1. ## Taylor Series (1+x)^(1/2)

I'm looking to find an expression for the $\displaystyle n^{th}$ term of the Taylor Series for $\displaystyle (1+x)^{\frac{1}{2}}$.

I have found that:

$\displaystyle (1+x)^{\frac{1}{2}} = 1 + \frac{1}{1!} \left( \frac{1}{2} \right) x +\frac{1}{2!} \left( \frac{1}{2} \right) \left( \frac{-1}{2} \right) x^2 + \frac{1}{3!} \left( \frac{1}{2} \right) \left(\frac{-1}{2} \right) \left( \frac{-3}{2} \right) x^3 + \ldots$

The first term (1) is the only term that I can't get to fit the suggested general term:

$\displaystyle \frac{1}{(n - 1)!} \displaystyle\prod_{i = 1}^n \left( \frac{5 - 2i}{2} \right)$

Any suggestions much appreciated, as this is really bugging me!

2. Originally Posted by Harry1W
I'm looking to find an expression for the $\displaystyle n^{th}$ term of the Taylor Series for $\displaystyle (1+x)^{\frac{1}{2}}$.

I have found that:

$\displaystyle (1+x)^{\frac{1}{2}} = 1 + \frac{1}{1!} \left( \frac{1}{2} \right) x +\frac{1}{2!} \left( \frac{1}{2} \right) \left( \frac{-1}{2} \right) x^2 + \frac{1}{3!} \left( \frac{1}{2} \right) \left(\frac{-1}{2} \right) \left( \frac{-3}{2} \right) x^3 + \ldots$

The first term (1) is the only term that I can't get to fit the suggested general term:

$\displaystyle \frac{1}{(n - 1)!} \displaystyle\prod_{i = 1}^n \left( \frac{5 - 2i}{2} \right)$

Any suggestions much appreciated, as this is really bugging me!
let $\displaystyle f(x)=\sqrt{1+x}.$ then $\displaystyle f^{(n)}(x)=\frac{(-1)^{n+1}(2n)!}{(2n-1)4^n n!} \sqrt{(1+x)^{1-2n}}.$ thus: $\displaystyle \frac{f^{(n)}(0)}{n!}=\frac{(-1)^{n+1}(2n)!}{(2n-1)4^n(n!)^2}.$ therefore: $\displaystyle \sqrt{1+x}=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(2n)!}{(2n-1)4^n(n!)^2} x^n.$

3. Thanks for that. I'm afraid that I might need to ask for some explanation of how you found that the coeffecient of $\displaystyle \sqrt{(1+x)^{1-2n}}$ is $\displaystyle \frac{ (-1)^{n + 1} (2n)! }{ (2n - 1) 4^n n! }$? I can see that it works, but I don't see why it works - if that makes sense! In other words, I can see that it generates the desired result - but I'm not sure how I would find it independently. Thanks in advance.