01 · First four terms
Write the first four terms in ascending powers of x of (1 − x/2)⁸.
Hint
Use 1, 8, 28, 56 and powers of −x/2.
Worked solution
1 − 4x + 7x² − 7x³ + …
Understand · explore · practise
Binomial approximations
Use a few binomial terms to estimate powers, compare errors and choose a suitable substitution. Includes percentage error, products, growth and independent-trial probability models.
Before you startFinite binomial expansion and percentages
01 / Exact or approximate?
A finite expansion is exact; dropping terms changes that.
For a positive integer n, the full expansion of (1 + u)ⁿ is valid for every real u. There is no small-u restriction on the identity itself.
(1 + u)⁸ = 1 + 8u + 28u² + 56u³
+ 70u⁴ + 56u⁵ + 28u⁶ + 8u⁷ + u⁸
When |u| is small, later terms may be small enough to ignore. Retaining only some terms gives an approximation, written with ≈. Use the control to include terms through degree k and compare with the full expression.
At k = 8, every term is included. At u = 0, the constant alone already gives the exact value.
(1 + u)⁸Choose and compare
For u = 0.05, (1 + u)⁸ is about 1.47745544. Retaining through u² gives 1.47, an underestimate. Values on the chart are rounded to eight decimal places.
Watch partial sums approach the full value
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / What counts as small?
Inspect whole terms, including their coefficients.
For 0 < |u| < 1, powers |u|ʳ get smaller as r increases. But a binomial term also contains C(n,r). Small powers alone do not guarantee a small omitted term.
(1 + u)ⁿ begins 1 + nu + n(n − 1)u²/2 + …
For n = 1000 and u = 0.001, the first three terms are 1, 1 and 0.4995. Even this small u does not make the quadratic term negligible. The exponent and the accuracy needed both matter.
In the model, try u = 0.2 and retain only the cubic terms. The first omitted term is 70(0.2)⁴ = 0.112, already large enough to affect the estimate substantially.
Negative inputs
For negative u, terms alternate in sign. Partial sums can lie above or below the full value. Do not assume every extra term increases the answer, or that the approximation must be positive just because the full expression is positive.
03 / Choose the substitution
Match the base exactly before substituting.
To estimate 0.99⁸ using (1 − x/2)⁸, solve 1 − x/2 = 0.99. This gives x = 0.02; the relative change u = −x/2 is −0.01.
Estimate with four termsWorked example
(1 − x/2)⁸
= 1 − 4x + 7x² − 7x³ + …
Use r = 0, 1, 2, 3.
x = 0.02
0.99⁸ ≈ 1 − 0.08 + 0.0028 − 0.000056
Substitute after simplifying the coefficients.
0.99⁸ ≈ 0.922744
The next omitted term is positive. The full calculator value is about 0.9227446944.
04 / A base other than 1
Factor out the constant to identify the relative change.
For (a + bx)ⁿ with a ≠ 0, write aⁿ(1 + bx/a)ⁿ. The relevant small quantity is u = bx/a, not necessarily x itself.
(2 + x)⁵ = 32(1 + x/2)⁵
= 32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵
To estimate 2.02⁵, set x = 0.02 and retain through x²:
2.02⁵ ≈ 32 + 80(0.02) + 80(0.02)²
= 33.632
The full value is 33.6323216032. Factoring out 2 without raising it to the fifth power would lose the multiplier 32.
05 / Approximate products
Keep every contribution through the requested degree.
To keep terms through x² in a product, each factor only needs terms through x², provided neither contains negative powers. Multiply, collect, then discard terms of degree 3 or above.
(2 − x)(1 + 3x)⁴
= (2 − x)(1 + 12x + 54x² + …)
= 2 + 23x + 96x² + …
The x² coefficient includes both 2 × 54 and −1 × 12. Dropping the latter contribution gives the wrong approximation.
Keeping only linear terms
(1 + x)(1 − 2x)⁵
= (1 + x)(1 − 10x + …)
≈ 1 − 9x
The x × (−10x) product is quadratic, so it is dropped in a linear approximation.
06 / Error and rounding
Rounding an estimate does not certify its accuracy.
If A is an approximation and E is the exact value, the absolute error is |A − E|. When E ≠ 0, percentage error is:
Percentage error = 100 × |A − E|/|E|
Use the unrounded calculator values in this calculation. For 1.02⁸, the cubic estimate is 1.171648, while the full value is about 1.1716593810022656. Its percentage error is about 0.000971%.
A useful caution: the quadratic estimate for 1.01⁸ is 1.0828. The full value is about 1.0828567056, which rounds to 1.0829 at four decimal places. The estimate has four written decimals but is not accurate to four decimal places.
Can I tell whether an estimate is too high or too low?
For u > 0, every omitted term in a positive-integer expansion of (1 + u)ⁿ is positive. A truncated sum is therefore an underestimate.
For 0.99⁸, the omitted terms after the cubic alternate in sign with decreasing magnitudes, starting with +70(0.01)⁴ = 0.0000007. Pairing these decreasing terms bounds the remaining sum strictly between 0 and 0.0000007. The full value lies between 0.922744 and 0.9227447, so 0.9227 to four decimal places is justified here.
07 / Models in context
Keep the model’s assumptions separate from the approximation.
A quantity initially 1000 that increases by p percent per period for four periods is modelled by 1000(1 + p/100)⁴. Keeping terms through p² gives:
1000(1 + p/100)⁴
≈ 1000 + 40p + 0.6p²
Here p is the percentage number: p = 2 means 2%, so the relative change is 0.02. The exact model assumes the same proportional change each period.
An independent-trial probability model
Suppose 80 independent trials each have failure probability p. The probability of no failures is P = (1 − p)⁸⁰ for 0 ≤ p ≤ 1. This exact formula does not require p to be small; a short expansion does.
P ≈ 1 − 80p + 3160p²
On the proposed small-p interval 0 ≤ p ≤ 0.001, asking this approximation to be at least 0.95 gives 3160p² − 80p + 0.05 ≥ 0. The small root is:
p = [80 − √5768]/6320
≈ 0.000641242
The other root is about 0.0246752, outside that interval. The approximate condition is therefore p ≤ 0.000641242 within the proposed interval.
Checking the exact model gives p ≤ 1 − 0.95¹ᐟ⁸⁰ ≈ 0.000640961, slightly smaller. A threshold inferred from a truncated model should not be presented as an exact guarantee.
08 / Your turn
State the substitution and the retained degree.
Use ≈ for a truncated numerical answer. Keep extra calculator digits until the final comparison or rounding.
02 · A small substitution
Use those four terms to estimate 0.995⁸.
Hint
1 − x/2 = 0.995 gives x = 0.01.
Worked solution
0.995⁸ ≈ 1 − 0.04 + 0.0007 − 0.000007
= 0.960693
03 · Factor the constant
Use terms through x² in (2 + x)⁵ to estimate 2.02⁵.
Hint
Retain 32 + 80x + 80x² and set x = 0.02.
Worked solution
32 + 1.6 + 0.032 = 33.632
The omitted terms make the full value slightly larger.
04 · A product
Find a + bx + cx² approximating (2 − x)(1 + 3x)⁴ when higher powers can be ignored.
Hint
(1 + 3x)⁴ begins 1 + 12x + 54x².
Worked solution
a = 2
b = 2 × 12 − 1 = 23
c = 2 × 54 − 12 = 96
05 · Percentage error
Compare the cubic estimate 1.171648 with 1.02⁸. Find the percentage error to three significant figures.
Hint
Divide the absolute difference by the full calculator value, then multiply by 100.
Worked solution
100 × |1.171648 − 1.02⁸| / 1.02⁸
≈ 0.000971%
This is a percentage, so do not multiply by 100 again.
06 · A growth model
A quantity starts at 500 and grows by g percent per period for three periods. Give a quadratic approximation in g, then estimate the result for g = 2.
Hint
Expand 500(1 + g/100)³.
Worked solution
≈ 500 + 15g + 0.15g²
At g = 2: 500 + 30 + 0.6 = 530.6
The complete finite expansion gives 530.604.
07 · Probability approximation
For 40 independent trials with equal failure probability p, approximate the probability of no failures through p². Evaluate it at p = 0.001.
Hint
Expand (1 − p)⁴⁰.
Worked solution
P ≈ 1 − 40p + 780p²
At p = 0.001: P ≈ 0.96078
The exact probability is about 0.960770211. The approximation is slightly too high.
08 · Check the accuracy claim
A student says 1.0828 is accurate to four decimal places for 1.01⁸ because the quadratic estimate has four decimals. Explain the mistake.
Hint
Compare with 1.01⁸ ≈ 1.0828567056.
Worked solution
The exact value rounds to 1.0829 at four decimal places. Writing or rounding four decimals does not bound the error of a truncated expansion.
09 · Large coefficients
Why is retaining only 1 + nu unsafe for (1.001)¹⁰⁰⁰ without an error check?
Hint
Calculate the first omitted term.
Worked solution
C(1000,2)(0.001)²
= 499500 × 0.000001 = 0.4995
This omitted term alone is substantial compared with the estimate 2. Small u does not cancel the effect of a large exponent.
09 / Recap
An approximation is useful only at the accuracy you need.
- The full finite identity holds without a small-input restriction.
- For truncation, inspect the relative change u and the whole omitted terms.
- State the substitution explicitly.
- In products, retain every contribution through the required degree.
- Use the full value as the denominator in percentage error.
- Check accuracy before claiming correct decimal places or a threshold.
Section 1 of 9 · Exact or approximate?