# 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.

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Hint: $(1+x)^y=e^{y\ln(1+x)}$. –  Raskolnikov Dec 15 '10 at 12:44
You mean $\sum_{k=0}^\infty a_kx^k$ in the last equation? –  mpiktas Dec 15 '10 at 12:44
Hint for a different approach: Stirling numbers and the falling factorial. –  Raskolnikov Dec 15 '10 at 12:46
@mpiktas, no, I'm considering the power series in terms of $y$. –  Neves Dec 15 '10 at 12:50
Neves, @Raskolnikov gave a pretty transparent hint. Do the appropriate substitutions into a certain famous power series... –  Ｊ. Ｍ. Dec 15 '10 at 15:53

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:

1. Set $f(x) = \displaystyle \sum\limits_{k = 0}^\infty {\alpha \choose k} {{x^k}}$ and prove the radius of convergence is 1.
2. Show that $f(x)$ is the solution to the ODE $$y' - \frac{\alpha }{{x + 1}}y = 0$$ with initial condition $f(0)=1$.
3. 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.)

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