f(x) =ln (1-x)
a) Compute f'(x), f''(x), f'''(x). Spot the pattern and give an expression for f ^(n) (x) [the n-th derivative of f(x)]
b) Compute the MacLaurin series of f(x) (i.e. the Taylor series of f(x) around x=0)
c) Compute the radius of convergence and determine the interval of convergence of the series in b).
d) Determine the Taylor series of f'(x) around x=0. Can you do so without using b)?
e) How would you have computed part b) if you had first done part d)?
for part a) i got, please check.
f'(x) = -1/(1-x)
f''(x) = -1/(1-x)^2
f'''(x) = -2/(1-x)^3
f^(n) (x) = -((n-1)!)/(1-x)^n
for n = 1,2,3,...
This is how far i got to, please help
f^(n)(0) = -(n-1)!
so Sum_{n=0 to infinity} x^n/n! f^(n)(0)=
-Sum_{n=1 to infinity} x^n/n
Here we have used that the zero-th derivative (i.e. the function itself) is zero for x=0.
The radius of convergence, R, is given by:
R = 1/L
where L is the limit of the absolute value of a_{n+1}/a_{n} as n--> infinity.
Inour case a_{n} = 1/n and you see that L = 1.
f'(x) = -1/(1-x). You can use that the series expoansion of this function is given by the geometric series:
1/(1-x) = Sum_{n=0 to infinity} x^(n)
If you integrate this term by term you obtain the series expansion of -log(1-x).
3 answers