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Approximations using binomial expansion

Approximate roots, reciprocals and quotients with binomial expansion. Choose a small input, calculate absolute and percentage errors and justify rounding with worked practice.

Before you startGeneralised binomial expansion, products and validity

01 / Use a short polynomial near a known value

Small input helps, but check the resulting accuracy.

A binomial expansion can turn a difficult numerical calculation into a few multiplications. The key is to rewrite the target so that the substituted quantity is small, then retain enough terms.

Compare the errors in the model. A valid input near the edge of the interval can need many terms. A small-looking error bar still has a numerical value; read it before deciding how many digits to report.

How much error is left?Explore

√(1 + u), with u = 0.2

P₂(u) = 1 + u/2 − u²/8.

Exact value ≈ 1.095445115; polynomial = 1.095; absolute error ≈ 0.000445115; percentage error ≈ 0.040633255%.

The vertical scale updates with the input and function. Gold marks the selected degree. Read the numbers when a bar is too small to see.

All offered inputs satisfy |u| < 1. Values and errors are calculated before display rounding; very small nonzero values use scientific notation.

Function Input u 0.2Highest power retained 2

Watch a geometric approximation approach its exact value

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Choose the small input

Factor a nearby value whose root you know exactly.

Prepare √50 for approximationWorked example

50 = 49(1 + 1/49), so √50 = 7(1 + 1/49)^(½)

The known square 49 makes u = 1/49 small.

Use √(1 + u) ≈ 1 + u/2 − u²/8

Keep through degree 2 in u.

√50 ≈ 7 + 1/14 − 1/2744 = 19403/2744

Apply the outside factor 7 to every term.

≈ 7.071064140, compared with √50 ≈ 7.071067812

The absolute error is about 0.000003672.

Writing √50 as (1 + 49)^(½) gives u = 49, outside the standard interval. A correct algebraic rewrite is not automatically a useful binomial substitution.

03 / A target already close to one

Keep the substitution exact until the final evaluation.

Approximate √1.06 through the cubic termWorked example

Set u = 0.06 in 1 + u/2 − u²/8 + u³/16

Here |u| < 1.

1 + 0.03 − 0.00045 + 0.0000135 = 1.0295635

Every power uses u = 0.06.

The exact reference is about 1.029563014

Absolute error is about 0.000000486.

Keeping only through u² gives 1.02955

That shorter truncation has error about 0.000013014.

A higher degree changes the approximation, but you still need an accuracy check before claiming a particular number of decimal places.

04 / Cube roots and reciprocal roots

Rewrite the target and keep the sign of the input.

Approximate ∛7.76 through degree 2Worked example

∛7.76 = 2(1 − 0.03)^(⅓)

Factor the nearby cube 8: 7.76/8 = 0.97.

Use (1 + u)^(⅓) ≈ 1 + u/3 − u²/9

Here u = −0.03.

2[1 − 0.01 − 0.0001] = 1.9798

The square term stays negative despite u being negative.

Approximate 1/√0.98 through degree 3Worked example

Use (1 + u)^(−½) with u = −0.02

The coefficients are 1, −½, 3/8, −5/16.

1 + 0.01 + 0.00015 + 0.0000025 = 1.0101525

Both negative signs cancel in the cubic contribution.

Outside constants, such as the factor 2 in the cube-root example, also scale the approximation error. Do not report the error inside the bracket as the final target’s error.

05 / A quotient can use a product expansion

Translate the target into the same variable everywhere.

Approximate √1.04 / 0.98Worked example

Use √(1 + 2x)/(1 − x) with x = 0.02

This choice matches both the numerator and denominator.

The expansion is 1 + 2x + (3/2)x² + 2x³ + …

It is guaranteed for |x| < ½.

Substitute x = 0.02: 1 + 0.04 + 0.0006 + 0.000016 = 1.040616

Keep through x³.

The reference value is about 1.040616227

The absolute error is about 0.000000227.

Do not independently choose two different x values for the two brackets in an expansion derived using one common x. If the target does not match that structure, derive the appropriate expansion instead.

06 / Measure the error

Use the exact reference in the percentage denominator.

Absolute error = |approximation − exact value|
Percentage error = absolute error ÷ |exact value| × 100%, when exact value ≠ 0.

Approximate 1/1.04 by four geometric termsWorked example

Use (1 + u)⁻¹ ≈ 1 − u + u² − u³, with u = 0.04

The approximation is 0.961536.

The reference is 25/26 ≈ 0.961538462

Calculate errors using the unrounded values.

Absolute error = 1/406250 ≈ 0.000002462

The exact geometric remainder gives this value.

Percentage error = 0.000256%

Divide by 25/26, not by the approximation.

Errors calculated from two prematurely rounded numbers can be badly misleading. Keep guard digits or exact fractions, then round the reported error.

07 / The first omitted term is not always the error

A bound needs a reason.

For 1/(1 − u), retain terms through uᵐWorked example

Pₘ = 1 + u + … + uᵐ = (1 − uᵐ⁺¹)/(1 − u)

This is a finite geometric sum, for u ≠ 1.

Exact remainder = 1/(1 − u) − Pₘ = uᵐ⁺¹/(1 − u)

The entire tail includes more than its first term.

At u = 0.5, m = 2: remainder = 0.5³/0.5 = 0.25

The first omitted term alone is 0.125: only half the error.

For some alternating series with decreasing term magnitudes, the first omitted term supplies an error bound. Those conditions need checking; they do not follow merely from using a binomial expansion. A finite truncation is an approximation unless the remaining terms vanish.

08 / Justify the digits you report

A rounding boundary matters as well as the error size.

Justify √1.06 = 1.02956 to five decimal placesWorked example

The rounding cell for 1.02956 is from 1.029555 to 1.029565

These are the halfway boundaries between neighbouring five-decimal values.

1.029555² = 1.059983498025 < 1.06

Both the lower bound and the root are positive.

1.029565² = 1.060004089225 > 1.06

Squaring preserves the order for positive numbers.

Therefore 1.029555 < √1.06 < 1.029565

The exact root lies strictly inside that rounding cell.

Approximate 1/0.97 with 1 + u + u² + u³, where u = 0.03. Give an exact expression for the absolute error.

Hint

The omitted geometric tail is u⁴/(1 − u).

Worked solution

The approximation is 1.030927. Error = (0.03)⁴/0.97 = 81/97000000 ≈ 0.000000835. It is positive because all omitted terms are positive.

10 / Recap

Choose well, calculate exactly, then judge the error.

Factor a nearby known square, cube or convenient constant.
Keep the transformed input small and inside the guaranteed region.
Include all outside factors.
Use unrounded values to calculate errors.
Distinguish a first omitted term from a proven bound.
Check rounding boundaries before claiming decimal accuracy.

Section 1 of 10 · Use a short polynomial near a known value