Unit 7, lesson 110

Binomial Series

How the binomial theorem extends to any real exponent as an infinite series, where that series converges including the endpoints, and how the finite binomial and multinomial theorems fit in.

  • Theory part 219
  • Practice part 220
  • Taylor and Maclaurin Series

You know that (1+x)2=1+2x+x2(1 + x)^2 = 1 + 2x + x^2 and (1+x)3=1+3x+3x2+x3(1 + x)^3 = 1 + 3x + 3x^2 + x^3. What about (1+x)1/2(1 + x)^{1/2} or (1+x)−1(1 + x)^{-1}? The same pattern of coefficients still makes sense for any exponent, but the expansion no longer stops. It becomes an infinite series, and it is the quickest way to approximate roots and powers by hand.

Key terms

Generalized binomial coefficients

Definition For a real number α\alpha and an integer k≥0k \ge 0, the binomial coefficient is

(α0)=1,(αk)=α(α−1)(α−2)⋯(α−k+1)k!for k≥1.\binom{\alpha}{0} = 1, \qquad \binom{\alpha}{k} = \frac{\alpha(\alpha - 1)(\alpha - 2) \cdots (\alpha - k + 1)}{k!} \quad \text{for } k \ge 1.

The numerator has exactly kk factors. When α\alpha is a nonnegative integer this agrees with the usual counting coefficient.

In plain words Start at α\alpha, multiply kk numbers counting down by one, then divide by k!k!. Nothing requires α\alpha to be a whole number.

Example For α=12\alpha = \frac{1}{2},

(1/21)=12,(1/22)=12⋅(−12)2=−18,(1/23)=12⋅(−12)⋅(−32)6=116.\binom{1/2}{1} = \frac{1}{2}, \quad \binom{1/2}{2} = \frac{\frac{1}{2} \cdot \left(-\frac{1}{2}\right)}{2} = -\frac{1}{8}, \quad \binom{1/2}{3} = \frac{\frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{3}{2}\right)}{6} = \frac{1}{16}.

For α=−1\alpha = -1 every coefficient is (−1k)=(−1)k\binom{-1}{k} = (-1)^k.

Binomial series theorem

Definition For every real α\alpha and every xx with ∣x∣<1\lvert x \rvert < 1,

(1+x)α=∑k=0∞(αk)xk=1+αx+α(α−1)2!x2+α(α−1)(α−2)3!x3+⋯ .(1 + x)^\alpha = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^k = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!} x^2 + \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!} x^3 + \cdots.

This is the Maclaurin series of (1+x)α(1 + x)^\alpha, since

dkdxk(1+x)α∣x=0=α(α−1)⋯(α−k+1).\frac{d^k}{dx^k}(1 + x)^\alpha \bigg\vert_{x = 0} = \alpha(\alpha - 1) \cdots (\alpha - k + 1).

If α\alpha is not a nonnegative integer, the radius of convergence is exactly 11.

In plain words Any power of 1+x1 + x can be expanded like the familiar binomial, as long as xx is between −1-1 and 11.

Example For α=12\alpha = \frac{1}{2},

1+x=1+x2−x28+x316−⋯ ,∣x∣<1.\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \cdots, \qquad \lvert x \rvert < 1.

At x=0.1x = 0.1 four terms give

1+0.05−0.00125+0.0000625=1.0488125,1 + 0.05 - 0.00125 + 0.0000625 = 1.0488125,

and 1.1=1.0488088…\sqrt{1.1} = 1.0488088\ldots. Choose 1+x\sqrt{1 + x} in the Taylor polynomial explorer with degree 33 and x=0.1x = 0.1 to see both numbers side by side.

Convergence inside the radius and at the endpoints

Definition Let α\alpha not be a nonnegative integer. The ratio of consecutive terms is

∣(αk+1)xk+1(αk)xk∣=∣α−k∣k+1∣x∣→∣x∣,\left\lvert \frac{\binom{\alpha}{k+1} x^{k+1}}{\binom{\alpha}{k} x^k} \right\rvert = \frac{\lvert \alpha - k \rvert}{k + 1} \lvert x \rvert \to \lvert x \rvert,

so the series converges absolutely for ∣x∣<1\lvert x \rvert < 1 and diverges for ∣x∣>1\lvert x \rvert > 1. At the endpoints the behavior depends on α\alpha.

ExponentAt x=1x = 1At x=−1x = -1
α>0\alpha > 0converges absolutelyconverges absolutely
−1<α<0-1 < \alpha < 0converges conditionallydiverges
α≤−1\alpha \le -1divergesdiverges

Wherever the series converges at an endpoint, Abel’s theorem together with the continuity of (1+x)α(1 + x)^\alpha at that endpoint shows its sum is (1+x)α(1 + x)^\alpha there. For example, at x=−1x = -1 with α>0\alpha > 0 the sum is 0α=00^\alpha = 0.

In plain words Inside (−1,1)(-1, 1) the series always works. At the ends, a positive exponent helps and a very negative one hurts.

Example For α=−12\alpha = -\frac{1}{2} the coefficients are 1,−12,38,−516,…1, -\frac{1}{2}, \frac{3}{8}, -\frac{5}{16}, \ldots. At x=1x = 1 the series

1−12+38−516+⋯1 - \frac{1}{2} + \frac{3}{8} - \frac{5}{16} + \cdots

alternates with terms decreasing to 00, and its sum is 2−1/2=12≈0.70712^{-1/2} = \frac{1}{\sqrt{2}} \approx 0.7071. At x=−1x = -1 all terms are positive, 1+12+38+516+⋯1 + \frac{1}{2} + \frac{3}{8} + \frac{5}{16} + \cdots, and they shrink only like a constant times 1k\frac{1}{\sqrt{k}}, so the series diverges.

Integer termination

Definition If α=m\alpha = m is a nonnegative integer, then (mk)=0\binom{m}{k} = 0 for every k>mk > m, because the factor m−m=0m - m = 0 appears in the numerator. The series stops and becomes the finite binomial theorem

(1+x)m=∑k=0m(mk)xk,and more generally(a+b)m=∑k=0m(mk)am−kbk,(1 + x)^m = \sum_{k=0}^{m} \binom{m}{k} x^k, \qquad \text{and more generally} \qquad (a + b)^m = \sum_{k=0}^{m} \binom{m}{k} a^{m-k} b^k,

valid for all real xx, aa and bb.

In plain words For whole number exponents the infinite series is secretly a polynomial, so there is no convergence question at all.

Example For m=3m = 3 the coefficients are 1,3,3,11, 3, 3, 1 and then

(34)=3⋅2⋅1⋅024=0.\binom{3}{4} = \frac{3 \cdot 2 \cdot 1 \cdot 0}{24} = 0.

So (1+x)3=1+3x+3x2+x3(1 + x)^3 = 1 + 3x + 3x^2 + x^3, true even at x=5x = 5, where both sides equal 216216.

Finite multinomial theorem

Definition For a nonnegative integer mm and real numbers x1,…,xrx_1, \ldots, x_r,

(x1+x2+⋯+xr)m=∑k1+k2+⋯+kr=mm!k1! k2!⋯kr! x1k1x2k2⋯xrkr,(x_1 + x_2 + \cdots + x_r)^m = \sum_{k_1 + k_2 + \cdots + k_r = m} \frac{m!}{k_1! \, k_2! \cdots k_r!} \, x_1^{k_1} x_2^{k_2} \cdots x_r^{k_r},

where the sum runs over all rr-tuples of nonnegative integers with sum mm. For r=2r = 2 this is the binomial theorem.

In plain words To find a coefficient, count how many ways the mm factors can supply the required numbers of each variable.

Example The coefficient of x2yzx^2 y z in (x+y+z)4(x + y + z)^4 is

4!2! 1! 1!=12.\frac{4!}{2! \, 1! \, 1!} = 12.

As a check, setting x=y=z=1x = y = z = 1 adds up all coefficients and gives 34=813^4 = 81.

Check yourself

  1. Write the first four terms of the series for 11−x\frac{1}{\sqrt{1 - x}}.
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Use α=−12\alpha = -\frac{1}{2} and replace xx by −x-x. The result is

11−x=1+x2+3x28+5x316+⋯for ∣x∣<1.\frac{1}{\sqrt{1 - x}} = 1 + \frac{x}{2} + \frac{3x^2}{8} + \frac{5x^3}{16} + \cdots \qquad \text{for } \lvert x \rvert < 1.
  1. Find the coefficient of x3x^3 in the series for (1+x)1/3(1 + x)^{1/3}.
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The coefficient is

(1/33)=13⋅(−23)⋅(−53)6=10/276=581.\binom{1/3}{3} = \frac{\frac{1}{3} \cdot \left(-\frac{2}{3}\right) \cdot \left(-\frac{5}{3}\right)}{6} = \frac{10/27}{6} = \frac{5}{81}.
  1. Find the coefficient of a2b2ca^2 b^2 c in (a+b+c)5(a + b + c)^5.
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By the multinomial theorem it is

5!2! 2! 1!=1204=30.\frac{5!}{2! \, 2! \, 1!} = \frac{120}{4} = 30.

How this lesson connects

It also uses generalized binomial coefficients.

Resources for this lesson