Question

find the Mclaurin series for f(x) = (1-x)^(-2)

So I took the derivative of

f(x) = (1-x)^(-2)

f(x) = (1-x)^(-2)
f'(x) = 2(1-x)^(-3)
f''(x) = 6(1-x)^(-4)
f'''(x) = 24(1-x)^(-5)
f''''(x) = 120(1-x)^(-6)

then I took f(0) for all the derivatives

f(0) = 1
f''(0) = 2
f'''(0) = 6
f''''(0) = 24
f'''''(0) = 120

Can anyone show me how to get a series from this?

Answer 1

I think there is easier way but let me study it for a bit

Answer 2

I assume I use the formula

\[\frac{f^{n}(0)(x)^{n}}{n!}\]

Answer 3

I will teach you a better way

Answer 4

now , it is common knowlege(back of the book)

\[\frac{1}{1-x}=\sum _{n=0}^{\infty } x^n\]

Answer 5

I need to use Mclaurin series, I know that method at least I think I do I still need to seek help to ensure I'm doing it correclty

Answer 6

though

Answer 7

you want to find series for
\[\frac{1}{(1-x)^2}\]

Answer 8

yes but I have to use the Mclaurin series method

Answer 9

notice

\[\frac{D}{\text{Dx}}\left[\frac{1}{(1-x)}\right]=\frac{1}{(1-x)^2}\]

Answer 10

I think @91 is leading you to the easiest method.

Answer 11

remember
\[\frac{1}{(1-x)}=\sum _{n=0}^{\infty } x^n\]

so take derivative of other side to get
\[\frac{1}{(1-x)^2}=\sum _{n=0}^{\infty } n x^{n-1}\]

Answer 12

this question is more about learning how to use the Mclaurin series method than anything else

Answer 13

We should add that the convergence is only valid for |x|<1

Answer 14

I know the method he is using

Answer 15

Since the McLaurin Series is unique. It does not matter which method to use to find it.

Answer 16

ugh, I should have asked a question with a trig function or e^(x) something