Let $\Large f(x) = log(1-x)$

c) Show that the formula for the power series holds analogously for $\Large k=0$ (although the functions $\Large f^{(0)}(x)=f(x)$ and $\Large f^{(k)}(x)$ for $\Large k 0 $ are quite different).

Question

Let $\Large f(x) = log(1-x)$

c) Show that the formula for the power series holds analogously for $\Large k=0$ (although  the functions $\Large f^{(0)}(x)=f(x)$ and $\Large f^{(k)}(x)$ for $\Large k > 0 $ are quite different).

anonymous · Answer

http://www.wolframalpha.com/input/?i=expand+ln%281-x%29+at+x%3D0

anonymous · Answer

is there total solution or is it a  just part of solution there ?

anonymous · Answer

expand log(1-x) using taylor series at x=0 .. that must be equivalent to formula

anonymous · Answer

is it the blok which starts with "Series expansion at x=0:" ?

anonymous · Answer

$$
\frac 1 {1-x} =\sum_{n=0}^\infty x^n\
\int_{0}^t \frac 1 {1-x} =\sum_{n=0}^\infty \int_{0}^t x^n\
\ln(1-t) =\sum_{n=0}^\infty \frac{ t^{n+1}}{n+1}
$$

anonymous · Answer

Of course |t|<1

anonymous · Answer

:) thank you very much mr Elias, it was a suprise of you

anonymous · Answer

@hash.nuke  thank you too naturally

anonymous · Answer

yw

anonymous · Answer

np!!