how $\sqrt x \sqrt{1+\epsilon} \approx \sqrt x \left( 1+\frac{\epsilon}{2} +O(\epsilon ^2) \right)$

Question

I am looking for some explanation of how the following was completed. $$\sqrt{x(1+\epsilon)}= \sqrt x \sqrt{1+\epsilon}\approx \sqrt x \left( 1+\frac{\epsilon}{2} +O(\epsilon ^2) \right)$$

I understand $$\sqrt{x(1+\epsilon)}= \sqrt x \sqrt{1+\epsilon}$$

but I am confused how $$\sqrt x \sqrt{1+\epsilon} \approx \sqrt x \left( 1+\frac{\epsilon}{2} +O(\epsilon ^2) \right)$$

the above uses the taylor series

Answer 1

You also simply add a small quantity to get a complete square $$ \sqrt{1+ϵ}\approx\sqrt{1+ϵ+\frac{ϵ^2}4}=1+\fracϵ2. $$ To check the difference, use the binomial formula $(a-b)(a+b)=a^2-b^2$ in $$ \sqrt{1+ϵ}-(1+\fracϵ2)=\frac{(1+ϵ)-(1+ϵ+\frac{ϵ^2}4)}{\sqrt{1+ϵ}+(1+\fracϵ2)}\approx-\frac{ϵ^2}8. $$

Answer 2

This is just the formula of the asymptotic expansion of the function : $x \mapsto \sqrt{1+x}$ near $0$.

The taylor serie of $x \mapsto \sqrt{1+x}$ is: $$(\forall \mid x \mid < 1), \sqrt{1+x} = 1 + \sum_{k = 1}^\infty \binom{1/2}{k}x^k $$

But note that we have :

$$\sum_{k = n}^{\infty} \binom{1/2}{k}x^k = O_{x \to 0} (x^n)$$

Hence an asymptotic expansion of $x \mapsto \sqrt{1+x}$ at the order $n$ is : $$\sqrt{1+x} = 1 + \sum_{k = 1}^n \binom{1/2}{k}x^k + O(x^n)$$

in your case it's just the asymptotic expansion of order $2$ (so $n = 2$ in the above formula).