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May 25, 2015

Homework Help: Calculus!!!

Posted by Joan on Monday, March 19, 2007 at 9:19pm.

I need your help... I've posted this question a while ago but no one has answered it yet. Please please help me.

Which of the following series can be used to compute ln(.8)?
a) ln(x-1) expanded about x=0
b) lnx about x=0
c) ln x in powers of (x-1)
d) ln(x-1) in powers of (x-1)
e) none of the proceding

I am confused... please help!

I am not certain any of our volunteers have expandied series of log functions in the recent past, and would likely give you misleading info. Perhaps one of the non regular math profs will drop in and help you.

The Taylor Series says:
ln(x)== ln(a) + (x-a) / a - (x-a)2 / 2a2 + (x-a)3 / 3a3 - (x-a)4 / 4a4 + ...

where a is a value for which you know ln(a)
your x=.8, then I would choose a=1
and
ln(.8)=ln(1)+(-.2)/1+(-.2)^2/2+(-.2)^3/3...
=0+a series of negative terms.

I did this for 5 terms and got -.2233866
with the real value of ln(.8)=-.22314..

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