Generalized Binomial Expansion of $\left(1+x \right )^{y}$ I was trying to expand $\displaystyle \left(1+x\right)^{y}$ as a power series in terms of $y$ using the Generalized Binomial Theorem, 
$\displaystyle \left(1+x\right)^{y}=\sum_{k=0}^{\infty}\binom{y}{k}x^{k}$
Assume $x,y \in \mathbb{R}$ and $\left | x \right | < 1$. With this in mind I would get something like 
$\displaystyle \left(1+x\right)^{y}=\sum_{k=0}^{\infty}a_{k}\cdot y^{k}$
My question is, what is the general form of the $a_{k}$? 
thanks.
 A: EDIT: Didn't realize you asked for $y^k$, by bad. I just read past and assumed the obvious.
You can do the same as my first response:
$${\left( {1 + x} \right)^y} = f(y)$$
Since the $k$th derivative will be ${{{\log }^k}\left( {1 + x} \right)}$ you have
$${\left( {1 + x} \right)^y} = \sum\limits_{k = 0}^\infty  {\frac{{{{\log }^k}\left( {1 + x} \right)}}{{k!}}} {y^k}$$
I don't think you can go any further without spending a long time with some pen and paper.

Assume
$${\left( {1 + x} \right)^{\alpha}} = \sum\limits_{k = 0}^\infty  {{a_k}{x^k}} $$
Since if two series
$$\eqalign{
  & \sum {{a_k}{{\left( {x - a} \right)}^k}}   \cr 
  & \sum {{b_k}{{\left( {x - a} \right)}^k}}  \cr} $$
sum up to the same function then
$${a_k} = {b_k} = \frac{{{f^{\left( k \right)}}\left( a \right)}}{{k!}}$$
for every $k \leq 0$, we can assume:
$$a_k = \dfrac{f^{(k)}(0)}{k!}$$ 
Putting $y = {\left( {1 + x} \right)^{\alpha}}$ we get
$$y'(0) = \alpha$$
$$y''(0) = \alpha(\alpha-1)$$
$$y'''(0) = \alpha(\alpha-1)(\alpha-2)$$
$$y^{{IV}}(0) = \alpha(\alpha-1)(\alpha-2)(\alpha-3)$$
We can prove in general that
$$y^{(k)}= \alpha(\alpha-1)\cdots(\alpha-k+1)$$
or put in terms of factorials
$$y^{(k)}(0)= \frac{\alpha!}{(\alpha-k)!}$$
This makes
$$a_k =  \frac{\alpha!}{k!(\alpha-k)!}$$ 
which is what we wanted.
$${\left( {1 + x} \right)^\alpha } = \sum\limits_{k = 0}^\infty {\alpha \choose k}  {{x^k}} $$
You can prove this in a more rigorous manner by differential equations:


*

*Set $f(x) = \displaystyle \sum\limits_{k = 0}^\infty {\alpha \choose k}  {{x^k}}$ and prove the radius of convergence is 1.

*Show that $f(x)$ is the solution to the ODE $$y' - \frac{\alpha }{{x + 1}}y = 0$$ with initial condition $f(0)=1$. 

*By the theorem that the solution to the linear equation


$$y'+P(x)y=R(x)$$
with initial conditions $f(a) = b$ is unique, you can prove the assertion. (prove that 
$y = {\left( {1 + x} \right)^{\alpha}}$ also satifies the equation and you're done.)
