1.04c, 1.04d - Rational binomial expansions and approximation | OCR A-Level Mathematics A

A short polynomial can approximate a square root or reciprocal surprisingly well near a suitable input. You will build rational binomial expansions, factor the constant correctly, state where the expansion is valid and distinguish that condition from numerical accuracy.

Expanding a rational power

Can four easy terms approximate $\sqrt{4.08}$ to better than one millionth? Since $4.08=4(1+0.02)$ , the real question is how $\sqrt{1+u}$ behaves when $u$ is small. We will build the polynomial first and return to the number at the end.

You already know

For positive integer powers, the binomial coefficients terminate: $(1+u)^2=1+2u+u^2$ . Fractional and negative powers obey the index laws on suitable real domains; for example, $(1+u)^{-1}=1/(1+u)$ .

The descending-factor pattern still works for rational powers, but it need not stop. A half-power gives coefficient factors $1/2,-1/2,-3/2,\ldots$ ; none becomes zero.

Rational binomial expansion

For rational $n$ and $|u|<1$ ,

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

For positive integer $n$ , the expansion terminates and is exact for every real input. For $n=0$ , the value is $1$ when the base is non-zero; $u=-1$ would give the undefined expression $0^0$ . For other rational powers the displayed finite part is a truncation of an infinite expansion. The sheet gives this descending-factor pattern.

In the formula sheet.

You need to calculate specified terms, rather than memorise a general term or prove convergence. Keep fractions exact until the numerical approximation stage.

Worked example 1: a square-root expansion

Question. Expand $(1+x)^{1/2}$ through the term in $x^3$ .

Plan. Substitute $n=1/2$ into the first four coefficient factors.

\begin{aligned} (1+x)^{1/2}&=1+\frac{x}{2}+\frac{(1/2)(-1/2)}{2}x^2+\frac{(1/2)(-1/2)(-3/2)}{6}x^3+\cdots &&\text{descending factors}\\ &=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+\cdots &&\text{simplify fractions} \end{aligned}

Why is the cubic coefficient positive? Two of its three coefficient factors are negative.

Answer. $1+x/2-x^2/8+x^3/16+\cdots$ , valid for $|x|<1$ .

Check. Squaring this polynomial gives constant and first three power terms $1+x+0x^2+0x^3$ , as a square-root expansion should.

Change only the exponent to $-1/2$ . Do not assume that every rational expansion has the same signs.

Worked example 2: the reciprocal square root

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

Plan. Use the same input, changing the descending coefficient factors.

\begin{aligned} (1+x)^{-1/2}&=1-\frac{x}{2}+\frac{(-1/2)(-3/2)}2x^2+\frac{(-1/2)(-3/2)(-5/2)}6x^3+\cdots &&\text{use }n=-1/2\\ &=1-\frac{x}{2}+\frac{3x^2}{8}-\frac{5x^3}{16}+\cdots &&\text{simplify} \end{aligned}

Why retain every sign? Here the quadratic has two negative factors and the cubic has three.

Answer. $1-x/2+3x^2/8-5x^3/16+\cdots$ , for $|x|<1$ .

Check. Multiplying the two worked polynomials gives $1+0x+0x^2+0x^3$ , matching $(1+x)^{1/2}(1+x)^{-1/2}=1$ to this degree.

A scaled input must also be powered: $(1+2x)^{-1}=1-2x+4x^2-8x^3+\cdots$ , for $|x|<1/2$ . It is the whole $2x$ that is squared and cubed.

Watch out

Wrong: $(1+x)^{1/2}=1+x/2-x^2/8+x^3/16$ for general $x$ . Right: retain the ellipsis for the infinite expansion, or use $\approx$ for the finite polynomial. Equality to a truncation would claim that the omitted terms are zero.

Keep the input and degree, changing only the exponent again.

Factor out the constant before expanding

The formula begins with $1$ inside the bracket. If the constant is $4$ , factor it out before expanding, then apply the exponent to that constant too.

Normalising the bracket

For $a>0$ , a real input in the required domain and rational $n$ ,

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

This is an index-law rearrangement used with the supplied expansion; it is not separately printed as a formula.

Not in the formula sheet: learn it.

The outside factor scales every term, not just the constant term.

Worked example 3: factor out a positive constant

Question. Expand $\sqrt{4+2x}$ through $x^3$ .

Plan. Write the expression as $2(1+x/2)^{1/2}$ , then substitute into the half-power expansion.

\begin{aligned} \sqrt{4+2x}&=2\left(1+\frac{x}{2}\right)^{1/2} &&\text{factor }4\text{ and take its root}\\ &=2\left[1+\frac{x}{4}-\frac{x^2}{32}+\frac{x^3}{128}+\cdots\right] &&\text{use }u=x/2\\ &=2+\frac{x}{2}-\frac{x^2}{16}+\frac{x^3}{64}+\cdots &&\text{multiply every term by }2 \end{aligned}

Why is the outside factor $2$ , not $4$ ? It is $4^{1/2}$ .

Answer. $2+x/2-x^2/16+x^3/64+\cdots$ , for $|x|<2$ .

Check. At $x=0$ , the constant is $2=\sqrt4$ . Squaring the polynomial gives $4+2x$ through cubic degree.

Change only the sign of the variable term. The outside scale remains $2$ .

Worked example 4: a negative inner input

Question. Expand $\sqrt{4-2x}$ through $x^3$ .

Plan. Use $u=-x/2$ , keeping its sign inside each power.

\begin{aligned} \sqrt{4-2x}&=2\left(1-\frac{x}{2}\right)^{1/2} &&\text{normalise the bracket}\\ &=2\left[1-\frac{x}{4}-\frac{x^2}{32}-\frac{x^3}{128}+\cdots\right] &&\text{substitute }u=-x/2\\ &=2-\frac{x}{2}-\frac{x^2}{16}-\frac{x^3}{64}+\cdots &&\text{restore the outside scale} \end{aligned}

Why is the cubic negative now? Its positive half-power coefficient multiplies a negative cube.

Answer. $2-x/2-x^2/16-x^3/64+\cdots$ , for $|x|<2$ .

Check. Replacing $x$ by $-x$ in example 3 gives these same signs.

Negative exponents need the outside power too: $(3-x)^{-2}=1/9+2x/27+x^2/27+4x^3/243+\cdots$ , for $|x|<3$ . The outside factor is $3^{-2}=1/9$ .

For a negative constant, check the real domain first. Near zero, $\sqrt[3]{-8+2x}=-2(1-x/4)^{1/3}$ uses the real cube root. The square root of $-8+2x$ is not real there. Factoring cannot repair an invalid real input.

Watch out

Wrong: $(3-x)^{-2}=3(1-x/3)^{-2}$ . Right: $(3-x)^{-2}=3^{-2}(1-x/3)^{-2}$ . The exponent applies to the factor taken out.

Change the positive constant and its associated scale in another square-root bracket.

Validity and combining truncated expansions

Validity applies to the entire inner input: $|bx/a|<1$ . For example, $(9-6x)^{-1/2}=\tfrac13(1-2x/3)^{-1/2}$ has the condition $-3/2<x<3/2$ , not merely $x<3/2$ .

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1.04c, 1.04d - Rational binomial expansions and approximation

Expanding a rational power

Factor out the constant before expanding

Validity and combining truncated expansions

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