- #1
the1ceman
- 28
- 0
Hi.
I need to show that for all [tex]|z|\leq1 [/tex]:
[tex] (3-e)|z|\leq|e^{z}-1|\leq(e-1)|z|[/tex]
Now
[tex]|e^{z}-1|=\left|\sum^{\infty}_{n=1}\frac{z^{n}}{n!}\right|\leq\sum^{\infty}_{n=1}\left|\frac{z^{n}}{n!}\right|=|z|\sum^{\infty}_{n=1}\frac{\left|z\right|^{n-1}}{n!}\leq|z|\sum^{\infty}_{n=1}\frac{1}{n!}=\left(\sum^{\infty}_{n=0}\frac{1}{n!}-1\right)|z|=(e-1)|z|[/tex]
So the upper bound is done. I have no idea how to do the lower bound. I have tried different techniques but just can't seem to get it.
Please someone help!
I need to show that for all [tex]|z|\leq1 [/tex]:
[tex] (3-e)|z|\leq|e^{z}-1|\leq(e-1)|z|[/tex]
Now
[tex]|e^{z}-1|=\left|\sum^{\infty}_{n=1}\frac{z^{n}}{n!}\right|\leq\sum^{\infty}_{n=1}\left|\frac{z^{n}}{n!}\right|=|z|\sum^{\infty}_{n=1}\frac{\left|z\right|^{n-1}}{n!}\leq|z|\sum^{\infty}_{n=1}\frac{1}{n!}=\left(\sum^{\infty}_{n=0}\frac{1}{n!}-1\right)|z|=(e-1)|z|[/tex]
So the upper bound is done. I have no idea how to do the lower bound. I have tried different techniques but just can't seem to get it.
Please someone help!