Answer to Question #234343 in Calculus for Sha

Question

Expand (2+x)^(-1/2) in ascending powers of x, up to the term x^3 . Hence approximate (2/3)^(1/2) correct to 3 significant figures

Answer 1

The general formula is given by

$(1+x)^n=1+nx+{n(n-1)x^2\over 2!}+{n(n-1)(n-2)x^3\over 3!}+...$

We are given

$(2+x)^{-{1\over 2}}$

Factor out $2$ in $(2+x)^{-{1\over 2}}$ to get

$2^{-{1\over 2}} (1+{x\over 2})^{-{1\over 2}}$

Expand $(1+{x\over 2})^{-{1\over 2}}$ in ascending powers of $x$ , up to the term $x^3$

$(1+x)^n=1+nx+{n(n-1)x^2\over 2!}+{n(n-1)(n-2)x^3\over 3!}+...$

$\implies (1+{x\over 2})^{-{1\over 2}}=1+(-{1\over 2})({x\over 2})+{(-{1\over 2})((-{1\over 2})-1)({x\over 2})^2\over 2!}+{(-{1\over 2})((-{1\over 2})-1)((-{1\over 2})-2)({x\over 2})^3\over 3!}+...$

$\approx 1+(-{1\over 2})({x\over 2})+{(-{1\over 2})((-{1\over 2})-1)({x\over 2})^2\over 2!}+{(-{1\over 2})((-{1\over 2})-1)((-{1\over 2})-2)({x\over 2})^3\over 3!}$

$= 1-{x\over 4}+0.09375x^2-0.0390625x^3$ up to the term $x^3$

$\implies (2+x)^{-{1\over 2}} \approx 2^{-{1\over 2}}( 1-{x\over 4}+0.09375x^2-0.0390625x^3)$

$= 0.707106781( 1-{x\over 4}+0.09375x^2-0.0390625x^3)$

$= 0.707106781-0.176776695x+0.06629126x^2-0.027621358x^3$

$({2\over 3})^{1\over 2}=({3\over 2})^{-{1\over 2}}=(1+{1\over 2})^{-{1\over 2}}$

Since $(1+{x\over 2})^{-{1\over 2}} \approx 1-{x\over 4}+0.09375x^2-0.0390625x^3$

Then $(1+{1\over 2})^{-{1\over 2}} \approx 1-{1\over 4}+0.09375(1)^2-0.0390625(1)^3 \approx 0.8046875$

$\therefore ({2\over 3})^{1\over 2} \approx 0.8046875$