## 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

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

2. Australopithecus

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

3. 91

I will teach you a better way

4. 91

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

5. Australopithecus

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

though

7. 91

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

8. Australopithecus

yes but I have to use the Mclaurin series method

9. 91

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

10. eliassaab

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

11. 91

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

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

13. eliassaab

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

14. Australopithecus

I know the method he is using

15. eliassaab

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

16. Australopithecus

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