<s>[INST] How prove this equation(18)? my book have this:let $f(x)=\sqrt{nx},n\in N,0<x<1$, then use Taylor we have 
$$f(x)=1+f'(x_{0})(x-x_{0})-\dfrac{g(x)}{2}$$
where $x_{0}=\dfrac{1}{n},g(x)=(\sqrt{nx}-1)^2$
my question: $g(x)=(\sqrt{nx}-1)^2$? why?
 [/INST] $$f(x)=\sqrt{nx}, \quad f'(x)=\frac{1}{2}\cdot\frac{1}{\sqrt{nx}}\cdot n = \frac{n}{2\sqrt{nx}}$$
$$x_0=1/n, \quad f(x_0)=\sqrt{n\cdot(1/n)}=1, \quad f'(x_0)=\frac{n}{2\sqrt{n\cdot(1/n)}}=\frac{n}{2}$$
$$\sqrt{nx} = 1 + \frac{n}{2} \cdot (x-1/n) - \frac{g(x)}{2}$$
$$2\sqrt{nx} = 2 + (nx-1) - g(x)$$
$$g(x) = 2 + (nx-1) - 2\sqrt{nx} = nx - 2 \sqrt{nx} - 1 = (\sqrt{nx}-1)^2$$
</s>