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 (Nod)

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- Apr 27th 2008, 03:24 PMcassie00Taylor 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 (Nod) - Apr 27th 2008, 04:02 PMPaulRS
Try proving the general case.

Given a real number $\displaystyle u $ and a natural number $\displaystyle n$ we define: $\displaystyle \binom{u}{n}=\frac{u\cdot{(u-1)\cdot{...\cdot{(u-n+1)}}}}{n!}$

**The Binomial Theorem states that:**

$\displaystyle (1+x)^u=\sum_{n=0}^{\infty}{\binom{u}{n}\cdot{x^n} }$

$\displaystyle |x|<1$

Note that if u is natural then the formula reduces to the famous: $\displaystyle (1+x)^u=\sum_{n=0}^{u}{\binom{u}{n}\cdot{x^n}}$

- Apr 27th 2008, 04:05 PMTheEmptySetPlease don't double post
http://www.mathhelpforum.com/math-he...or-series.html

Please don't double post

It just makes more work for everyone.

I gave you the formual for the nth derivative(The long way)

Try it and see what happens - Apr 27th 2008, 04:26 PMcassie00