Simplifying quotients involving powers of e

**halfnormalled** · August 14th 2011, 07:00 AM

Hello,

I'm struggling to see the method of how this

$\frac{3}{1-e^{-4s}}\left\{\frac{1-e^{-2s}}{s}\right\}$ becomes $\frac{3}{s(1+e^{-2s})}$

It's the simplification of $\frac{1-e^{-2s}}{1-e^{-4s}}$ becoming $\frac{1}{1+e^{-2s}}$ that I'm not following...

Can anyone offer any pointers please? Thanks!

**Prove It** · August 14th 2011, 07:21 AM

$\displaystyle \begin{align*} \left(\frac{3}{1 - e^{-4s}}\right)\left(\frac{1 - e^{-2s}}{s}\right) &= \left[\frac{3}{\left(1 - e^{-2s}\right)\left(1 + e^{-2s}\right)}\right]\left[\frac{1 - e^{-2s}}{s}\right] \\ &= \frac{3}{s\left(1 + e^{-2s}\right)} \end{align*}$

**Siron** · August 14th 2011, 07:25 AM

Notice you can write by using $a^2-b^2=(a-b)\cdot(a+b)$
$1-e^{-4s}=1-\left(\frac{1}{e}\right)^{4s}=\left[1-\left(\frac{1}{e}\right)^{2s}\right]\cdot \left[1+\left(\frac{1}{e}\right)^{2s}\right]=\left(1-e^{-2s}\right)\cdot\left(1+e^{-2s}\right)$

and so:
$\frac{1-e^{-2s}}{1-e^{-4s}}=\frac{1-e^{-2s}}{(1-e^{-2s})\cdot(1+e^{-2s})}=\frac{1}{1+e^{-2s}}$

**halfnormalled** · August 14th 2011, 08:11 AM

Yes, I see! Thank you very much guys, really appreciate your effort

That certainly clears it up for me.

It's stupid of me too. I know $a^2 - b^2 = (a-b)\cdot(a+b)$ It just didn't click into my head to apply that here! I'm trying to get as much practice as I can so these things start to become more intuitive

It's certainly interesting!

**Siron** · August 14th 2011, 08:27 AM

You're welcome! It's not stupid, It's good you recognize this 'special product' now. When you're doing other exercices next time you'll definitely remember this one en that's important now.

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