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Math Help - Simplifying quotients involving powers of e

  1. #1
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    Simplifying quotients involving powers of e

    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!
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  2. #2
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    Re: Simplifying quotients involving powers of e

    \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*}
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Simplifying quotients involving powers of e

    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}}
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  4. #4
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    Re: Simplifying quotients involving powers of e

    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!
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Simplifying quotients involving powers of e

    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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