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Math Help - exponential function

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by DontKnoMaff View Post
    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|.
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