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Textbook by: Lawrence Baggett. E-mail the author

# The General Binomial Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: 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.

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.

Definition 1:

Let αα be a complex number, and let kk be a nonnegative integer. We define the general binomial coefficientαkαk by

α k = α ( α - 1 ) ... ( α - k + 1 ) k ! . α k = α ( α - 1 ) ... ( α - k + 1 ) k ! .
(1)

If αα is itself a positive integer and kα,kα, then αkαk agrees with the earlier definition of the binomial coefficient, and αk=0αk=0 when k>α.k>α. However, if αα is not an integer, but just an arbitrary complex number, then every αk0.αk0.

## Exercise 1

Estimates for the size of binomial coefficients. Let αα be a fixed complex number.

1. Show that
|αk|j=1k(1+|α|j)|αk|j=1k(1+|α|j)
(2)
for all nonnegative integers k.k. HINT: Note that
|αk||α|(|alpha|+1)(|alpha|+2)...(|α|+k-1)k!.|αk||α|(|alpha|+1)(|alpha|+2)...(|α|+k-1)k!.
(3)
2. Use part (a) to prove that there exists a constant CC such that
|αk|C2k|αk|C2k
(4)
for all nonnegative integers k.k. HINT: Note that (1+|α|/j)<2(1+|α|/j)<2 for all j>|α|.j>|α|.
3. Show in fact that for each ϵ>0ϵ>0 there exists a constant CϵCϵ such that
|αk|Cϵ(1+ϵ)k|αk|Cϵ(1+ϵ)k
(5)
for all nonnegative integers k.k.
4. Let h(t)h(t) be the power series function given by h(t)=k=0αktk.h(t)=k=0αktk. Use the ratio test to show that the radius of convergence for hh equals 1.

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

( x + y ) α = k = 0 α k x α - k y k . ( x + y ) α = k = 0 α k x α - k y k .
(6)

The problem is to determine when this infinite series converges, i.e., for what values of the three variables x,y,x,y, and αα does it converge. It certainly is correct if x=0,x=0, so we may as well assume that x0,x0, in which case we are considering the validity of the formula

( x + y ) α = x α ( 1 + t ) α = x α k = 0 α k t k , ( x + y ) α = x α ( 1 + t ) α = x α k = 0 α k t k ,
(7)

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

k = 0 α k t k k = 0 α k t k
(8)

equal

( 1 + t ) α . ( 1 + t ) α .
(9)

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

## Theorem 1

Let α=a+biα=a+bi be a fixed complex number. Then

( 1 + t ) α = k = 0 α k t k ( 1 + t ) α = k = 0 α k t k
(10)

for all t(-1/2,1/2).t(-1/2,1/2).

### Proof

Of course, this theorem is true if αα 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.t. For a general complex number α,α, we have only defined xαxα for positive xx's, so that (1+t)α(1+t)α is not even defined for t<-1.t<-1.

Now, for a general α=a+bi,α=a+bi, consider the function g:(-1/2,1/2)Cg:(-1/2,1/2)C defined by g(t)=(1+t)α.g(t)=(1+t)α. Observe that the nnth derivative of gg is given by

g ( n ) ( t ) = α ( α - 1 ) ... ( α - n + 1 ) ( 1 + t ) n - α . g ( n ) ( t ) = α ( α - 1 ) ... ( α - n + 1 ) ( 1 + t ) n - α .
(11)

Then gC((-1/2,1/2)).gC((-1/2,1/2)). (Of course, gg is actually in C(-1,1),C(-1,1), but the present theorem is only concerned with tt's in (-1/2,1/2).(-1/2,1/2).)

For each nonnegative integer kk define

a k = g ( k ) ( 0 ) / k ! = α ( α - 1 ) ... ( α - k + 1 ) k ! = α k , a k = g ( k ) ( 0 ) / k ! = α ( α - 1 ) ... ( α - k + 1 ) k ! = α k ,
(12)

and set hh equal to the power series function given by h(t)=k=0aktk.h(t)=k=0aktk. According to part (d) of the preceding exercise, the radius of convergence for the power seriesaktkaktk is 1. The aim of this theorem is to show that g(t)=h(t)g(t)=h(t) for all -1/2<t<1/2.-1/2<t<1/2. In other words, we wish to show that gg agrees with this power series function at least on the interval (-1/2,1/2).(-1/2,1/2). It will suffice to show that the sequence {Sn}{Sn} of partial sums of the power series function hh converges to the function g,g, at least on (-1/2,1/2).(-1/2,1/2). We note also that the nnth partial sum of this power series is just the nnth Taylor polynomial TgnTgn for g.g.

S n ( t ) = k = 0 n α k t k = k = 0 n g ( k ) ( 0 ) k ! t k . S n ( t ) = k = 0 n α k t k = k = 0 n g ( k ) ( 0 ) k ! t k .
(13)

Now, fix a tt strictly between -1/2-1/2 and 1/2,1/2, and let r<1r<1 be as in part (c) of Exercise 1. That is, |t/(1+y)|<r|t/(1+y)|<r for every yy between 0 and t.t. (This is an important inequality for our proof, and this is one place where the hypothesis that t(-1/2,1/2)t(-1/2,1/2) is necessary.) Note also that, for any y(-1/2,1/2),y(-1/2,1/2), we have |(1+y)α|=(1+y)a,|(1+y)α|=(1+y)a, and this is trapped between (1/2)a(1/2)a and (3/2)a.(3/2)a. Hence, there exists a number MM such that |(1+y)α|M|(1+y)α|M for all y(-1/2,1/2).y(-1/2,1/2).

Next, choose an ϵ>0ϵ>0 for which β=(1+ϵ)r<1.β=(1+ϵ)r<1. We let CϵCϵ be a constant satisfying the inequality in Part (c) of Exercise 1. So, using Taylor's Remainder Theorem, we have that there exists a yy between 0 and tt for which

| g ( t ) - k = 0 n a k t k | = | g ( t ) - ( T ( g , 0 ) n ( t ) | = | g ( n + 1 ) ( y ) ( n + 1 ) ! t n + 1 | = | α ( α - 1 ) ... ( α - n ) ( n + 1 ) ! ( 1 + y ) n + 1 - α t n + 1 | | α n + 1 | | ( 1 + y ) α | | t 1 + y | n + 1 C ϵ ( 1 + ϵ ) n + 1 M | t 1 + y | n + 1 C ϵ ( 1 + ϵ ) n + 1 M r n + 1 C ϵ M β n + 1 , . | g ( t ) - k = 0 n a k t k | = | g ( t ) - ( T ( g , 0 ) n ( t ) | = | g ( n + 1 ) ( y ) ( n + 1 ) ! t n + 1 | = | α ( α - 1 ) ... ( α - n ) ( n + 1 ) ! ( 1 + y ) n + 1 - α t n + 1 | | α n + 1 | | ( 1 + y ) α | | t 1 + y | n + 1 C ϵ ( 1 + ϵ ) n + 1 M | t 1 + y | n + 1 C ϵ ( 1 + ϵ ) n + 1 M r n + 1 C ϵ M β n + 1 , .
(14)

Taking the limit as nn tends to ,, and recalling that β<1,β<1, shows that g(t)=h(t)g(t)=h(t) for all -1/2<t<1/2,-1/2<t<1/2, which completes the proof.

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