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Original question & working

Multiply two binomial series · source reference

[2009/IJC/Prelim/I/7] · Question pp. 2 · Working pp. 2

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The working and teaching notes are included in the interactive lesson.

Both Topic 09 PDFs are identical and include full solutions beside the questions; opening source pages reveals answers. Source p.2 has a product-line typo (1−2x) in place of (1+2x); the actual binomial expansion and result use the plus sign.

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Full original question, working & notes

Printed question

Find the expansion of (1−x1+2x)1/3\left(\frac{1-x}{1+2x}\right)^{1/3} in ascending powers of xx, up to and including the term in x2x^2. State the range of values of xx for which the expansion is valid. [5]

Without performing any calculations, explain why putting x=16x=\frac16 into the result gives a better approximation to 53\sqrt[3]{5} than putting x=−411x=-\frac4{11}. [1]

Hence by putting x=16x=\frac16, find an approximation for 53\sqrt[3]{5}, expressing your answer as a fraction in its lowest terms. [2]

[2009/IJC/Prelim/I/7]

Printed working

(1−x)1/3=1+13(−x)+(1/3)(−2/3)2!(−x)2+⋯=1−x/3−x2/9+⋯ ,(1+2x)−1/3=1−13(2x)+(−1/3)(−4/3)2!(2x)2+⋯=1−2x/3+8x2/9+⋯ .\begin{aligned}(1-x)^{1/3}&=1+\tfrac13(-x)+\tfrac{(1/3)(-2/3)}{2!}(-x)^2+\cdots=1-x/3-x^2/9+\cdots,\\(1+2x)^{-1/3}&=1-\tfrac13(2x)+\tfrac{(-1/3)(-4/3)}{2!}(2x)^2+\cdots=1-2x/3+8x^2/9+\cdots.\end{aligned}

Printed product-line typo: (1−x)1/3(1−2x)−1/3(1-x)^{1/3}(1-2x)^{-1/3}; correct denominator sign is (1+2x)−1/3(1+2x)^{-1/3}.

(1−x/3−x2/9+⋯ )(1−2x/3+8x2/9+⋯ )=1−x+x2+⋯(1-x/3-x^2/9+\cdots)(1-2x/3+8x^2/9+\cdots)=1-x+x^2+\cdots.

∣−x∣<1|-x|<1, ∣2x∣<1⇒∣x∣<1|2x|<1\Rightarrow |x|<1, ∣x∣<1/2⇒−1/2<x<1/2|x|<1/2\Rightarrow-1/2<x<1/2.

Both substitutions lie in the validity range. 0<∣1/6∣<∣−4/11∣0<|1/6|<|-4/11|, hence the first is closer to zero. At x=1/6x=1/6, ((1−1/6)/(1+2/6))1/3=(5/8)1/3((1-1/6)/(1+2/6))^{1/3}=(5/8)^{1/3} and 53≈2(1−1/6+1/36)=31/18\sqrt[3]5\approx2(1-1/6+1/36)=31/18.

Teacher note 1

Since both expansions of (1−x)1/3(1-x)^{1/3} and (1+2x)−1/3(1+2x)^{-1/3} must be valid at the same time, we take the intersection of the 2 regions.

Diagram 09-02-validity

Intersection of the two validity intervals

Both source number-line rows have open endpoints. Intersection is open.

Mathematical review

(1−x)1/3=1−x/3−x2/9+O(x3)(1-x)^{1/3}=1-x/3-x^2/9+O(x^3) and (1+2x)−1/3=1−2x/3+8x2/9+O(x3)(1+2x)^{-1/3}=1-2x/3+8x^2/9+O(x^3). Their product is 1−x+x2+O(x3)1-x+x^2+O(x^3); both converge for ∣x∣<1/2|x|<1/2.

Both substitutions lie inside this range, and 1/61/6 is closer to zero, so the truncated higher powers are smaller. At x=1/6x=1/6, the exact expression is 53/2\sqrt[3]{5}/2. Thus 53≈2(1−1/6+1/36)=31/18\sqrt[3]{5}\approx2(1-1/6+1/36)=31/18. At −4/11-4/11 the expression is 53\sqrt[3]{5} directly; the printed comparison is a truncation heuristic, not an error bound.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 2. Working page(s): 2. The two Series source files are byte-identical; their shared visible content is expressed once and mapped to both page sets. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.