I want to show that the Taylor expansion of $\frac{R\vec{e_1}-\vec{y}}{|| R\vec{e_1}-\vec{y} ||^3}$ at $\vec{y}=0$ is equal to $\frac {\vec{e_1}}{R^2}+\frac{3y_1 \vec{e_1}-\vec{y}}{R^3} + O(y^2)$. I think I should begin with calculating the Taylor expansion of $\frac {\vec{x}}{||\vec{x}||^3}$ at $R\vec{e_1}$ and then set $\vec{x}=R\vec{e_1}-\vec{y}$. I wrote out the equation explicitly and calculated for each term of $\vec{x}$, I understand where the $\frac{-\vec{y}}{R^3}$ comes from, but then I found $\frac{3y_1 \vec{e_1}}{R^4}$ with $R^4$ instead of $R^3$ in the denominator. Can someone show me explicitly how to expand this Taylor polynomial?
2 Answers
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Give R and y units of length and you'll see that your $\frac1{R^4}$ term gives the wrong units. Trace back your calculation to the step where the units stop making sense (every term in the expansion should have the same units as the original expression). That will probably help you find the mistake.
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I would suggest taking a look at this https://www.youtube.com/watch?v=Ijb-UI0gMJM.
You can use the binomial series approximation $(1+x)^n\approx 1+nx+\cdots$ $$\frac{1}{\|R\vec{e}_1-\vec{y}\|^ 3}=\frac{1}{(R^ 2-2R\vec{y}\cdot\vec{e}_1+\|\vec{y}\|^ 2)^{3/2}}\approx\frac{1}{(R^ 2-2R\vec{y}\cdot\vec{e}_1)^{3/2}}=\frac{1}{R^3\left(1-\frac{2\vec{y}\cdot\vec{e}_1}{R}\right)^{3/2}}\approx\frac{1}{R^3}\left(1+\frac{3}{2}\frac{2\vec{y}\cdot\vec{e}_1}{R}\right)=\frac{1}{R^3}+\frac{3\vec{y}\cdot\vec{e}_1}{R^4}.$$ Multiplying this by $R\vec{e}_1-\vec{y}$, up to $\mathcal{O}(\|\vec{y}\|^2)$ we have $$\frac{\vec{e}_1}{R^2}-\frac{\vec{y}}{R^3}+\frac{3\vec{y}\cdot\vec{e}_1}{R^4}R\vec{e}_1=\frac{\vec{e}_1}{R^2}+\frac{3\vec{y}\cdot\vec{e}_1\vec{e}_1-\vec{y}}{R^3}.$$
