Ms3–Ms5 - Rational binomial expansions and approximation | OCR A-Level Mathematics B (MEI)

Turn a rational power into a polynomial that is easy to evaluate near a chosen input. You will expand fractional and negative powers, normalise a bracket, combine valid expansions and judge what a truncated approximation actually tells you.

Expanding a rational power

Can four binomial terms calculate $1/2.02$ with an error below $0.000000005$ ? Writing $2.02=2(1+0.01)$ gives a promising small input. First we need an expansion that works when the exponent is negative.

You already know

A negative power is a reciprocal: $(1+x)^{-2}=1/(1+x)^2$ . In an integer binomial expansion the descending coefficient factors eventually contain zero, so the expansion ends.

With $n=-2$ , the factors $-2,-3,-4,\ldots$ never reach zero. Rational binomial expansions therefore need a validity condition and usually have infinitely many terms.

Binomial expansion for a rational power

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.

A positive integer exponent gives a finite exact identity for all real $u$ . At exponent $0$ , the value is $1$ for a non-zero base; exclude $u=-1$ , which would give $0^0$ . Otherwise stopping after a few terms gives an approximation. Use the displayed descending factors to find specified terms; a general term is not required here.

In the formula sheet.

Coefficient signs come from both the exponent factors and the powers of the entire input.

Worked example 1: a negative integer exponent

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

Plan. Substitute $n=-2$ and retain the first four terms.

\begin{aligned} (1+x)^{-2}&=1-2x+\frac{(-2)(-3)}2x^2+\frac{(-2)(-3)(-4)}6x^3+\cdots &&\text{descending factors}\\ &=1-2x+3x^2-4x^3+\cdots &&\text{simplify} \end{aligned}

Why does it not stop after the quadratic? A negative integer exponent does not make a descending factor zero.

Answer. $1-2x+3x^2-4x^3+\cdots$ , for $|x|<1$ .

Check. Multiplying this polynomial by $(1+x)^2=1+2x+x^2$ gives $1$ through cubic degree.

Change only the sign of the inner input. Keep the exponent and the degree unchanged.

Worked example 2: reversing the input sign

Question. Expand $(1-x)^{-2}$ through $x^3$ .

Plan. Substitute $u=-x$ into the same coefficient pattern.

\begin{aligned} (1-x)^{-2}&=1+(-2)(-x)+3(-x)^2-4(-x)^3+\cdots &&\text{power the signed input}\\ &=1+2x+3x^2+4x^3+\cdots &&\text{simplify signs} \end{aligned}

Why are all four terms positive? The odd-power sign reversals cancel the negative odd coefficients.

Answer. $1+2x+3x^2+4x^3+\cdots$ , for $|x|<1$ .

Check. Multiplying by $(1-x)^2=1-2x+x^2$ gives $1+0x+0x^2+0x^3$ .

Fractional powers use the same pattern: $(1+x)^{1/2}=1+x/2-x^2/8+x^3/16+\cdots$ . Those signs are different again; they follow from $1/2,-1/2,-3/2$ , rather than from a rule that all rational expansions alternate.

Watch out

Wrong: $(1-x)^{-2}=1+2x+3x^2+4x^3$ for all $x$ . Right: write the ellipsis for the infinite expansion, or use $\approx$ for the truncated polynomial, and keep $|x|<1$ .

Now change only the magnitude of the negative input.

Factor out the constant before expanding

An input close to a positive constant can be rescaled into a bracket close to $1$ . The outside scale must receive the exponent too.

Rescaling a binomial bracket

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

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

This index-law rearrangement prepares the supplied expansion; it is not itself printed as a separate formula.

Not in the formula sheet: learn it.

For example, the outside factor of $(3-x)^{-2}$ is $3^{-2}=1/9$ . It is neither $3$ nor $-6$ .

Worked example 3: retaining the outside power

Question. Expand $(3-x)^{-2}$ through $x^3$ .

Plan. Factor out $3$ and use $u=-x/3$ in the negative-square expansion.

\begin{aligned} (3-x)^{-2}&=\frac19(1-x/3)^{-2} &&\text{raise the factor }3\text{ to }-2\\ &=\frac19\left(1+\frac{2x}{3}+\frac{x^2}{3}+\frac{4x^3}{27}+\cdots\right) &&\text{substitute }u=-x/3\\ &=\frac19+\frac{2x}{27}+\frac{x^2}{27}+\frac{4x^3}{243}+\cdots &&\text{scale every term} \end{aligned}

Why multiply the quadratic by $1/9$ too? The outside factor multiplies the entire bracket.

Answer. $1/9+2x/27+x^2/27+4x^3/243+\cdots$ , for $|x|<3$ .

Check. At $x=0$ , the constant is $1/9=3^{-2}$ .

Double the entire bracket, changing its outside scale while keeping the inner input $-x/3$ .

Worked example 4: a rescaled bracket

Question. Expand $(6-2x)^{-2}$ through $x^3$ .

Plan. Use the same inner expansion, now with outside factor $6^{-2}=1/36$ .

\begin{aligned} (6-2x)^{-2}&=\frac1{36}(1-x/3)^{-2} &&\text{factor out }6\\ &=\frac1{36}\left(1+\frac{2x}{3}+\frac{x^2}{3}+\frac{4x^3}{27}+\cdots\right) &&\text{reuse the inner input}\\ &=\frac1{36}+\frac{x}{54}+\frac{x^2}{108}+\frac{x^3}{243}+\cdots &&\text{multiply by }1/36 \end{aligned}

Why is every coefficient a quarter of example 3? The whole expression has been multiplied by $2^{-2}=1/4$ .

Answer. $1/36+x/54+x^2/108+x^3/243+\cdots$ , for $|x|<3$ .

Check. The constant is $6^{-2}$ , and each coefficient is a quarter of its counterpart in example 3.

The same care applies to roots: $\sqrt{4+2x}=2+x/2-x^2/16+x^3/64+\cdots$ , for $|x|<2$ , since $4^{1/2}=2$ . If the factored constant is negative, check the real root convention first: $\sqrt[3]{-8+2x}=-2(1-x/4)^{1/3}$ near zero is real, but the square root of $-8+2x$ is not real there.

Watch out

Wrong: take out the constant, then leave it to its first power. Right: keep $a^n$ visible until every term has been multiplied by it. Substitution of $x=0$ catches a wrong constant immediately.

Scale the same bracket once more, so only the outside multiplier changes.

Validity and combining truncated expansions

Use $|u|<1$ on the whole inner input. Thus $(9-6x)^{-1/2}=\tfrac13(1-2x/3)^{-1/2}$ has the condition $-3/2<x<3/2$ . A negative scale does not turn this into a one-sided restriction.

If two expansions are combined, use the intersection of their validity intervals. When multiplying, keep every way to make a required degree, including products of linear terms. A convergence proof is not needed; the open intervals used here do not settle special endpoint behaviour.

Worked example 5: cancellation of linear terms

Question. Expand $1/[(1-x)\sqrt{1+2x}]$ through $x^2$ , and state the common validity interval.

Plan. Expand the reciprocal and reciprocal square root separately, then collect all terms through degree two.

\begin{aligned} (1-x)^{-1}&=1+x+x^2+\cdots &&\text{requires }|x|<1\\ (1+2x)^{-1/2}&=1-x+\frac{3x^2}{2}+\cdots &&\text{requires }|x|<1/2\\ \frac1{(1-x)\sqrt{1+2x}}&=1+(1-1)x+(1-1+3/2)x^2+\cdots &&\text{collect all contributions} \end{aligned}

Why is the linear term zero? The $+x$ from one factor cancels the $-x$ from the other; neither should be omitted before multiplying.

Answer. $1+3x^2/2+\cdots$ , for $-1/2<x<1/2$ , the interval contained in both component ranges.

Check. Squaring the expression gives $1/[(1-x)^2(1+2x)]$ . Its denominator is $1-3x^2+2x^3$ , consistent with a reciprocal starting $1+3x^2+\cdots$ .

Change only the first denominator sign. Now the two linear contributions reinforce rather than cancel.

Worked example 6: reinforcing linear contributions

Question. Expand $1/[(1+x)\sqrt{1+2x}]$ through $x^2$ , and state the common validity interval.

Plan. Replace the reciprocal factor by $1-x+x^2+\cdots$ , retaining the reciprocal square-root factor.

\begin{aligned} \frac1{(1+x)\sqrt{1+2x}}&=(1-x+x^2+\cdots)(1-x+3x^2/2+\cdots) &&\text{expand both factors}\\ &=1-2x+(1+1+3/2)x^2+\cdots &&\text{include }(-x)(-x)\\ &=1-2x+\frac{7x^2}{2}+\cdots &&\text{simplify} \end{aligned}

Why does the quadratic include $+x^2$ from a cross-product? The product of the two negative linear terms is positive.

Answer. $1-2x+7x^2/2+\cdots$ , for $|x|<1/2$ . The restrictions $|x|<1$ and $|2x|<1$ still apply.

Check. Multiplying by $1+x$ gives $1-x+3x^2/2$ through degree two, recovering the reciprocal square root.

A linear coefficient of zero is useful information. Do not replace “first three terms” with “through $x^2$ ” automatically when powers cancel; read which truncation the question asks for.

Exam tip

Show enough of the multiplication to identify the constant, the two linear contributions and the three quadratic contributions. This makes a valid method visible even if a later coefficient has an arithmetic slip.

Change only the scale in the first denominator factor.

Using the polynomial as an approximation

A valid binomial expansion need not have an accurate low-degree truncation. Compare $P(x)=1+x/2-x^2/8$ with $\sqrt{1+x}$ : at $x=0.1$ , the values are $1.04875$ and $1.048808848\ldots$ ; at $x=0.8$ , they are $1.32$ and $1.341640786\ldots$ . Both inputs satisfy the condition, but the second gap is much larger.

[DIAGRAM: asset_name: Valid inputs can have different truncation errors; caption: Valid inputs can have different truncation errors; asset_slug: h640-binomial-approximation; description: The exact square-root curve and its quadratic binomial polynomial are plotted for inputs within minus 1 to 1, with the small discrepancy at 0.1 and larger discrepancy at 0.8 identified]

For the opening reciprocal, take out $2$ , then use a particularly simple polynomial: $(1+u)^{-1}=1-u+u^2-u^3+\cdots$ . This follows from the supplied binomial formula with $n=-1$ ; it is not a separate formula-sheet entry.

Worked example 7: a four-term reciprocal

Question. Use the expansion through $u^3$ to approximate $1/2.02$ , and compare with a calculator value.

Plan. Write $1/2.02=\tfrac12(1+0.01)^{-1}$ .

\begin{aligned} \frac1{2.02}&=\frac12(1+0.01)^{-1} &&\text{normalise the denominator}\\ &\approx\frac12(1-0.01+0.0001-0.000001) &&\text{retain four binomial terms}\\ &=0.4950495 &&\text{evaluate} \end{aligned}

Why is the outside factor $1/2$ ? The power $-1$ applies to the factor $2$ as well.

Answer. The polynomial gives $0.4950495$ , compared with $0.49504950495\ldots$ . It is low by about $4.95\times10^{-9}$ .

Check. Multiplying the approximation by $2.02$ gives $0.99999999$ , very close to $1$ .

Change only the target denominator to $2.04$ . Keep the outside factor and the cubic truncation.

Worked example 8: a larger small input

Question. Use the same expansion to approximate $1/2.04$ , and compare with a calculator.

Plan. Set $u=0.02$ in $\tfrac12(1+u)^{-1}$ .

\begin{aligned} \frac1{2.04}&=\frac12(1+0.02)^{-1} &&\text{normalise}\\ &\approx\frac12(1-0.02+0.0004-0.000008) &&\text{same four terms}\\ &=0.490196 &&\text{evaluate} \end{aligned}

Why compare with a calculator? The validity condition $|0.02|<1$ does not itself tell us the error after four terms.

Answer. $0.490196$ , compared with $0.49019607843\ldots$ ; it is low by about $7.84\times10^{-8}$ .

Check. $2.04(0.490196)=0.99999984$ , which is below $1$ , consistent with an underestimate.

There is a pleasing exact check for these particular reciprocals:

(1+u)(1-u+u^2-u^3)=1-u^4.

Subtracting the cubic polynomial from $(1+u)^{-1}$ therefore gives $u^4/(1+u)$ , when $u\ne-1$ . This identity is derived by multiplication, not a general error rule for all rational powers. At $u=0.01$ , the error after the outside factor is exactly $0.01^4/[2(1.01)]<5\times10^{-9}$ , settling the opening puzzle.

Watch out

Wrong: “the first omitted term always bounds the error”. Right: a bound needs justification. The identity above justifies one reciprocal error; it does not establish the same claim for every rational exponent.

Use the same cubic reciprocal polynomial for a new denominator.

Quick Recap

Rational powers use descending coefficient factors; most give non-terminating expansions.
Normalise a bracket to $1+u$ , including the correct outside power.
Keep every contribution to the required degree and intersect all validity intervals.
A polynomial approximation has a truncation error; justify any claimed bound.

Four terms give $1/2.02\approx0.4950495$ . The exact reciprocal identity shows its error is less than $0.000000005$ .

Last Accessed

Pure Mathematics

Statistics

Mechanics

Ms3–Ms5 - Rational binomial expansions and approximation

Expanding a rational power

Factor out the constant before expanding

Validity and combining truncated expansions

Using the polynomial as an approximation

Quick Recap

Learn faster with Nova