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

Mathematics
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\[\frac{ 1+x }{ \sqrt{1-x} }\]
idk
expand in a power series i take it

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Other answers:

about zero i hope first term is \(f(0)=1\)
first derivative is \[f'(x)=\frac{3-x}{2\sqrt{1-x}^3}\] so second term is \[f'(0)x=\frac{3}{2}x\]
yikes, now we need the derivative of that beast
do i have to do that too?
Yes, please?
grrr
lets cheat
looks like it is \[f''(x)=\frac{7-x}{4\sqrt{1-x}^5}\]
making the next term \[\frac{f''(0)}{2}x^2=\frac{7}{8}x^2\]
one more?
So I find the derivative of that too?
\[f'^{(3)}(x)=\frac{33-3x}{8\sqrt{1-x}^7}\]
i think
if so then \(\frac{f^{(3)}(0)}{3!}x^3=\frac{33}{16}x^3\)
hmm let me check if this is right
no messed up \(3!=6\) doh
third term is \(\frac{11}{16}x^3\)
k lets check the whole dam thing and see if it is right
http://www.wolframalpha.com/input/?i=%281%2Bx%29%2Fsqrt%281-x%29 yup !!
Okay, thank you so much! :)
yw

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