If x is so small that x^3 and higher powers of x may be neglected, then [( 1 + x )^3 2 - (…

Question

If $x$ is so small that ${x^3}$ and higher powers of $x$ may be neglected, then $${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$ may be approximated as

Accepted Answer

$${\left( {1 + x} \right)^{{3 \over 2}}}$$ = 1 + $${3 \over 2}x + {{{3 \over 2}.{1 \over 2}} \over {1.2}}{x^2} + ...$$

= 1 + $${3 \over 2}x + {3 \over 8}{x^2}$$ (As $x$ is so small, so ${x^3}$ and higher powers of $x$ neglected)

$${{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}} \over {{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$$

= $${{\left( {1 + {3 \over 2}x + {3 \over 8}{x^2}} \right) - \left( {1 + {}^3{C_0}.{x \over 2} + {}^3{C_1}.{{\left( {{x \over 2}} \right)}^2}} \right)} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{{3 \over 8}{x^2} - {3 \over 4}{x^2}} \over {{{\left( {1 - x} \right)}^2}}}$$

= $${{x^2}\left( {{3 \over 8} - {3 \over 4}} \right){{\left( {1 - x} \right)}^{ - 2}}}$$

= $${ - {3 \over 8}{x^2}\left( {1 - {1 \over 2}\left( { - x} \right) + ....} \right)}$$

= $$ - {3 \over 8}{x^2} - {3 \over {16}}{x^3}$$

[ As x3 is so small we can ignore $$-{3 \over {16}}{x^3}$$]

= $$ - {3 \over 8}{x^2}$$

Solutions · 1