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!
To determine which series can be used to compute ln(0.8), we need to consider the Taylor series expansion of the ln(x) function.
The Taylor series expansion of ln(x) about x = a is given by:
ln(x) = [f(a) / (x-a)] + [f'(a) / (2!)] * (x-a) + [f''(a) / (3!)] * (x-a)^2 + [f'''(a) / (4!)] * (x-a)^3 + ...
Now, let's examine each option:
a) ln(x-1) expanded about x=0:
The expansion of ln(x-1) about x = 0 would involve evaluating derivatives of ln(x-1) at x = 0. But since ln(x) is undefined for x ≤ 0, this expansion is not valid for computing ln(0.8). Therefore, option a) is not the correct choice.
b) lnx about x=0:
The expansion of lnx about x = 0 is the standard natural logarithm series expansion. In this case, we can use the Taylor series expansion of ln(x) to compute ln(0.8). Therefore, option b) is a possible choice.
c) ln x in powers of (x-1):
The expansion of ln x in powers of (x-1) implies that we are expanding ln x using the Taylor series expansion, but centered around x = 1. Since ln(x) is typically expanded around x = 0, option c) is not the correct choice.
d) ln(x-1) in powers of (x-1):
The expansion of ln(x-1) in powers of (x-1) is another way of expressing the Taylor series expansion of ln(x). However, it is important to note that for this expansion, we need to evaluate ln(x-1) at x = 1. Since we are interested in computing ln(0.8), option d) is not the correct choice.
e) none of the preceding:
Based on our analysis, the correct choice would be b) lnx about x=0. This expansion allows us to compute ln(0.8) using the Taylor series expansion of ln(x) around x = 0.
Therefore, the correct option is b) lnx about x=0.