Compute the Taylor series of (1-x)^(-1/2) by Taylor's formula.
I used Taylor's formula to get the first few terms but couldn't see a reasonable pattern to generalize...
Any help at all is appreciated
If you know the binomial theorem it would make this problem alot faster?
$\displaystyle f(x)=(1-x)^{-1/2}$
$\displaystyle f'(x)=(\frac{1}{2})(1-x)^{-3/2}$
$\displaystyle f''(x)=(\frac{1}{2})(\frac{3}{2})(1-x)^{-5/2}$
$\displaystyle f'''(x)=(\frac{1}{2})(\frac{3}{2})(\frac{5}{2})(1-x)^{-7/2}$
if we notice the pattern for n > 1 we get
$\displaystyle f^{(n)}(x)=\frac{1\cdot 3\cdot 5 \cdot \cdot \cdot (2n-1)}{2^n}(1-x)^{2n+1}$
This should get you started