Question

Let \(\Large f(x) = log(1-x)\)
b) Expand \(\Large f^{(k)}(x)\) in a power series around \(\Large x=0\) and determine its radius of convergence.

Answer 1

the superscript k is the order of derivative or the number of times the function is composed
\[ f^k(x)\neq f^{(k)}(x) \]
you should be careful with notation

Answer 2

i have edited thx ;)

Answer 3

ok. so
\[ \LARGE f'(x)=\frac{-1}{1-x} \]

Answer 4

then
\[ \LARGE f''(x)=\frac{-(-1)(-1)}{(1-x)^2}=\frac{-1}{(1-x)^2} \]
agree?

Answer 5

hmm it looks good..

Answer 6

i have question but i want to ask when you finish..

Answer 7

then
\[ \LARGE f'''(x)=\frac{-(-1)2(1-x)(-1)}{(1-x)^4}=\frac{-2}{(1-x)^3} \]
\[ \LARGE f^{(4)}(x)=\frac{-(-2)3(1-x)^2(-1)}{(1-x)^6}=\frac{-6}{(1-x)^4} \]

Answer 8

i think you can infer the general formula for the k-th derivative from all this

Answer 9

which general formula you mean, can you post a link ?

Answer 10

i mean
\[ \LARGE f^{(k)}(x)=\frac{-(k-1)!}{(1-x)^k} \]
for k>=1