Friday, September 16, 2011

How to Guess the Binomial Theorem for any index


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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
\begin{equation}\label{achoosek}
\frac{n(n-1)\cdots (n-k+1)}{k!},
\end{equation}
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 $a=-1$.

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