## Australopithecus Group Title 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? 2 years ago 2 years ago

1. 91 Group Title

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

2. Australopithecus Group Title

I assume I use the formula $\frac{f^{n}(0)(x)^{n}}{n!}$

3. 91 Group Title

I will teach you a better way

4. 91 Group Title

now , it is common knowlege(back of the book) $\frac{1}{1-x}=\sum _{n=0}^{\infty } x^n$

5. Australopithecus Group Title

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

6. Australopithecus Group Title

though

7. 91 Group Title

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

8. Australopithecus Group Title

yes but I have to use the Mclaurin series method

9. 91 Group Title

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

10. eliassaab Group Title

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

11. 91 Group Title

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

12. Australopithecus Group Title

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

13. eliassaab Group Title

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

14. Australopithecus Group Title

I know the method he is using

15. eliassaab Group Title

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

16. Australopithecus Group Title

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