# Taylor seires

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,...

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).

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