## Friday, September 16, 2011

### How to Guess the Binomial Theorem for any index

Newton extended the Binomial Theorem to the case where the index is no longer a non-negative integer. Newton did not provide a proof of the general case, where the index is a real number. We too will not provide a proof, but will motivate Newton's Binomial Theorem by showing some of the clues that lead to the statement of the general case.

We wish to generalize the identity
$$(1+x)^n=\sum_{k=0}^n {n\choose k} x^k$$
by replacing $n$ by a real number $a$. On the LHS, there is no problem, since the product $(1+x)^a$ makes sense for $a$ a real number. But on the RHS, there are two problems:

1. The Binomial Coefficient ${n\choose k}$ is defined only when $n$ is a non-negative integer.
2. The index of summation goes from $0$ to $n$, and thus $n$ has to be a non-negative integer.

The problems are easily solved. Note that ${n\choose k}$ may be written as
\label{achoosek}
\frac{n(n-1)\cdots (n-k+1)}{k!},

and \eqref{achoosek} makes sense if we replace $n$ by $a$.
Further, note that when $k>n$, then \eqref{achoosek} reduces to $0$. So we may as well write the Binomial Theorem as
$$(1+x)^n=\sum_{k=0}^{\infty} \frac{n(n-1)\cdots (n-k+1)}{k!} x^k.$$
Since all the terms of this series where $k$ is bigger than $n$ reduce to $0$, the series reduces to the finite sum of the familiar Binomial Theorem for non-negative integral index.

However, if we replace $n$ by a real number $a$, we may have to deal with an infinite series, and we need conditions for it to converge. It turns out the series converges whenever $|x|<1$. So finally, we are ready to state the Binomial Theorem for real index.
\begin{align}
(1+x)^a&=&\sum_{k=0}^{\infty} \frac{a(a-1)\cdots (a-k+1)}{k!} x^k, \text{ for $|x|<1$}\label{binseries} \\
&=& 1+ax+\frac{a(a-1)}{2!}x^2+\frac{a(a-1)(a-2)}{3!}x^3+\cdots\notag
\end{align}

The conditions we need on $x$ are motivated by an example of the Binomial Theorem for real index that we have already seen. Recall the formula
$$\sum_{k=0}^\infty {x^k} = \frac{1}{1-x}, \text{ for |x|<1. }$$
for the sum of the geometric series with first term $1$ and common ratio $x$. This formula is a special case of \eqref{binseries}, where $a=-1$.

### The q-analog of the Gamma Function

I have begun reading Bruce Berndt's "Ramanujan's Notebooks", Part III. Here is a small morsel from Ramanujan's table: Entry 1(ii) of Chapter 16 of his Notebooks. Its a discovery proof of the limit of the $q$-Gamma function, as $q$ goes to 1. In my humble opinion, this is easier than the usual proof (due to Gosper) which appears in Gasper and Rahman.

The $q$-analog of the Gamma Function

The objective of this note is to show how to arrive at the definition of the $q$-analog of the Gamma function. To do so, we "discover" the limit:
\label{entry1ii} \newcommand{\qrfac}[2]{{\left({#1}; q\right)_{#2}}} \lim_{q\to 1} \frac{\qrfac{q}{\infty}}{(1-q)^x \qrfac{q^{x+1}}{\infty}}= \Gamma (x+1).

Recall the limit definition of the Gamma function (from, for example Rainville [5, p. 11]):
$$\Gamma(x+1):=\lim_{n\to \infty} \frac{n! n^x}{(x+1)(x+2)\cdots (x+n)}.$$
To derive \eqref{entry1ii}, we find a $q$-analog of this limit. To that end, we use:
1.  $\displaystyle \lim_{q\to 1} \frac{\qrfac{q}{n}}{(1-q)^n} = n!$
2. $\displaystyle \lim_{q\to 1} \left(\frac{1-q^n}{1-q}\right)^x =n^x$
3. $\displaystyle \lim_{q\to 1} \frac{\qrfac{q^{x+1}}{n}}{(1-q)^n}=(x+1)(x+2)\cdots (x+n)$
Thus, we have
\begin{align*}\require{cancel} \Gamma(x+1)&= \lim_{n\to \infty} \frac{n! n^x}{(x+1)(x+2)\cdots (x+n)}\cr
& = \lim_{n\to \infty} \lim_{q\to 1} \frac{(1-q)^n}{\qrfac{q^{x+1}}{n}}\cdot \frac{\qrfac{q}{n}}{(1-q)^n} \cdot \left(\frac{1-q^n}{1-q}\right)^x\cr &= \lim_{q\to 1}\lim_{n\to \infty} \frac{\cancel{(1-q)^n}}{\qrfac{q^{x+1}}{n}}\cdot \frac{\qrfac{q}{n}}{\cancel{(1-q)^n}} \cdot \left(\frac{1-q^n}{1-q}\right)^x\cr &= \lim_{q\to 1} \frac{\qrfac{q}{\infty}}{\qrfac{q^{x+1}}{\infty}} \frac{1}{(1-q)^x}. \end{align*}
Here, we assume that the limits can be interchanged, and $|q|<1$. This completes the derivation of \eqref{entry1ii}.

Given the relation \eqref{entry1ii}, we can define the $q$-Gamma function, for $|q|<1$, as \label{qgammadef} \Gamma_q (x)= \frac{\qrfac{q}{\infty}}{(1-q)^{x-1} \qrfac{q^{x}}{\infty}}.

Remarks. The proof by Gosper, reported by Andrews [1] and reproduced in Gasper and Rahman [4] uses Euler's Product definition of the Gamma Function. Equation \eqref{entry1ii} is Entry 1(ii) in Berndt [2, ch.16]. The limit definition is entry 2293 in Carr's book [3], so Ramanujan had access to it.

References
1. G. E. Andrews, $q$-Series: Their development and application in analysis, number theory, combinatorics, physics and computer algebra, NSF CBMS Regional Conference Series, 66 1986.
2. B. C. Berndt, Ramanujan's Notebooks, Part III, Springer Verlag, New York, 1991.
3. G. S. Carr, Formulas and Theorems of Pure Mathematics, 2nd ed., Chelsea, NY, 1970.
4. G. Gasper and M. Rahman, Basic Hypergeometric Series, Encyclopedia of Mathematics And Its Applications 35, Cambridge University Press, Cambridge, 1990; Second Ed. (2004).
5. E. D. Rainville, Special Functions, Chelsea, NY (1960).