If $x$ is very small, then prove the following:

Question

If $x$ is very small, then prove that: $$\bigg[\frac{1-x}{1+x}\bigg]^{1/2}=1-x+\frac{x^2}{2}$$

I wrote it as $[1-x]^{1/2} [1+x]^{-1/2} $ and expanded and neglected $x^3$ and higher powers. But expression I am getting is $1-\frac{x^2}{4}$. How would I get required R.H.S.?

Answer 1

$$(1-x)^{1/2}(1+x)^{-1/2}$$

$$=\left(1+\dfrac12(-x)+\dfrac{1(-1)}4(-x)^2+\cdots\right)\left(1-\dfrac 12x+x^2\dfrac{(-1)(-3)}4+\cdots\right)$$

$$=1+x\left(-\dfrac12-\dfrac12\right)+x^2\left(\dfrac34-\dfrac14+\dfrac14\right)+O(x^3)$$

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Alternatively,

$$\sqrt{\dfrac{1-x}{1+x}}=(1-x)(1-x^2)^{-1/2}=(1-x)\left(1+\dfrac{x^2}2+O(x^4)\right)=1-x+\dfrac{x^2}2+O(x^3)$$