{left( {frac{a}{{a + x}}} right)^{frac{1}{2}} | vedclass

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Question

(A) We have the expression: ${\left( {1 + \frac{x}{a}} \right)^{-1/2}} + {\left( {1 - \frac{x}{a}} \right)^{-1/2}}$.Using the binomial expansion $(1+z

Vedclass · Accepted Answer

$2 + \frac{{3{x^2}}}{{4{a^2}}} + \dots$

Vedclass · Answer

(A) We have the expression: ${\left( {1 + \frac{x}{a}} \right)^{-1/2}} + {\left( {1 - \frac{x}{a}} \right)^{-1/2}}$.Using the binomial expansion $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \dots$,we expand both terms:${\left( {1 + \frac{x}{a}} \right)^{-1/2}} = 1 - \frac{1}{2}\left( \frac{x}{a} \right) + \frac{(-1/2)(-3/2)}{2}\left( \frac{x}{a} \right)^2 + \dots = 1 - \frac{x}{2a} + \frac{3x^2}{8a^2} + \dots$${\left( {1 - \frac{x}{a}} \right)^{-1/2}} = 1 - \frac{1}{2}\left( -\frac{x}{a} \right) + \frac{(-1/2)(-3/2)}{2}\left( -\frac{x}{a} \right)^2 + \dots = 1 + \frac{x}{2a} + \frac{3x^2}{8a^2} + \dots$Adding these two expansions,the odd terms (involving $x/a$) cancel out:$(1 + 1) + (-\frac{x}{2a} + \frac{x}{2a}) + (\frac{3x^2}{8a^2} + \frac{3x^2}{8a^2}) + \dots = 2 + \frac{6x^2}{8a^2} + \dots = 2 + \frac{3x^2}{4a^2} + \dots$

${\left( {\frac{a}{{a + x}}} \right)^{\frac{1}{2}}} + {\left( {\frac{a}{{a - x}}} \right)^{\frac{1}{2}}} = $

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