Question

Urgent help needed:
Expand the following up to the term in x^3:

Answer 1

\[\frac{ 1+x }{ \sqrt{1-x} }\]

Answer 2

idk

Answer 3

expand in a power series i take it

Answer 4

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

Answer 5

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

Answer 6

yikes, now we need the derivative of that beast

Answer 7

do i have to do that too?

Answer 8

Yes, please?

Answer 9

grrr

Answer 10

lets cheat

Answer 11

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

Answer 12

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

Answer 13

one more?

Answer 14

So I find the derivative of that too?

Answer 15

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

Answer 16

i think

Answer 17

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

Answer 18

hmm let me check if this is right

Answer 19

no messed up \(3!=6\) doh

Answer 20

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

Answer 21

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

Answer 22

http://www.wolframalpha.com/input/?i=%281%2Bx%29%2Fsqrt%281-x%29
yup !!

Answer 23

Okay, thank you so much! :)

Answer 24

yw