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..

To solve this question, you need to use the Taylor series expansion for the natural logarithm function ln(x).

The Taylor series expansion for ln(x) is:

ln(x) = ln(a) + (x - a) / a - (x - a)^2 / 2a^2 + (x - a)^3 / 3a^3 - (x - a)^4 / 4a^4 + ...

In this case, you want to compute ln(0.8). You need to choose a value for which you know ln(a). In this case, it's a good choice to let a = 1.

Substituting the values into the equation, we get:

ln(0.8) = ln(1) + (0.8 - 1) / 1 - (0.8 - 1)^2 / 2(1)^2 + (0.8 - 1)^3 / 3(1)^3 - (0.8 - 1)^4 / 4(1)^4 + ...

Now, you can simplify and compute the answer. Take the terms one by one and evaluate them:

ln(0.8) = 0 + (-0.2) / 1 - (-0.2)^2 / 2 + (-0.2)^3 / 3 - (-0.2)^4 / 4 + ...

Continue evaluating the terms until you reach the desired level of accuracy or until the pattern becomes clear.

For example, evaluating the first 5 terms, we get:

ln(0.8) = 0 + (-0.2) / 1 - (-0.2)^2 / 2 + (-0.2)^3 / 3 - (-0.2)^4 / 4
= 0 - 0.2 - 0.02 / 2 + 0.008 / 3 - 0.00032 / 4
= -0.2 + 0.01 - 0.0026667 + 0.00008
= -0.1915867

Comparing this with the actual value of ln(0.8) ≈ -0.22314, you can see that it is a close approximation.

Therefore, the correct answer is e) none of the proceeding, as none of the given series can be directly used to compute ln(0.8), but it can be approximated using the Taylor series expansion.