let's look at the last one:
ln (1 + x) = x − x^2/2 + x^3/3 - x^4/4 + x^5/5 .....
of course x ≥ -1
this should be in your text, in your class notes, or you can easily look it up
For the other two, how about just looking at the general binomial expansion:
2/(1+x)^3
= 2(1+x)^-3
= 2(1 + (-3)x + (-3)(-4)/2! x^2 + (-3)(-4)(-5)/3! x^3 ......)
simplify and state the first three terms
in 1/1+x^2 , I will assume you meant 1/(1+x^2)
= (1 + x^2)^-1
= ......
Determine the first three terms of the power series representations for the following functions:
a) 2/(1+x)^3
b) 1/1+x^2
c) ln(1+x)
2 answers
correction:
for ln (1 + x) = x − x^2/2 + x^3/3 - x^4/4 + x^5/5 .....
x is between -1 and +1
e.g. let x = .46
on my calculator: ln(1.46) = .378436...
using the above formula:
ln(1.46) = .46 - (.46)^2/2 + (.46)^3/3 - (.46)^4/4 ....
= .37545... using 4 terms
for ln (1 + x) = x − x^2/2 + x^3/3 - x^4/4 + x^5/5 .....
x is between -1 and +1
e.g. let x = .46
on my calculator: ln(1.46) = .378436...
using the above formula:
ln(1.46) = .46 - (.46)^2/2 + (.46)^3/3 - (.46)^4/4 ....
= .37545... using 4 terms