1
$\begingroup$

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.

  • 0
    Using Stirling numbers is not the easiest way to obtain the $a_k$, nevertheless it would be interesting to see if the $a_k$ is $\frac{\ln^{k}(1+x)}{k!}$ as expected.2010-12-15

1 Answers 1

1

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.)