Two ways of approximating a function?

Question

Consider the potential $$V(x)=\frac{x^3}{3}-l^2x$$ I would like to approximate this potential around a equilibrium position $x_0$. The first way to do it, and the most rigorous one, is to use the formula : $$V(x_0+h)\approx V(x_0)+V'(x_0)h+\frac{1}{2}V''(x_0)h^2$$ with $h=x-x_0$. I find that $$V(x_0+h)\approx \frac{x_0^3}{3}-l^2x_0+x_0h^2$$ But, sometimes, in physics, in order to approximate potential, we approximate directly in the formula that gives the potential. In this case, we have : $$V(x_0+h)=\frac{1}{3}(x_0+h)^3-l^2(x_0+h)=\frac{1}{3}x_0^3(1+\frac{h}{x_0})^3-l^2x_0-l^2h$$ By using the Taylor approximation : $$(1-x)^{\alpha}\approx 1+\alpha x + \frac{\alpha(\alpha-1)}{2!}x^2\Rightarrow V(x_0+h)=\frac{x_0^3}{3}(1-3\frac{h}{x_0}+\frac{6}{2}\frac{h^2}{x_0^2})-l^2x_0-l^2h$$ Hence $$V(x_0+h)=\frac{x_0^3}{3}-x_0^2h+x_0h^2-l^2x_0-l^2h$$ we is not the result that I found using the formula given in the second equation $V(x_0+h)=\frac{x_0^3}{3}-l^2x_0+x_0h^2$ I don't really know why I get this difference. Any help would be appreciated,

Answer 1

By direct computation,

$$V(x_0+h)=\frac{(x_0+h)^3}3-l^2(x_0+h)=V_0(x_0)+x_0^2h-l^2h+x_0h^2+\frac{h^3}3$$

and by Taylor,

$$V(x_0+h)=V(x_0)+(x_0^2-l^2)h+2x_0\frac{h^2}2+2\frac{h^3}{3!}.$$

The two expression coincide.