# Chaitanya's Random Pages

## December 25, 2013

### The binomial theorem for non-positive-integer indices

Filed under: mathematics — ckrao @ 9:16 pm

If $n$ is a positive integer, the expansion of $(x+y)^n$ has each term being a product of $n$ variables of the form $x^k y^{n-k}$ where $k$ ranges from 0 to n. The coefficient of $x^k y^{n-k}$ is precisely the number of ways we can choose $k$ of the $n$ variables to be $x$, which is $\binom{n}{k}$. Hence we have the binomial theorem:

$\displaystyle (x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}. \quad \quad(1)$

But what if $n$ is not a positive integer? This post is about how we extend this formula and why it still holds. Interestingly the same formula holds with minor modifications. Firstly we change the upper limit of the sum to infinity – the sum will then only converge in particular circumstances. Secondly we extend the definition of the binomial coefficient to complex values of $n$ via $\binom{n}{k} := \frac{n(n-1)\ldots (n-k+1)}{k!}$.

Note that if $x$ is a complex number then $x^k$ is defined for any non-negative integer $k$ – we simply do repeated multiplication, and define $x^0 = 1$ (even for $x = 0$). However a non-integer power of a complex number is not straightforward to define unless the number is a positive real. If $x > 0$ we can define $x^n := \exp(n \alpha)$ where $\alpha$ is the unique real solution to $\exp(\alpha) = x$. Extending the definition to other complex numbers leads to issues with multifunctions or discontinuities (e.g. $(-1)^{1/2}$ could be one of two values $i$ or $-i$). Hence we are going to restrict ourselves to the case $x$ and $y$ real.

While the term $x^k$ is fine, we need to take care with $y^{n-k}$ if $n$ is no longer an integer. Hence we shall restrict ourselves to non-negative $y$. The binomial theorem is valid when $y = 0$ so consider $y > 0$. When does the infinite series converge? By the ratio test, the sum $\sum_{k=0}^{\infty}a_k$ converges if the limit of $|a_{k+1}|/|a_k|$ converges to a number less than 1 as $k \rightarrow \infty$. (The sum does not converge if the limit is greater than 1 and if the limit either does not converge or is equal to 1, then the test is inconclusive.) Applying this test to our case of $a_k = \binom{n}{k} x^k y^{n-k}$ gives us

\begin{aligned}\frac{|a_{k+1}|}{|a_k|} &= \frac{n(n-1)\ldots (n-(k+1)+1) x^{k+1}y^{n-(k+1)}}{(k+1)!} \frac{k!}{n(n-1)\ldots (n-k+1) x^k y^{n-k}}&= \frac{|n-k|}{k+1}\frac{|x|}{y}. \end{aligned}

This has limit less than 1 as $k \rightarrow \infty$ provided $|x| < y$. We can thus state the following.

If $n$ is a complex number, $x$ is real, $y$ is positive  and $|x| < y$, then the sum $\displaystyle\sum_{k=0}^{\infty} \binom{n}{k} x^k y^{n-k}$ converges.

The reason this sum is equal to $(x+y)^n$ is a consequence of the Taylor series expansion of $(1+x)^n$ about $x = 0$ and the fact that the identity $\displaystyle (x+y)^n = y^n (x/y + 1)^n$ is valid when $y > 0$. Also note that the condition $|x| < y$ implies $x+y > 0$ so $(x+y)^n$ is well defined.

We can also use a differential equations approach to proving this. Fix $y$ and consider $\displaystyle f(x) =\sum_{k=0}^{\infty} \binom{n}{k} x^k y^{n-k}$ as a power series in $x$ valid for $|x| < y$. The power series is differentiable term by term within this interval of convergence and so

$\displaystyle f'(x) = \sum_{k=1}^{\infty} \binom{n}{k} k x^{k-1}y^{n-k}. \quad \quad (2)$

We may also write this as

\begin{aligned} f'(x) &=\sum_{j=0}^{\infty} \binom{n}{j+1} (j+1) x^{j}y^{n-(j+1)}\\ &= \sum_{j=0}^{\infty} \frac{n(n-1)\ldots (n-j) (j+1) x^{j }y^{n-(j+1)}}{(j+1)!}\\ &= \sum_{k=0}^{\infty} \binom{n}{k} (n-k) x^{k}y^{n-k-1}.\quad \quad (3) \end{aligned}

Adding $x$ times (2) to $y$ times (3),

\begin{aligned} (x + y)f'(x) &= \sum_{k=0}^{\infty} \binom{n}{k} k x^{k}y^{n-k} + \sum_{k=0}^{\infty} \binom{n}{k} (n-k) x^{k}y^{n-k}\\ &= n \sum_{k=0}^{\infty}x^{k}y^{n-k}\\ &= nf(x). \quad\quad (4)\end{aligned}

Hence by (4)

\begin{aligned} d/dx [(x+y)^{-n} f(x) ] &= (x+y)^{-n} f'(x) -n (x+y)^{-n-1} f(x)\\ &= (x+y)^{-n-1} [(x+y) f'(x) - nf(x)]\\ &= 0,\end{aligned}

so we deduce that $(x+y)^{-n} f(x)$ is constant. For $x=0$ this is $y^{-n} f(0) = y^{-n}y^n = 1$, and we conclude that $f(x) = (x+y)^n$.  Summarising, we have the following result.

If $n$ is a complex number, $x$ is real, $y$ is positive  and $|x| < y$, then

$\displaystyle (x+y)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k y^{n-k}.$

A particularly attractive special case of this formula is for $y=1, n = -1/2$ and $x$ replaced with $-x$:

\begin{aligned} \frac{1}{\sqrt{1-x}} &= \sum_{k=0}^{\infty} \binom{-1/2}{k} (-x)^k \\&= \sum_{k=0}^{\infty} \frac{(-1/2)(-3/2)\ldots (-1/2 - k + 1)}{k!} (-x)^k\\ &= \sum_{k=0}^{\infty} (-1)^k \frac{(1)(3)(5)\ldots (2k-1)}{k!}\frac{(-x)^k}{2^k}\\ &= \sum_{k=0}^{\infty} \frac{(2k)!}{(2)(4)\ldots (2k) k!}\frac{x^k}{2^k}\\ &= \sum_{k=0}^{\infty} \frac{(2k)!}{k! k!} \frac{x^k}{4^k} \\ &= \sum_{k=0}^{\infty} \binom{2k}{k} \frac{x^k}{4^k}, \quad |x| < 1.\end{aligned}