Question

Need help w/ a Maclaurin series:

f(x)=(1-x)^(-2)
I know how to do the derivation and write out the series. I mostly need help w/ understanding and writing the summation.

Answer 1

\[f(x)=\frac{1}{(1-x)^2}\]

well, the geometric series looks like
\[\sum_{0}^{\infty} x^n=\frac{1}{1-x}\]
the derivative of tfe right hand side is
\[\frac{1}{(1-x)^2}=\sum_{1}^{\infty} nx^{n-1}\]
and there you go, just take derivative of summation with respect to x, and change the lower index by +1.

Answer 2

Feel free to ask if in doubt of anything.

Answer 3

Okay, so it is like doing a power series? And so the summation above would be the final summation? Or do I still need to manipulate it?

Answer 4

And thank you btw for answering my question! :D

Answer 5

Oh wait, I think the only thing I am confused about is where the "n" comes from (before the x^(n-1)) in the last summation above.

Answer 6

So "n" is the last power, so for example
\[(\frac{1}{2})^{0}+(\frac{1}{2})^{1}+(\frac{1}{2})^{2}+(\frac{1}{2})^{3}+(\frac{1}{2})^{4}+....+(\frac{1}{2})^{n-2}+(\frac{1}{2})^{n-1}+(\frac{1}{2})^{n}=\sum_{k=0}^{n}(\frac{1}{2})^k\]
The "k" is just the powers, so 0,1,2,3,4.... all the way up to the last one which we call "n".So n is just the last power, so in our series the "n" goes to infinity.

Answer 7

I get it now :) Thanks again!