6A - Using a binomial approximation

A short binomial expansion can turn a root or reciprocal into ordinary arithmetic. You will choose a useful small variable, calculate an estimate from a prescribed number of terms, and combine short expansions without losing contributions to the powers you need.

Choosing a useful small variable

How could a short polynomial estimate $\sqrt{4.12}$ without using a square-root key?

The number $4.12$ is close to the convenient square $4$ . Factor out that square:

\sqrt{4.12}=2\sqrt{1.03}=2(1+0.03)^{1/2}.

The binomial variable is now only $0.03$ . In contrast, writing $4.12=1+3.12$ would give $u=3.12$ , outside the condition $|u|<1$ .

You already know

For a rational non-integer power, put the expression into $C(1+u)^n$ . The binomial series uses the entire $u$ , and requires $|u|<1$ . The constant $C$ multiplies every term.

If a question supplies an expansion in $x$ , equate its original expression to the target to find $x$ . The value of $x$ and the small binomial variable $u$ need not be the same. Check both the substitution and the condition before calculating.

Worked example 1: a value above a convenient square

Question. A binomial expansion for $\sqrt{4+x}$ is available. Find the value of $x$ needed to estimate $\sqrt{4.12}$ , and identify the binomial variable $u$ .

Plan. Match the expressions under the roots, then factor out $4$ .

\begin{aligned} 4+x&=4.12 &&\text{match the target}\\ x&=0.12 &&\text{subtract }4\\ \sqrt{4+x}&=2\left(1+\frac{x}{4}\right)^{1/2} &&\text{put the bracket in standard form}\\ u&=\frac{0.12}{4}=0.03 &&\text{use the whole substituted expression} \end{aligned}

Answer. Use $x=0.12$ ; the binomial variable is $u=0.03$ , with $|u|<1$ .

Check. $4+0.12=4.12$ , and $4(1+0.03)=4.12$ . Both substitutions recover the intended target exactly.

Keep the same convenient square, but move the target the same distance below it. The small variable changes sign.

Worked example 2: a value below a convenient square

Question. Find the values of $x$ and $u$ needed to use the same expansion of $\sqrt{4+x}$ to estimate $\sqrt{3.88}$ .

Plan. Match $4+x$ to $3.88$ , then use $u=x/4$ .

\begin{aligned} 4+x&=3.88 &&\text{match the new target}\\ x&=-0.12 &&\text{the target is below }4\\ u&=\frac{-0.12}{4}=-0.03 &&\text{normalise by }4 \end{aligned}

Answer. Use $x=-0.12$ and $u=-0.03$ , which still satisfies $|u|<1$ .

Check. $4(1-0.03)=3.88$ . Also, $2\sqrt{0.97}$ must be below $2$ , matching the target's size.

A small variable makes successive powers small: $0.03^2=0.0009$ and $0.03^3=0.000027$ . This helps explain why a few terms can be useful, although validity alone does not guarantee a particular accuracy.

Watch out

Substitute the value of the variable, not the target root itself. To estimate $\sqrt{3.88}$ from an expansion of $\sqrt{4+x}$ , use $x=-0.12$ , not $x=\sqrt{3.88}$ .

Choose the variable before attempting any binomial arithmetic.

Calculating an approximation

A prescribed approximation uses exactly the requested terms, with the chosen variable substituted into every power.

For a square root, the first four terms give

(1+u)^{1/2}\approx1+\frac{u}{2}-\frac{u^2}{8}+\frac{u^3}{16}.

The constant counts as the first term. “Up to and including $u^3$ ” therefore means four terms here. Retain extra calculator digits in the working and round the resulting estimate only at the end.

Truncated expansion

A truncated expansion stops after a specified term. The omitted terms still contribute to the exact value, so write $\approx$ when replacing the full expression by the finite sum.

Use the expansion requested in the question even if the calculator can evaluate the target directly. Direct evaluation is useful afterwards as a check; it is a different calculation from the specified estimate.

Worked example 3: using four terms

Question. Use the first four terms of the binomial series to estimate $\sqrt{4.12}$ . Give the estimate to 6 decimal places.

Plan. Use $\sqrt{4.12}=2(1+0.03)^{1/2}$ , substitute into all four terms, then round.

The condition holds because $|0.03|<1$ .

\begin{aligned} \sqrt{4.12}&\approx2\left(1+\frac{0.03}{2}-\frac{0.03^2}{8}+\frac{0.03^3}{16}\right)\\ &=2(1+0.015-0.0001125+0.0000016875) &&\text{evaluate each term}\\ &=2.029778375 &&\text{retain digits before rounding} \end{aligned}

Answer. The four-term estimate is $2.029778$ to 6 decimal places.

Check. Direct evaluation gives $\sqrt{4.12}\approx2.029778313$ , close to the unrounded estimate. The result is just above $2$ , as expected.

Use the same four-term expression for the target below $4$ . Only the sign of $u$ changes, so the quadratic term stays negative and the cubic term becomes negative.

Worked example 4: substituting a negative value

Question. Use the first four terms to estimate $\sqrt{3.88}$ . Give the estimate to 6 decimal places.

Plan. Use $u=-0.03$ in $2(1+u)^{1/2}$ , retaining the signs of its powers.

Again, $|u|=0.03<1$ .

\begin{aligned} \sqrt{3.88}&\approx2\left(1+\frac{-0.03}{2}-\frac{(-0.03)^2}{8}+\frac{(-0.03)^3}{16}\right)\\ &=2(1-0.015-0.0001125-0.0000016875) &&\text{square positive, cube negative}\\ &=1.969771625 &&\text{evaluate the finite sum} \end{aligned}

Answer. The four-term estimate is $1.969772$ to 6 decimal places.

Check. Direct evaluation gives $\sqrt{3.88}\approx1.969771560$ , close to the unrounded estimate and below $2$ .

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Number

Algebra

Series

Geometry

Trigonometry

Calculus

6A - Using a binomial approximation

Choosing a useful small variable

Calculating an approximation

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