## anonymous 2 years ago Urgent help needed: Expand the following up to the term in x^3:

1. anonymous

$\frac{ 1+x }{ \sqrt{1-x} }$

2. anonymous

idk

3. anonymous

expand in a power series i take it

4. anonymous

about zero i hope first term is $$f(0)=1$$

5. anonymous

first derivative is $f'(x)=\frac{3-x}{2\sqrt{1-x}^3}$ so second term is $f'(0)x=\frac{3}{2}x$

6. anonymous

yikes, now we need the derivative of that beast

7. anonymous

do i have to do that too?

8. anonymous

9. anonymous

grrr

10. anonymous

lets cheat

11. anonymous

looks like it is $f''(x)=\frac{7-x}{4\sqrt{1-x}^5}$

12. anonymous

making the next term $\frac{f''(0)}{2}x^2=\frac{7}{8}x^2$

13. anonymous

one more?

14. anonymous

So I find the derivative of that too?

15. anonymous

$f'^{(3)}(x)=\frac{33-3x}{8\sqrt{1-x}^7}$

16. anonymous

i think

17. anonymous

if so then $$\frac{f^{(3)}(0)}{3!}x^3=\frac{33}{16}x^3$$

18. anonymous

hmm let me check if this is right

19. anonymous

no messed up $$3!=6$$ doh

20. anonymous

third term is $$\frac{11}{16}x^3$$

21. anonymous

k lets check the whole dam thing and see if it is right

22. anonymous
23. anonymous

Okay, thank you so much! :)

24. anonymous

yw