1. ## Taylor Series

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

2. Originally Posted by cassie00
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
Since you are concerned with $\displaystyle f(0), f'(0), f''(0)$ etc., keep in mind that $\displaystyle 1^n = 1$ for all n.

So you are having trouble generalizing the product $\displaystyle \frac{-1}{2}*\frac{-3}{2}*\frac{-5}{2}...$?

3. Originally Posted by icemanfan
Since you are concerned with $\displaystyle f(0), f'(0), f''(0)$ etc., keep in mind that $\displaystyle 1^n = 1$ for all n.

So you are having trouble generalizing the product $\displaystyle \frac{-1}{2}*\frac{-3}{2}*\frac{-5}{2}...$?

I thought the summation also involved dividing by n!.

4. Originally Posted by cassie00
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