Approximation using a binomial expansion

A power such as \((1.045)^7\) can be estimated without a calculator by substituting a small value of \(x\) into the first few terms of an expansion such as \((1+3x)^7\).

Why: when \(x\) is small, each further term carries another factor of \(x\), so the terms shrink quickly and the terms left out are too small to change the answer at the accuracy asked for.

Method

1. Set the bracket equal to the number and solve for \(x\).
2. Use the terms the question asks for. A quick check: if the first term left out is smaller than half a unit in the last place you need (\(0.0005\) for 3 decimal places), the answer is usually safe; if it is not, use one more term. The check can fail when the total is very close to a rounding boundary.
3. Substitute, add the terms with all their digits, and round only at the end.

Example

Expand \((1+3x)^7\) up to the term in \(x^3\), and use it to estimate \((1.045)^7\) correct to 3 decimal places.

\((1+3x)^7=1+7(3x)+21(3x)^2+35(3x)^3+\cdots=1+21x+189x^2+945x^3+\cdots\)

\(1+3x=1.045\), so \(3x=0.045\) and \(x=0.015\).

\((1.045)^7\approx1+21(0.015)+189(0.015)^2+945(0.015)^3=1+0.315+0.042525+0.003189375=1.360714375\approx1.361\)

The next term, \(2835x^4\approx0.00014\), is smaller than \(0.0005\), so the quick check says four terms are enough. Three terms give \(1.357525\approx1.358\), which is wrong in the third decimal place.