z|^{n-1}}{n!} \overset{\text{?}}\le 2 $$ $$ \sum_{n\ge 2} \frac{|z|^{n-1}}{n!} \overset{\text{?}}\le 1 $$ $$ \frac{|z|} 2 + \underbrace{\frac{|z|^2} 6 + \frac{|z|^3} {24} + \frac{|z|^4}{120} + \cdots} \overset{\text{?}}\le 1 $$ Can one show that the part over the $\underbrace{\text{underbrace}}$ is bounded above by a geometric series with first term $1/6$ and common ratio $1/4$? The sum of that series is $2/9$ and the first term above is $\le 1/2$.

$\endgroup$
1
$\begingroup$

$g(z)=\frac{e^z-1}{z}$ is a holomorphic map in the unit disk, hence by the Schwarz lemma: $$\forall z\in\mathbb{C}:|z|\leq 1,\qquad |g(z)-1|\leq |z|\tag{1} $$ that implies: $$ \forall z\in\mathbb{C}:|z|\leq 1,\qquad |g(z)|\leq 1+|z|\tag{2} $$ and finally: $$ \forall z\in\mathbb{C}:|z|\leq 1,\qquad |e^z-1|\leq |z|+|z|^2\tag{3} $$ that is stronger than needed.

$\endgroup$

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.