# How to expand this equation? $H_{1}=\frac{e^{2}}{R}+\frac{e^{2}}{R+x_{1}+x_{2}}-\frac{e^{2}}{R+x_{1}}-\frac{e^{2}}{R+x_{2}}$ [closed]

$$H_{1}=\frac{e^{2}}{R}+\frac{e^{2}}{R+x_{1}+x_{2}}-\frac{e^{2}}{R+x_{1}}-\frac{e^{2}}{R+x_{2}}$$ in the approximation $$\left |x_{1}\right |,\left |x_{2}\right |\ll R$$ we expand to obtain in lowest order $$H_{1} \simeq \frac{2e^{2}x_{1}x_{2}}{R^{3}}$$

Introduction to Solid State Physics,Charles Kittel, 8va edition,pag 55

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## 1 Answer

Use the approximation that $$1/(1-x)\simeq1+x+x^2$$ for small $$x$$. To get you started, I'll do the first two terms. Let us define $$H' =\frac{e^2}{R}+\frac{e^2}{R+(x_1-x_2)}.$$ Now define $$z\equiv(x_2-x_1)/R$$ and we have that $$|z|\ll1$$. We find then that \begin{align} H' &\equiv\frac{e^2}{R}\bigg(1+\frac{1}{1-z}\bigg) \\ &\simeq \frac{e^2}{R}\bigg(1+1+z+z^2+\cdots\bigg). \end{align} Try doing the same for the remaining two terms. You should find that expanding to first order is insufficient due to cancellations, therefore a second order expansion is required. Let me know if you get stuck.