Elementary Calculus: Binomial Series

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Binomial Series

THE BINOMIAL SERIES FOR (1+x)^p

Let us first consider the case where p is a nonnegative integer m, whence (1 + x)^m is a polynomial. The Binomial Theorem states that for nonnegative integers m,

Setting a = 1 and b = x we obtain a finite power series for (f + x)^m,

When p < 0, and when p > 0 but p is not an integer, we shall see that (1 + x)^pis the sum of a similar power series but with infinitely many terms. Let g(x) = (1 + x)^p.

Step 1

By differentiation we see that

g'(x) = p(i + x^)p-1g"(x) = p (p - 1)(1 + x)^p-2,

g⁽ⁿ⁾(x) = p (p - 1) ... (p - n + 1)(1 + x)^p-n.

Thus at x = 0, g(0) = 1, g'(0) = p, g"(0) = p (p - 1), g⁽ⁿ⁾(0) = p(p - l)...(p - n + 1).

Step 2

The MacLaurin series is

We use the Ratio Test.

Therefore the series converges for |x| < 1, diverges for |x| > 1, and has radius of convergence r = 1. We denote the sum by f(x).

Step 3

We wish to show that the sum f(x) is equal to (1 +x)^pfor|x| < 1. In this case, the MacLaurin Formula does not give the needed information (see Problem 27 at the end of this section). Instead we show that the quotient f(x)/(l + x)^phas derivative zero for |x| < 1. We have

It suffices to show that f'(x)(1 + x) = pf(x) or f'(x) + xf'(x) = pf(x). Let us compute f'(x) and xf'(x).

Adding the power series, we have

Thus f'(x) + xf'(x) = pf(x), We conclude that for some constant C, f(x)(l + x)^-p = C.

At x = 0, f(x) = 1 = (1 + x)^-p. Hence C = 1. This shows that (1 + x)^p = f(x) for |x| < 1. Thus we have the binomial series

|x| < 1.

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Last Update: 2006-11-08