Expanding (a + bx) to a Rational Power

Make the bracket start with 1

The binomial series ${\left(1+x\right)}^n=1+nx+\frac{n\left(n-1\right)}{2!}{x}^2+\dots$(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + … is written for a bracket whose first term is 1. To expand ${\left(4+x\right)}^{\frac{1}{2}}$(4 + x)^½, first make the bracket start with 1 by taking the 4 out as a factor: $4+x=4\left(1+\frac{x}{4}\right)$4 + x = 4(1 + x/4).

The power applies to both factors: ${\left(4\left(1+\frac{x}{4}\right)\right)}^{\frac{1}{2}}=4^{\frac{1}{2}}\times {\left(1+\frac{x}{4}\right)}^{\frac{1}{2}}=2{\left(1+\frac{x}{4}\right)}^{\frac{1}{2}}$(4(1 + x/4))^½ = 4^½ × (1 + x/4)^½ = 2(1 + x/4)^½, because $4^{\frac{1}{2}}=\sqrt{4}=2$4^½ = √4 = 2.

Expand with $\frac{x}{4}$x/4 in place of x

The series for ${\left(1+x\right)}^{\frac{1}{2}}$(1 + x)^½ is $1+\frac{1}{2}x-\frac{1}{8}{x}^2+\dots$1 + ½x − ⅛x² + …. Write $\frac{x}{4}$x/4 everywhere that series has x: ${\left(1+\frac{x}{4}\right)}^{\frac{1}{2}}=1+\frac{1}{2}\left(\frac{x}{4}\right)-\frac{1}{8}{\left(\frac{x}{4}\right)}^2+\cdots =1+\frac{x}{8}-\frac{x^2}{128}+\dots$(1 + x/4)^½ = 1 + ½(x/4) − ⅛(x/4)² + … = 1 + x/8 − x²/128 + ….

Then multiply the 2 back through every term: ${\left(4+x\right)}^{\frac{1}{2}}=2\left(1+\frac{x}{8}-\frac{x^2}{128}+\dots \right)=2+\frac{x}{4}-\frac{x^2}{64}+\dots$(4 + x)^½ = 2(1 + x/8 − x²/128 + …) = 2 + x/4 − x²/64 + ….

Check at x = 1, where the left side is $\sqrt{5}=2.236068$√5 = 2.236068 to 6 decimal places. The three terms give 2 + 0.25 − 0.015625 = 2.234375. The next term of the series is $2\times \left(\frac{1}{16}\right){\left(\frac{x}{4}\right)}^3=\frac{1}{512}=0.001953$2 × (1/16)(x/4)³ = 1/512 = 0.001953 at x = 1, and adding it gives 2.236328, closer still.

The range moves with the substitution

The series for ${\left(1+u\right)}^{\frac{1}{2}}$(1 + u)^½ is valid only for |u| < 1. Here u is $\frac{x}{4}$x/4, so the condition is $\frac{|x}{4|}<1$|x/4| < 1. Multiply through by 4: |x| < 4, which is −4 < x < 4.

So the range for ${\left(4+x\right)}^{\frac{1}{2}}$(4 + x)^½ is |x| < 4, not |x| < 1. Taking the constant out divided x by 4 inside the bracket, and that let x be four times as large before the series stops working.

The general rule

For any bracket ${\left(a+bx\right)}^n$(a + bx)ⁿ, take a out: ${\left(a+bx\right)}^n=a^n{\left(1+\frac{bx}{a}\right)}^n$(a + bx)ⁿ = aⁿ(1 + bx/a)ⁿ. Expand ${\left(1+u\right)}^n$(1 + u)ⁿ with $u=\frac{bx}{a}$u = bx/a, then multiply every term by $a^n$aⁿ.

The condition is $\frac{|bx}{a|}<1$|bx/a| < 1, which is $|x|<\frac{|a}{b|}$|x| < |a/b|. The range is set by the ratio of the constant to the coefficient of x. For ${\left(6+3x\right)}^{-1}=6^{-1}{\left(1+\frac{x}{2}\right)}^{-1}$(6 + 3x)⁻¹ = 6⁻¹(1 + x/2)⁻¹, the condition is $\frac{|x}{2|}<1$|x/2| < 1, so $|x|<\frac{6}{3}=2$|x| < 6/3 = 2.

A graph of the moved range

Take $\frac{1}{\left(2+x\right)}={\left(2+x\right)}^{-1}=\frac{1}{2}{\left(1+\frac{x}{2}\right)}^{-1}$1/(2 + x) = (2 + x)⁻¹ = ½(1 + x/2)⁻¹. With $u=\frac{x}{2}$u = x/2, the series ${\left(1+u\right)}^{-1}=1-u+u^2-u^3+\dots$(1 + u)⁻¹ = 1 − u + u² − u³ + … gives $\frac{1}{2}\left(1-\frac{x}{2}+\frac{x^2}{4}-\frac{x^3}{8}+\dots \right)=\frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+\dots$½(1 − x/2 + x²/4 − x³/8 + …) = ½ − x/4 + x²/8 − x³/16 + …, valid for $\frac{|x}{2|}<1$|x/2| < 1, that is |x| < 2.

At x = 1, inside the range, the four terms give 0.5 − 0.25 + 0.125 − 0.0625 = 0.3125, and $\frac{1}{3}=0.333333$1/3 = 0.333333. At x = 3, outside it, they give 0.5 − 0.75 + 1.125 − 1.6875 = −0.8125, while $\frac{1}{5}=0.2$1/5 = 0.2.

The gold curve is $y=\frac{1}{\left(2+x\right)}$y = 1/(2 + x), which has no value at x = −2. The plain curve is the four terms $\frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}$½ − x/4 + x²/8 − x³/16. Near x = 0 the two almost coincide. The whole series, with all its terms, equals $\frac{1}{\left(2+x\right)}$1/(2 + x) for every x between −2 and 2, twice as far each way as the series for $\frac{1}{\left(1+x\right)}$1/(1 + x). At x = 3 the dots show them apart: 0.2 on the gold curve and −0.8125 on the plain one.

Two more expansions

Find the first two terms of ${\left(9+x\right)}^{\frac{1}{2}}$(9 + x)^½. Take out the 9: ${\left(9+x\right)}^{\frac{1}{2}}=9^{\frac{1}{2}}{\left(1+\frac{x}{9}\right)}^{\frac{1}{2}}=3\left(1+\frac{x}{18}+\dots \right)=3+\frac{x}{6}+\dots$(9 + x)^½ = 9^½(1 + x/9)^½ = 3(1 + x/18 + …) = 3 + x/6 + …. The 3 multiplies the bracket’s $\frac{x}{18}$x/18, giving $\frac{x}{6}$x/6. Check at x = 1: $3+\frac{1}{6}=3.166667$3 + 1/6 = 3.166667, and $\sqrt{10}=3.162278$√10 = 3.162278; the next term, $3\times \left(-\frac{1}{8}\right){\left(\frac{1}{9}\right)}^2=-0.004630$3 × (−⅛)(1/9)² = −0.004630, brings it to 3.162037.

Expand ${\left(2-3x\right)}^{-2}$(2 − 3x)⁻² as far as $x^2$x². Take out the 2: ${\left(2-3x\right)}^{-2}=2^{-2}{\left(1-\frac{3x}{2}\right)}^{-2}=\frac{1}{4}{\left(1+u\right)}^{-2}$(2 − 3x)⁻² = 2⁻²(1 − 3x/2)⁻² = ¼(1 + u)⁻² with $u=-\frac{3x}{2}$u = −3x/2. The series ${\left(1+u\right)}^{-2}=1-2u+3u^2-\dots$(1 + u)⁻² = 1 − 2u + 3u² − … gives $1+3x+\frac{27x^2}{4}$1 + 3x + 27x²/4, so ${\left(2-3x\right)}^{-2}=\frac{1}{4}+\frac{3x}{4}+\frac{27x^2}{16}+\dots$(2 − 3x)⁻² = ¼ + 3x/4 + 27x²/16 + …. It is valid for $\frac{|3x}{2|}<1$|3x/2| < 1, that is $|x|<\frac{2}{3}$|x| < 2/3.

Check at x = 0.1: $\frac{1}{1.7^2}=\frac{1}{2.89}=0.346021$1/1.7² = 1/2.89 = 0.346021, and the three terms give 0.25 + 0.075 + 0.016875 = 0.341875. The $x^3$x³ term, $\frac{1}{4}\times \left(-4\right)u^3=0.003375$¼ × (−4)u³ = 0.003375 at x = 0.1, brings it to 0.345250.

The usual mistakes

Expanding as if the bracket started with 1. ${\left(4+x\right)}^{\frac{1}{2}}$(4 + x)^½ is not $1+\frac{1}{2}x-\frac{1}{8}{x}^2+\dots$1 + ½x − ⅛x² + …; that series forgets the 4 entirely.

Forgetting to multiply the outside factor through. The first two terms of ${\left(9+x\right)}^{\frac{1}{2}}$(9 + x)^½ are $3+\frac{x}{6}$3 + x/6, not $3+\frac{x}{18}$3 + x/18: $\frac{x}{18}$x/18 is the bracket’s own second term, before the 3 multiplies it.

Leaving $a^n$aⁿ as a. The factor outside is $4^{\frac{1}{2}}=2$4^½ = 2 for ${\left(4+x\right)}^{\frac{1}{2}}$(4 + x)^½, and $2^{-2}=\frac{1}{4}$2⁻² = ¼ for ${\left(2-3x\right)}^{-2}$(2 − 3x)⁻².

Keeping the range |x| < 1. Once the constant is taken out, the condition is on $\frac{bx}{a}$bx/a, so ${\left(6+3x\right)}^{-1}$(6 + 3x)⁻¹ is valid for |x| < 2.

Taking the coefficient of x as the bound. For ${\left(6+3x\right)}^{-1}$(6 + 3x)⁻¹ the bound is $\frac{6}{3}=2$6/3 = 2, the ratio $\frac{a}{b}$a/b, not 3.

A ferry and a current

In the application below, a crossing takes $\frac{100}{\left(50+v\right)}$100/(50 + v) hours with a current of v km/h. Taking the 50 out gives $2{\left(1+\frac{v}{50}\right)}^{-1}$2(1 + v/50)⁻¹, and the series is valid for −50 < v < 50: any current slower than the ferry.