How to expand $(x+y)^{-n}$?

Question

How to expand $(x+y)^{-n}$ by binomial theorem, where $n$ is a positive integer? Is there any limitation for $x$ and $y$? If it can be expanded, how to compute the coefficients? Many thanks in advance.

Answer 1

It's not really the binomial theorem, but rather the binomial series.

$$ (x + y)^{-n} = x^{-n} (1 + y/x)^{-n} = x^{-n} \sum_{j=0}^\infty {{-n} \choose j} (y/x)^j$$ (convergent if $|y| < |x|$)

Answer 2

See:

http://mathworld.wolfram.com/NegativeBinomialSeries.html

A limitation would be (x+y) <> 0

Answer 3

Hint : $$(x+y)^{-n}=\frac{1}{y^n}\sum_{i=0}^{\infty}\binom{n+i-1}{i}(x/y)^i$$

Answer 4

This has a singularity along the line $y = -x$. The expansion in terms of powers of $y$ and $x$ will differ depending on whether you are above or below this line.

No binomial theorem, but note that $$ \sum_{n = 0}^\infty z^n = \frac{1}{1-z} $$ whenever $|z| < 1$.

If we relabel $y = -w$, and look at $(x - w)^{-1}$, we get $$ \frac{1}{x-w} = \frac{1}{x}\frac{1}{1-\frac{w}{x}} $$ and $$ \frac{1}{x-w} = -\frac{1}{w}\frac{1}{1-\frac{x}{w}} $$ which converge separately for $x > w$ and $w > x$ in each case (above and below the line x = w respectively).

But then $(x - w)^{-1}$ can be written as $$ \sum_{n = 0}^{\infty} x^{-n -1} w^n $$ and $$ - \sum_{n = 0}^{\infty} x^n w^{-n-1} $$ in each region.

Now, $$ \partial_x^{n} (x - w)^{-1} = (-1)^{n} n! (x - w)^{-n-1} $$ where I have written $\partial_x$ for the partial derivative with respect to $x$.

So $$ (x - w)^{-n-1} = \frac{(-1)^n}{n!} \partial_x^n \left(\sum_{k > 0} x^{-k-1} w^k\right) \\ = \frac{(-1)^n}{n!} \sum_{k = 0}^\infty (-k-1)(-k-2)\cdots (-k - n) x^{-k-1-n}w^k \\ = (-1)^n \sum_{k = 0}^\infty (-1)^n \left(\begin{array}{c} k+n \\ k \end{array}\right) x^{-k-1-n}w^k \\ = x^{-n} \sum_{k = 0}^\infty \left(\begin{array}{c} k+n \\ k \end{array}\right) x^{-k-1}w^k $$ for the first case, and similarly, $$ (x - w)^{-n-1} = \frac{(-1)^{n+1}}{n!} \partial_x^n \left(\sum_{k > 0} x^k w^{-k-1}\right) \\ = \frac{(-1)^{n+1}}{n!}\sum_{k \geq n} (k)(k-1)\cdots(k-n+1)x^{k-n} w^{-k-1} \\ = (-1)^{n+1} \sum_{k \geq n} \left(\begin{array}{c} k \\ n \end{array}\right)x^{k-n} w^{-k-1}\\ = (-1)^{n+1}w^{-n}\sum_{k = 0}^\infty \left(\begin{array}{c} k+n \\ n \end{array}\right)x^k w^{-k-1} $$ for the second.

Note that this works for any values $x, y$ $(= -w)$ in $\mathbb{C}$.