show-that-x-is-small-enough-for-its-cube-and-higher-power-to-be-neghected-1-x-1-x-1-x-1-2-x-2-by-putting-1-8-show-that-7-2-83-128-

Question Number 9576 by j.masanja06@gmail.com last updated on 18/Dec/16

show that x is small enough for its cube and higher power to be neghected

\sqrt{\frac{1 - x}{1 + x}} = 1 - x + \frac{1}{2} x^{2} .

by putting = \frac{1}{8}, show that \sqrt{7} \approx 2 \frac{83}{128} .

Commented by prakash jain last updated on 20/Dec/16

From binomial theorem

{(1 - x)}^{1 / 2} = 1 - \frac{1}{2} x + \frac{\frac{1}{2} (\frac{1}{2} - 1)}{2!} x^{2} + \dots \approx 1 - \frac{1}{2} x - \frac{1}{8} x^{2}

similarly

{(1 + x)}^{- 1 / 2} \approx 1 - \frac{1}{2} x + \frac{(- \frac{1}{2}) (- \frac{1}{2} - 1)}{2!} x^{2} = 1 - \frac{1}{2} x + \frac{3}{8} x^{2}

{(1 - x)}^{1 / 2} {(1 + x)}^{- 1 / 2} = 1 - x + \frac{1}{4} x^{2} + \frac{3}{8} x^{2} - \frac{1}{8} x^{2}

{(1 - x)}^{1 / 2} {(1 + x)}^{- 1 / 2} = 1 - x + \frac{1}{2} x^{2}

put x = \frac{1}{8} to get the required result .