# exponential function

• Oct 23rd 2010, 10:37 AM
DontKnoMaff
exponential function
For the exponential function $exp(z) = \sum^{\infty}_{n=0} \frac{z^n}{n!}$, we know that $exp(z+w)=exp(z).exp(w)$, and that, if $x \geq 0$, then $exp(x) \geq 1+x$. Use these facts to prove that, if $1 \leq n \leq m$, then
$|\prod^n_{j=1} (1+a_j) - \prod^m_{j=1} (1+a_j)| \leq exp(\sum^n_{j=1} |a_j|) . (exp(\sum^m_{j=n+1} |a_j|)-1)$

I simplified to:
$|a_{n+1}+a_{n+2}+...+a_m+a_{n+1}a_{n+2}+...+a_{n+1 }a_{n+2}...a_m| \leq \sum^{\infty}_{n=0} \frac{(\sum^m_{j=1} |a_j|)^n}{n!} - \sum^{\infty}_{n=0} \frac{(\sum^n_{j=1} |a_j|)^n}{n!}$
from here I had the idea that I could break n=0 and n=1 out of the exponentials since n=0 doesnt yield anything and n=1 yields some individual terms i could subtract from the left side. I am thinking this is the wrong approach though
• Oct 24th 2010, 09:42 AM
Opalg
Quote:

Originally Posted by DontKnoMaff
For the exponential function $\exp(z) = \sum^{\infty}_{n=0} \frac{z^n}{n!}$, we know that $\exp(z+w)=\exp(z).\exp(w)$, and that, if $x \geq 0$, then $\exp(x) \geq 1+x$. Use these facts to prove that, if $1 \leq n \leq m$, then

$\left|\prod^n_{j=1} (1+a_j) - \prod^m_{j=1} (1+a_j)\right| \leq \exp\left(\sum^n_{j=1} |a_j|\right) . \left(\exp\left(\sum^m_{j=n+1} |a_j|\right)-1\right)$.

I think you should start by noticing that the left side of that inequality can be written as $\left|\prod^m_{j=1} (1+a_j).\left(\prod^n_{j=m+1} (1+a_j) - 1\right)\right|$.