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)?
I did part a) (hope its right)
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,...
I am not too sure for the other ones, thanks for any help.
The power series that we get is,*)
In order to find the radius of convergence we will use the generalized ratio test for power series:
Thus the reciprocal of that is also 1.
Meaning the radius of convergence is 1.
Thus,
you can also write
The series converges absolutely.
*)Which is in fact not the power series for .
i am kind of lost right now, sorry about that.
so far, we have found part a) which is the n-th derivative of f(x),
so what about MacLaurin series of f(x), is it (-x/1) +(x^2/2) - (x^3/3) +...?
and also for f''(x), i don't know why i kept getting f''(x) = -1/(1-x)^2, which you don't have a negative.
thanks.
To add, "If a function is infinitely differenciable at then the Taylor series is definied as a power series...."
But, the Taylor series of a function does not always represent the function. You need to show that the remainder term sequence is convergent to zero. Like here.