There is a lemma in my textbook that says that $\Delta E(x) = 0$, whereas $E(x)$ is the Newton Potential. It also says that $\nabla E(x) = \frac{1}{w_n}\cdot \frac{x}{||x||^n}$, and this is easily provable by just differentiating the formula for the Newton Potential, which is
$$E(x) = \frac{1}{(2-n)\,w_n\,||x||^{n-2}}$$
for $n$ not equal to $2$ or lower.
The definition of the Laplace is differentiating $E$ a second time, but doing that doesn't give me $0$. I don't understand why the second derivative of $E$ is supposed to be $0$. Any explanation for what I'm doing wrong in the differentiation would be much appreciated. What am I missing here?
I know $E$ is named something else in other texts, and the $w_n$ is supposed to be $2$ for $n=1$, $2 \pi$ for $n = 2$, etc.
Any help would be much appreciated!
Note the Newtonian potential is only harmonic away from 0. For $x \neq 0$ \begin{align*} D_i \vert x \vert^{2-n} &= (2-n) \vert x \vert^{1-n} \cdot \frac{x_i}{\vert x\vert} \\ &= (2-n) x_i \vert x \vert^{-n} \end{align*} and \begin{align*} D_{ii} \vert x \vert^{2-n} &= (2-n) \vert x \vert^{-n} - n(2-n) x_i \vert x \vert^{-n-1} \cdot \frac{x_i}{\vert x \vert} \\ &= (2-n) \vert x \vert^{-n} - n(2-n) x_i^2 \vert x \vert^{-n-2}. \end{align*} The Laplacian is not the "second derivative" of $E$ as you put it, rather it is the sum of pure second derivatives of $E$: \begin{align*} \Delta E(x) &= \frac 1 {(2-n) \omega_n}\sum_{i=1}^n D_{ii} \vert x \vert^{2-n} \\ &= \frac 1 { \omega_n}\sum_{i=1}^n \vert x \vert^{-n} - \frac n { \omega_n} \vert x \vert^{-n-2} \sum_{i=1}^n x_i^2 \\ &= \frac n {\omega_n} \vert x \vert^{-n} - \frac n {\omega_n} \vert x \vert^{-n} = 0. \end{align*}