4.9 The general binomial theorem

We use Taylor's Remainder Theorem to derive a generalization of the Binomial Theorem to nonintegral exponents. First we must generalize the definition of binomial coefficient.

Let $\alpha$ be a complex number, and let $k$ be a nonnegative integer. We define the general binomial coefficient $\left(\genfrac{}{}{0pt}{}{\alpha }{k}\right)$ by

$$\left(\genfrac{}{}{0pt}{}{\alpha }{k}\right)=\frac{\alpha (\alpha -1)...(\alpha -k+1)}{k!}.$$

If $\alpha$ is itself a positive integer and $k\le \alpha ,$ then $\left(\genfrac{}{}{0pt}{}{\alpha }{k}\right)$ agrees with the earlier definition of the binomial coefficient,and $\left(\genfrac{}{}{0pt}{}{\alpha }{k}\right)=0$ when $k>\alpha .$ However, if $\alpha$ is not an integer, but just an arbitrary complex number, then every $\left(\genfrac{}{}{0pt}{}{\alpha }{k}\right)\ne 0.$

REMARK The general Binomial Theorem, if there is one, should be something like the following:

The problem is to determine when this infinite series converges, i.e., for what values of the three variables $x,y,$ and $\alpha$ does it converge. It certainly is correct if $x=0,$ so we may as well assume that $x\ne 0,$ in which case we are considering the validity of the formula

where $t=y/x.$ Therefore, it will suffice to determine for what values of $t$ and $\alpha$ does the infinite series

equal

The answer is that, for n arbitrary complex number $\alpha ,$ this series converges to the correct value for all $t\in (-1,1).$ (Of course, $t$ must be larger than $-1$ for the expression ${(1+t)}^{\alpha }$ even to be defined.) However, the next theorem only establishes this equality for $t$ 's in the subinterval $(-1/2,1/2).$ As mentioned earlier, its proof is based on Taylor's Remainder Theorem. We must postpone the complete proof to [link] , where we will have a better version of Taylor's Theorem.

Let $\alpha =a+bi$ be a fixed complex number. Then

for all $t\in (-1/2,1/2).$

Of course, this theorem is true if $\alpha$ is a nonnegative integer, for it is then just the original Binomial Theorem,and in fact in that case it holds for every complex number $t.$ For a general complex number $\alpha ,$ we have only defined $x^{\alpha }$ for positive $x$ 's, so that ${(1+t)}^{\alpha }$ is not even defined for $t<-1.$

Now, for a general $\alpha =a+bi,$ consider the function $g:(-1/2,1/2)\to C$ defined by $g\left(t\right)={(1+t)}^{\alpha }.$ Observe that the $n$ th derivative of $g$ is given by

Then $g\in C^{\infty }\left((-1/2,1/2)\right).$ (Of course, $g$ is actually in $C^{\infty }(-1,1),$ but the present theorem is only concerned with $t$ 's in $(-1/2,1/2).$ )

For each nonnegative integer $k$ define

and set $h$ equal to the power series function given by $h\left(t\right)={\sum }_{k=0}^{\infty }{a}_k{t}^k.$ According to part (d) of the preceding exercise, the radius of convergence for the power series $\sum a_k{t}^k$ is 1. The aim of this theorem is toshow that $g\left(t\right)=h\left(t\right)$ for all $-1/2<t<1/2.$ In other words, we wish to show that $g$ agrees with this power series function at least on the interval $(-1/2,1/2).$ It will suffice to show that the sequence $\left\{S_n\right\}$ of partial sums of the power series function $h$ converges to the function $g,$ at least on $(-1/2,1/2).$ We note also that the $n$ th partial sum of this power series is just the $n$ th Taylor polynomial $T_g^n$ for $g.$

Now, fix a $t$ strictly between $-1/2$ and $1/2,$ and let $r<1$ be as in part (c) of [link] . That is, $|t/(1+y\left)\right|<r$ for every $y$ between 0 and $t.$ (This is an important inequality for our proof, and this is one place where the hypothesis that $t\in (-1/2,1/2)$ is necessary.) Note also that, for any $y\in (-1/2,1/2),$ we have ${|(1+y)}^{\alpha }{|=(1+y)}^a,$ and this is trapped between ${(1/2)}^a$ and ${(3/2)}^a.$ Hence, there exists a number $M$ such that ${|(1+y)}^{\alpha }|\le M$ for all $y\in (-1/2,1/2).$

Next, choose an $ϵ>0$ for which $\beta =(1+ϵ)r<1.$ We let $C_{ϵ}$ be a constant satisfying the inequality in Part (c) of [link] . So, using Taylor's Remainder Theorem, we have thatthere exists a $y$ between 0 and $t$ for which

Taking the limit as $n$ tends to $\infty ,$ and recalling that $\beta <1,$ shows that $g\left(t\right)=h\left(t\right)$ for all $-1/2<t<1/2,$ which completes the proof.