Question

For \(n \geq 1\) set \( f(x) = \frac{1}{(1-x)^{n+1} }\) . Calculate the power series expansion of \(f\) at \(x_{0} = 0\) by two different ways:

a) Apply the formula the binomial series.

Answer 1

for any n?

Answer 2

i think yes..

Answer 3

ok maybe it is not that bad

Answer 4

we have
\[\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k\] then
\[\frac{1}{(1-x)^2}\] is the derivative of the previous one
so it is
\[\sum_{k=1}^\infty kx^{k-1}\]

Answer 5

ok..

Answer 6

if you want to keep starting at \(k=0\) you can write it as
\[\sum_{k=0}^{\infty}(k+1)x^k\]
\[\frac{1}{(1-x)^3}\] is almost the derivative of the previous one, but not quite , it is \(\frac{1}{2}\) times that one so you get
\[\frac{1}{(1-x)^3}=\sum_{k=0}^{\infty}\frac{1}{2}(n+1)(n+2)x^n\]

Answer 7

i don't know the k became an n

Answer 8

but you get the idea
i guess you can arrive at a general formula but i am too befuddled to do it
can probably google it if you like

Answer 9

ok thanks for your effort satellite i made you very tired today :)

Answer 10

yeah a quick search gave me this
http://www.ucl.ac.uk/Mathematics/geomath/level2/series/ser8.html
you might come up with a better one, but i think the idea is right. take successive derivatives and adjust

Answer 11

oh ignore that it is not what you are doing sorry

Answer 12

i mean ignore the link

Answer 13

ok no probleem

Answer 14

thank you satellite