# Negative logarithm while finding limiting difference

• Apr 12th 2011, 11:40 PM
mathguy80
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?
• Apr 12th 2011, 11:56 PM
Moo
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...
• Apr 13th 2011, 12:53 AM
mathguy80
Quote:

Originally Posted by Moo
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.