3.1 - Binomial series with rational powers | Cambridge A-Level Mathematics

A positive integer power gives a finite binomial expansion; a negative or fractional power usually continues indefinitely. In this part of 3.1, you will calculate the first few terms, adapt the standard form and state its validity interval. You will also multiply short expansions and use them to approximate nearby numerical values.

An infinite binomial series

Can a few powers of $x$ give a useful approximation to $\sqrt{1+x}$ ? At $x=0$ , the value is exactly $1$ . Nearby, a polynomial beginning $1+x/2$ follows the square root closely; further terms improve the approximation. Unlike $(1+x)^3$ , the fractional power does not produce a finite polynomial identity.

For a positive integer $n$ , the familiar factors $n,n-1,n-2,\ldots$ eventually include zero. That makes the binomial expansion terminate. For $n=1/2$ or $n=-1$ , the factors never reach zero, so the series continues.

Binomial series for a rational power

For rational $n$ ,

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

Here $2!=2$ and $3!=6$ . The infinite series equals the function within this interval; truncating a non-terminating series after finitely many nonzero terms gives an approximation. In the formula booklet.

The notation $\cdots$ indicates further terms. When asked for terms “up to and including $x^3$ ”, retain the constant and all nonzero terms through that power. Do not replace $(n-1)$ by $(1-n)$ : the signs come from the actual factors.

Worked example 1: a square root

Question. Expand $(1+x)^{1/2}$ up to and including the term in $x^3$ , and state the standard validity interval.

Plan. Substitute $n=1/2$ in the supplied series and simplify each coefficient separately.

\begin{aligned} \frac{n(n-1)}2&=\frac{(1/2)(-1/2)}2=-\frac18\\ \frac{n(n-1)(n-2)}6&=\frac{(1/2)(-1/2)(-3/2)}6=\frac1{16} \end{aligned}

Answer. $1+\tfrac12x-\tfrac18x^2+\tfrac1{16}x^3+\cdots$ , for $-1<x<1$ .

Check. Squaring the displayed polynomial gives $1+x$ through the cubic term: the quadratic coefficient is $-1/4+1/4=0$ , and the cubic coefficient is $1/8-1/8=0$ . Terms of degree four and above are still present, so the polynomial itself is not exactly the square root.

Change only the sign of the exponent. This gives the reciprocal square root, with different signs and magnitudes in its coefficients.

Worked example 2: a reciprocal square root

Question. Expand $(1+x)^{-1/2}$ through the term in $x^3$ , and state the validity interval.

Plan. Use $n=-1/2$ , keeping the negative signs inside the coefficient products.

\begin{aligned} (1+x)^{-1/2} &=1-\frac12x +\frac{(-1/2)(-3/2)}2x^2 +\frac{(-1/2)(-3/2)(-5/2)}6x^3+\cdots\\ &=1-\frac12x+\frac38x^2-\frac5{16}x^3+\cdots \end{aligned}

Answer. $1-\tfrac12x+\tfrac38x^2-\tfrac5{16}x^3+\cdots$ , for $-1<x<1$ .

Check. Multiplying by the square-root expansion in example 1 gives $1$ through degree three. For instance, the quadratic coefficient is $-1/8-1/4+3/8=0$ , consistent with the two functions being reciprocals.

For the negative integer $n=-1$ , the formula produces $1-u+u^2-u^3+\cdots$ , the geometric series for $1/(1+u)$ . This links the new expansion to a series you already know.

Watch out

The condition concerns the whole added term $u$ : $|u|<1$ . A function may exist outside this interval without this series being valid there. Endpoint behaviour needs separate reasoning; do not include endpoints in the standard validity statement.

Keep exact fractional coefficients in this new expansion.

Extracting the constant factor

The supplied formula begins with $1+u$ , not $a+bx$ . First take out the constant and apply the power to that constant too. For positive $a$ ,

(a+bx)^n=a^n\left(1+\frac ba x\right)^n.

Now $u=(b/a)x$ , so the standard condition is $|(b/a)x|<1$ . For $b\ne0$ , this is $|x|<|a/b|$ . The interval comes from the expression inside the series, independently of the outside multiplier.

Worked example 3: a positive added term

Question. Expand $(4+x)^{-1/2}$ through $x^3$ , and state its validity interval.

Plan. Factor out $4$ , use the reciprocal-square-root series with $u=x/4$ , then multiply every coefficient by $1/2$ .

\begin{aligned} (4+x)^{-1/2} &=\frac12\left(1+\frac x4\right)^{-1/2}\\ &=\frac12\left[1-\frac12\left(\frac x4\right) +\frac38\left(\frac x4\right)^2-\frac5{16}\left(\frac x4\right)^3+\cdots\right] \end{aligned}

Answer. $\tfrac12-\tfrac1{16}x+\tfrac3{256}x^2-\tfrac5{2048}x^3+\cdots$ , for $-4<x<4$ .

Check. At $x=0$ , both the function and expansion give $4^{-1/2}=1/2$ . Squaring the displayed polynomial and multiplying by $4+x$ gives $1$ through degree three, checking the reciprocal-square-root relation.

Change only the sign before $x$ . Substitute $-x/4$ throughout; odd powers change sign and even powers do not.

Worked example 4: a negative added term

Question. Expand $(4-x)^{-1/2}$ through $x^3$ , and state its validity interval.

Plan. Keep the same outside factor, replacing $u$ by $-x/4$ .

\begin{aligned} (4-x)^{-1/2} &=\frac12\left(1-\frac x4\right)^{-1/2}\\ &=\frac12\left[1+\frac x8+\frac{3x^2}{128}+\frac{5x^3}{1024}+\cdots\right] \end{aligned}

Answer. $\tfrac12+\tfrac1{16}x+\tfrac3{256}x^2+\tfrac5{2048}x^3+\cdots$ , for $-4<x<4$ .

Check. Replacing $x$ by $-x$ in example 3 produces exactly these terms. Also, small positive $x$ decreases the denominator $\sqrt{4-x}$ , so the reciprocal should rise above $1/2$ , agreeing with the positive first correction.

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3.1 - Binomial series with rational powers

An infinite binomial series

Extracting the constant factor

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