Porisma

Worked solution

Approximate a cube root with a binomial series

(i) Expand 8+x3\sqrt[3]{8+x} through x2x^2 and state the range of validity.

(ii) Put x=1/27x=1/27 to approximate 2173\sqrt[3]{217} as p/qp/q in lowest terms, with p,q∈Z+p,q\in\mathbb Z^+.

(iii) We can also substitute x=1/217x=1/217 into the expansion found in (i) to get an approximate value of 2173\sqrt[3]{217}. Explain whether this approximation is better than the approximation in (ii).

[2007/YJC/Promo/1]

CJC · Discussion · Question 6

Answer

(i) 2+x/12−x2/288+O(x3)2+x/12-x^2/288+O(x^3), printed range ∣x∣<8|x|<8. (ii) 420551/69984420551/69984. (iii) The printed “better” claim lacks the transformation needed to approximate the requested cube root.

(i) Expand about 8

8+x3=2(1+x/8)1/3=2[1+13(x/8)+(1/3)(−2/3)2!(x/8)2+⋯ ]=2+x/12−x2/288+O(x3)\sqrt[3]{8+x}=2(1+x/8)^{1/3}=2[1+\tfrac13(x/8)+\tfrac{(1/3)(-2/3)}{2!}(x/8)^2+\cdots]=2+x/12-x^2/288+O(x^3). The printed MF27 validity condition is ∣x/8∣<1|x/8|<1, giving −8<x<8-8<x<8.

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(ii) Rescale the cube root

At x=1/27x=1/27, 217/273≈2+1/324−1/209952=420551/209952\sqrt[3]{217/27}\approx2+1/324-1/209952=420551/209952. Since 2173=3217/273\sqrt[3]{217}=3\sqrt[3]{217/27}, 2173≈420551/69984\sqrt[3]{217}\approx420551/69984, in lowest terms.

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(iii) Check what the substitution approximates

The printed solution answers “better” because 1/2171/217 is closer to zero. This compares local truncation for 8+x3\sqrt[3]{8+x}, but the stated substitution estimates 8+1/2173=1737/2173\sqrt[3]{8+1/217}=\sqrt[3]{1737/217}, not 2173\sqrt[3]{217}. The source supplies no transformation between these targets, so the claimed approximation of 2173\sqrt[3]{217} is not established as stated. A smaller expansion variable alone does not complete this comparison.

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Common mistakes

A smaller substitution improves a local approximation only after the target and scaling have been checked.

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