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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.

Neves
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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. ain't a mathematician Dec 15 '10 at 15:53
  • @Raskolnikov, @J.M. ok, the series would be $\sum_{k=0}^{\infty}\frac{(\ln(1+x))^{k}y^{k}}{k!}$, but would the $a_k$ be the same if I had followd directly from the binomial expansion? – Neves Dec 15 '10 at 16:20
  • Not sure what you mean by followed directly from the binomial expansion? You mean like in my hint for the second approach? – Raskolnikov Dec 15 '10 at 16:22
  • @Raskolnikov, In the binomial expansion the term $\binom{y}{k}x^{k}=\frac{y(y-1)(y-2)\ldots(y-k+1)}{k!}x^{k}$ could be expanded in terms of powers of $y$ and then I would get the $a_k$ term. If I proceeded this way would I conclude that $a_k=\frac{\ln^{k}(1+x)}{k!}$? – Neves Dec 15 '10 at 16:32
  • OK, that was what I was suggesting with the Stirling numbers. The factors with the product of y's is what is called a falling factorial and can be expressed as a sum of powers of y, the coefficents will be Stirling numbers. Now, the two expansions should be the same, but it takes some work to show that. I guess you could try expanding $\ln(1+x)$ and plug it in the other formula to see if it comes out. – Raskolnikov Dec 15 '10 at 16:40
  • 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. – Neves Dec 15 '10 at 17:47

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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.)

Pedro
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