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Thread: Negative logarithm while finding limiting difference

  1. #1
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    Negative logarithm while finding limiting difference

    Hi Guys,

    I hit a dead end while solving this problem. I haven't done Complex numbers so far, so I am not sure how to proceed.

    By using logarithms, find approximately the difference between $\displaystyle \dfrac{a}{1 - r}$ and $\displaystyle \dfrac{a(1 - r^n)}{1 - r}$ where,

    $\displaystyle a = 2, r = -0.7, n = 1001$

    I used the difference formula,
    $\displaystyle
    D = \dfrac{ar^n}{r - 1}
    $

    and substituting for a, r and n, I got,

    $\displaystyle D = \dfrac{2(-0.7)^{1001}}{1.7}$

    But for the -0.7 I could have used the logarithm identities to solve this. But I can't take $\displaystyle log -0.7$. The required answer is $\displaystyle -1.032 \times 10^{-155}$

    How do I proceed? Thanks again for your help!

    Edit:
    I went back and calculated $\displaystyle S$ and $\displaystyle S_{\infty}$ individually, They are both identical $\displaystyle =1.176$. Hence the difference would be 0. Is the textbook answer wrong?
    Last edited by mathguy80; Apr 13th 2011 at 06:51 AM. Reason: Fixed question
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  2. #2
    Moo
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    Hello,

    You are asked to compute $\displaystyle D-\frac{a}{r-1}$, not D alone.
    At the moment, I can't see why you're asked to use the logarithm... Anyway, $\displaystyle (-0.7)^{1001}=-(0.7)^{1001}$, if it can be of any help...
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    You are asked to compute $\displaystyle D-\frac{a}{r-1}$, not D alone.
    Sorry @Moo, I typed that in word form incorrectly, the question was difference of

    $\displaystyle \dfrac{a}{1 - r}$ and $\displaystyle \dfrac{a(1 - r^n)}{1 - r}$

    I've edited the question, It still has the negative logarithm.
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