Question

If \( x \) is so small that \( x^{3} \) and higher powers of \( x \) may be neglected show that \[ \frac{(4-7 x)^{1 / 2}}{(3+5 x)^{3}}=\frac{2}{27}\left(1-\frac{47}{8} x+\frac{7933}{384} x^{2}\right) \]

Answer 1

\(\frac{(4-7x)^{\frac{1}{2}}}{(3+5x)^3}=(4-7x)^{\frac{1}{2}}\) (3+5x) = 3

\(=\frac{2}{27}(1-\frac{7x}{4})^{\frac{1}{2}}(1+\frac{5x}{3})^{-3}\)

\(=\frac{2}{27}(1+\frac{1}{2}(\frac{-7x}{4})+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(\frac{-7x}{4})^2)\)

\((1+(-3)(\frac{5x}{3})+\frac{(-3)(-3-1)}{2!})(\frac{5x}{3}))^2\)

\(=\frac{2}{27}(1-\frac{7x}{8}-\frac{1}{8}(\frac{49x^2}{16}))(1-5x+\frac{50x^2}{3})\)

\(=\frac{2}{27}(1-5x-\frac{7x}{8}+\frac{50x^2}{3}+\frac{35x}{8}-\frac{49x^2}{128})\)

\(=\frac{2}{27}(1-\frac{47x}{8}+\frac{7933x^2}{384})\)

Hence proved