6A - Binomial series with rational powers

A square root cannot be expanded by multiplying half a bracket. The binomial series extends the coefficient pattern to fractional and negative powers; you will use it carefully, keep any constant factor, and state where the resulting series is valid.

Non-integer powers

Can the coefficient pattern for $(1+x)^5$ also tell us about $\sqrt{1+x}$ ?

For an integer power, the first coefficients are $1$ , $n$ , $n(n-1)/2!$ and $n(n-1)(n-2)/3!$ . Substituting $n=\tfrac12$ gives $1$ , $\tfrac12$ , $-\tfrac18$ and $\tfrac1{16}$ . The same falling-product pattern still works, but it no longer reaches a zero factor and stops.

Binomial series

A binomial series expresses a power $(1+u)^n$ as a sum in ascending powers of $u$ . For positive integer $n$ the expansion terminates. For a negative integer or a non-integer rational $n$ , it has infinitely many terms.

Use $u$ for the whole expression added to $1$ . It might be $x$ , $-2x$ , or $x^2$ . Writing $u$ first helps you raise the complete expression to the correct power.

General binomial series

For rational $n$ , the supplied series is

(1+u)^n=1+nu+\frac{n(n-1)}{2!}u^2+\frac{n(n-1)(n-2)}{3!}u^3+\cdots.

For an infinite series, use the validity condition $|u|<1$ . The displayed equality includes all the continuing terms; stopping after a finite number normally gives an approximation.

On the formulae sheet, including the validity condition.

The factorial denominators are still $2!=2$ , $3!=6$ , and so on. Do not try to use $n!/[r!(n-r)!]$ with a fractional or negative $n$ ; use the falling products in the supplied formula instead.

Worked example 1: a square-root series

Question. Expand $(1+x)^{1/2}$ in ascending powers of $x$ , up to and including the term in $x^3$ .

Plan. Set $n=\tfrac12$ and $u=x$ , keeping the signs in the falling products.

\begin{aligned} (1+x)^{1/2}&=1+\frac12x+\frac{(\tfrac12)(-\tfrac12)}{2}x^2\\ &\quad+\frac{(\tfrac12)(-\tfrac12)(-\tfrac32)}{6}x^3+\cdots \end{aligned}

The quadratic numerator is negative; the cubic numerator has two negative factors and is positive.

(1+x)^{1/2}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+\cdots.

Answer. The required terms are $1+\tfrac{x}{2}-\tfrac{x^2}{8}+\tfrac{x^3}{16}$ , from a series valid for $|x|<1$ .

Check. At $x=0.02$ , the four terms give $1.0099505$ , close to $\sqrt{1.02}\approx1.009950494$ . At $x=0$ , both equal $1$ .

Change only the exponent to $-\tfrac12$ . Each falling-product factor is now negative, so the signs alternate.

Worked example 2: a reciprocal square-root series

Question. Expand $(1+x)^{-1/2}$ up to and including the term in $x^3$ .

Plan. Use $n=-\tfrac12$ in the same supplied formula.

\begin{aligned} (1+x)^{-1/2}&=1-\frac12x+\frac{(-\tfrac12)(-\tfrac32)}{2}x^2\\ &\quad+\frac{(-\tfrac12)(-\tfrac32)(-\tfrac52)}{6}x^3+\cdots\\ &=1-\frac{x}{2}+\frac{3x^2}{8}-\frac{5x^3}{16}+\cdots \end{aligned}

Answer. The required terms are $1-\tfrac{x}{2}+\tfrac{3x^2}{8}-\tfrac{5x^3}{16}$ , for $|x|<1$ .

Check. At $x=0.02$ , these terms give $0.9901475$ , close to $1/\sqrt{1.02}\approx0.990147543$ . The value is below $1$ , as a reciprocal of a number above $1$ should be.

A negative exponent changes the coefficient products; it does not change the factorial denominators.

Watch out

For $n=-\tfrac12$ , $n-1=-\tfrac32$ , not $\tfrac12$ . Subtract $1$ at each step before multiplying, and simplify to exact fractions.

Use a different negative fractional exponent in the next check.

Putting the expression in standard form

The supplied formula starts with $1+u$ , so factor out a non-unit constant before expanding.

For example, $9+3x=9(1+x/3)$ , so $\sqrt{9+3x}=3(1+x/3)^{1/2}$ . The outside factor is $\sqrt9=3$ , not $9$ . You must multiply every term of the resulting series by it.

For a positive constant $a$ ,

(a+bx)^n=a^n\left(1+\frac{bx}{a}\right)^n.

This uses the index laws for positive bases. Within the condition $|bx/a|<1$ , the second base is positive. Identify three things before starting: the outside factor $a^n$ , the exponent $n$ , and the whole substituted expression $u=bx/a$ .

Worked example 3: retaining the outside square root

Question. Expand $\sqrt{9+3x}$ up to and including the term in $x^3$ .

Plan. Factor out $9$ , then use the square-root series with $u=x/3$ .

\begin{aligned} \sqrt{9+3x}&=3\left(1+\frac{x}{3}\right)^{1/2} &&\text{take the square root of }9\\ &=3\left[1+\frac12\left(\frac{x}{3}\right)-\frac18\left(\frac{x}{3}\right)^2+\frac1{16}\left(\frac{x}{3}\right)^3+\cdots\right]\\ &=3+\frac{x}{2}-\frac{x^2}{24}+\frac{x^3}{144}+\cdots &&\text{multiply every term by }3 \end{aligned}

Answer. $3+\tfrac{x}{2}-\tfrac{x^2}{24}+\tfrac{x^3}{144}+\cdots$ , with $|x/3|<1$ .

Check. At $x=0$ , the original expression is $3$ , matching the constant term. At $x=0.03$ , the displayed terms give $3.0149626875$ , close to $\sqrt{9.09}\approx3.014962686$ .

Keep the same bracket but put it in the denominator. The exponent and the outside factor both change.

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6A - Binomial series with rational powers

Non-integer powers

Putting the expression in standard form

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