<s>[INST] Why is this approximation correct? I came across an interesting approximation today in an indian olympiad paper (page 6). It essentially boils down to:
$$V(x)=\frac{k}{\sqrt{x^2+4/3}}+\frac{k}{\sqrt{(x+1)^2+1/3}}+\frac{k}{\sqrt{(x-1)^2+1/3}}$$
$$\approx k\sqrt{\frac 34}(3+\frac {9}{16}x^2)$$
Even before I got down to attempt to solve this step, I realized a problem. expression 1 decreases with $x$, since $x$ is in the denominator of all three terms in it. But expression 2 increases with $x$, since $x$ is in the numerator. But both are supposed to be approximately equal. How do we justify this?
 [/INST] Consider $$y(a,b)=\frac{1}{\sqrt{(x+a)^2+b}}$$ Using Taylor series around $x=0$, you get 
$$y(a,b)=\frac{1}{\sqrt{a^2+b}}-\frac{a x}{\left(a^2+b\right)^{3/2}}+\frac{x^2 \left(2
   a^2-b\right)}{2 \left(a^2+b\right)^{5/2}}+\frac{x^3 \left(3 a b-2 a^3\right)}{2
   \left(a^2+b\right)^{7/2}}+O\left(x^4\right)$$ Using this for the complete expression with the proper values of $a,b$
$$V(x)=\frac{ 3\sqrt 3}2k\left(1+\frac{3 x^2}{16}+O\left(x^4\right) \right)$$
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