Find the expansion of $(8 - x)^{\frac{1}{3}}$ in ascending powers of $x$ up to and including the

Question

Find the expansion of $(8 - x)^{\frac{1}{3}}$ in ascending powers of $x$ up to and including the term in $x^2$. State the set of values of $x$ for which this expansion is valid. By substituting $x=-1$, find an approximation of $\sqrt[3]{3}$, leaving your answer in the form $\frac{m}{n}$, where $m$ and $n$ are integers to be determined. The substitution $x=5$ can also be used to find an approximation for $\sqrt[3]{3}$. Explain whether this will give a better approximation than using $x=-1$.

Accepted Answer

(i) $2-\frac x{12}-\frac{x^2}{288}+\cdots$, valid for $-8<x<8$. (ii) $\sqrt[3]3\approx\frac{864}{599}$. (iii) No; $x=5$ gives a worse approximation.

Maclaurin Series and Power Series - Binomial Expansion as a Maclaurin Series

Maclaurin Series and Power Series - Binomial Expansion as a Maclaurin Series

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