Binomial Series for a Negative or Fractional Index

Why a whole-number power stops

The binomial theorem can be written as one formula that does not need Pascal’s triangle: ${\left(1+x\right)}^n=1+nx+\frac{n\left(n-1\right)}{2!}{x}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^3+\dots$(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + …. Each coefficient has one more factor on top than the one before, and one more factor in the factorial underneath.

Put n = 3. The coefficients are 3, then 3 × 2 ÷ 2! = 3, then 3 × 2 × 1 ÷ 3! = 1. The next one is 3 × 2 × 1 × 0 ÷ 4! = 0, because its top now has the factor 3 − 3 = 0. Every coefficient after it has that factor too, so they are all 0, and ${\left(1+x\right)}^3=1+3x+3x^2+x^3$(1 + x)³ = 1 + 3x + 3x² + x³, row 3 of the triangle.

That is why the expansion stops when n is a positive whole number: the factors n, n − 1, n − 2, … count down and reach 0.

A negative index never stops

Put n = −1 in the same formula. The factors are now −1, −2, −3, −4, …, which count down away from 0 and never reach it. The coefficients are −1, then (−1)(−2) ÷ 2! = 1, then (−1)(−2)(−3) ÷ 3! = −1, then (−1)(−2)(−3)(−4) ÷ 4! = 1, and so on forever.

So ${\left(1+x\right)}^{-1}=1-x+x^2-x^3+x^4-\dots$(1 + x)⁻¹ = 1 − x + x² − x³ + x⁴ − …, a series with no last term. Multiply the first four terms by 1 + x to see why it works: $\left(1+x\right)\left(1-x+x^2-x^3\right)=1-x^4$(1 + x)(1 − x + x² − x³) = 1 − x⁴. When x is small, $x^4$x⁴ is tiny, so the four terms give very nearly $\frac{1}{\left(1+x\right)}$1/(1 + x).

For n = −2 the coefficients are −2, then (−2)(−3) ÷ 2 = 3, then (−2)(−3)(−4) ÷ 6 = −4, so ${\left(1+x\right)}^{-2}=1-2x+3x^2-4x^3+\dots$(1 + x)⁻² = 1 − 2x + 3x² − 4x³ + …. The sign of the x term follows n, and the $x^2$x² term is positive because two negative factors multiply.

Close to the function near zero

At x = 0.1, the first four terms give 1 − 0.1 + 0.01 − 0.001 = 0.909, and $\frac{1}{1.1}=0.909091$1/1.1 = 0.909091 to 6 decimal places. At x = 0.5 they give 1 − 0.5 + 0.25 − 0.125 = 0.625, and $\frac{1}{1.5}=0.666667$1/1.5 = 0.666667, further off but still near.

Past x = 1 the series moves away. At x = 1.5 the four terms give 1 − 1.5 + 2.25 − 3.375 = −1.625, while $\frac{1}{2.5}=0.4$1/2.5 = 0.4. At x = 2 they give 1 − 2 + 4 − 8 = −5, while $\frac{1}{3}=0.333333$1/3 = 0.333333.

The gold curve is $y=\frac{1}{\left(1+x\right)}$y = 1/(1 + x), which has no value at x = −1. The plain curve is the first four terms, $y=1-x+x^2-x^3$y = 1 − x + x² − x³. Near x = 0 the two are almost on top of each other. At x = 1.5 the dots show them far apart: 0.4 on the gold curve and −1.625 on the plain one.

Why it needs |x| < 1

Adding more terms decides whether the series has a sum. At x = 0.5 the running totals are 1, 0.5, 0.75, 0.625, 0.6875, 0.65625, …, swinging above and below $\frac{2}{3}$2/3 by less each time, and closing in on $\frac{1}{1.5}=\frac{2}{3}$1/1.5 = 2/3.

At x = 2 the running totals are 1, −1, 3, −5, 11, −21, …. Each term is −2 times the one before, so the terms grow and the totals swing further each time. They never settle, so the series has no sum, and it cannot equal $\frac{1}{3}$1/3.

The terms of the n = −1 series are the powers of −x. They shrink when |x| < 1 and grow when |x| > 1, so the series is valid only for |x| < 1, which is −1 < x < 1.

The running totals of $1-x+x^2-x^3+\dots$1 − x + x² − x³ + … at x = 0.5, after 1, 2, 3, 4, 5 and 6 terms: 1, 0.5, 0.75, 0.625, 0.6875 and 0.65625. The plain straight line is at $\frac{2}{3}$2/3. The dots fall above and below it in turn, each one closer than the last.

A fractional index

The same formula works for a fraction. With n = ½, the coefficients are ½, then ½ × (−½) ÷ 2! = −⅛, then $\frac{1}{2}\times \left(-\frac{1}{2}\right)\times \left(-\frac{3}{2}\right)÷3!=\left(\frac{3}{8}\right)÷6=\frac{1}{16}$½ × (−½) × (−3/2) ÷ 3! = (3/8) ÷ 6 = 1/16. So ${\left(1+x\right)}^{\frac{1}{2}}=1+\frac{1}{2}x-\frac{1}{8}{x}^2+\left(\frac{1}{16}\right)x^3-\dots$(1 + x)^½ = 1 + ½x − ⅛x² + (1/16)x³ − …. The factors ½, −½, $-\frac{3}{2}$−3/2, … never reach 0, so this series never stops either.

Every index that is not a positive whole number gives a series that runs forever, and every one of them is valid only for |x| < 1.

Check at x = 0.1. The four terms give 1 + 0.05 − 0.00125 + 0.0000625 = 1.0488125, and $\sqrt{1.1}=1.0488088$√1.1 = 1.0488088 to 7 decimal places, so the two agree to 5 decimal places.

The first three terms of a negative power

Expand ${\left(1+x\right)}^{-3}$(1 + x)⁻³ as far as $x^2$x². Here n = −3, so nx = −3x, and $\frac{n\left(n-1\right)}{2}=\left(-3\right)\left(-4\right)/2=6$n(n − 1)/2 = (−3)(−4) / 2 = 6. So ${\left(1+x\right)}^{-3}=1-3x+6x^2-\dots$(1 + x)⁻³ = 1 − 3x + 6x² − ….

Check at x = 0.1: 1 − 0.3 + 0.06 = 0.76, and $\frac{1}{1.1^3}=0.751315$1/1.1³ = 0.751315. The next term is (−3)(−4)(−5) ÷ 3! = −10 times $x^3$x³, which is −0.01 here, and 0.76 − 0.01 = 0.75 is closer still.

The usual mistakes

Getting the sign of the x term wrong. For ${\left(1+x\right)}^{-3}$(1 + x)⁻³ the term is nx = −3x, not +3x: the sign follows n.

Squaring n for the $x^2$x² coefficient. It is $\frac{n\left(n-1\right)}{2}$n(n − 1)/2, so for n = −3 it is (−3)(−4) ÷ 2 = 6, not 9.

Forgetting the factorial. The $x^2$x² coefficient for n = ½ is ½ × (−½) ÷ 2 = −⅛, not −¼.

Using the series outside |x| < 1. The algebra can be written for any x, but past |x| = 1 the terms grow and the series has no sum. The range does not depend on n: it is |x| < 1 forevery index.

A pendulum on a hot day

In the application below, a pendulum grows 4% longer, and each swing takes $\sqrt{1.04}$√1.04 times as long. That is ${\left(1+x\right)}^{\frac{1}{2}}$(1 + x)^½ with x = 0.04, well inside |x| < 1. The same series cannot find $\sqrt{3}$√3 as ${\left(1+2\right)}^{\frac{1}{2}}$(1 + 2)^½, because x = 2 is outside the range.