Let $x=(x_1,x_2,...,x_n)$ be a point/vector on a ball with radius $R$, and $y=(y_1,y_2,...,y_n)$ be a point/vector inside the ball. Let $|x|$ denote the point $x$'s distance to the origin (or the euclidean module of vector $x$), which is $|x|=\sqrt {x_1^2 + x_2^2 + ... + x_n^2}$. Now why the following equation holds? It seems it cannot be proved through hard calculation (but I am not sure. I have been doing this for an hour and have not got any result).
To show this equation indeed works, suppose $R=1,x=(1,0),y=(1/2,1/2)$, then we have
then the left side of the equation is $\frac{{\frac{1}{2}}}{{\frac{1}{{{{(\sqrt 2 )}^n}}}}}$ and the right side of the equation is $\frac{{\frac{1}{2}}}{{\frac{1}{{{{(\sqrt 2 )}^n}}}}} - {(\sqrt 2 )^{n - 2}}\frac{0}{1}$, which is equal to the left side.
I am looking forward to an explanation on this equation. Thank you!
I'm pretty sure there are some geometric significance to this (looks like some sort of decomposition), but I don't see it immediately. Instead, I can algebra bash. Let us start with the second term on the right hand side:
$$\left( \frac{R}{|y|} \right)^{n-2}\frac{x \left(x-\frac{R^2}{|y|^2}y\right)}{\left|x-\frac{R^2}{|y|^2 }y\right|^n}$$
This is equivalent to
$$\left( \frac{R^{n-2}}{|y|^{n-2}} \right)\frac{\frac{x}{|y|^2} \left(x|y|^2-R^2y\right)}{\frac{1}{|y|^{2n}}\left|x|y|^2-R^2y\right|^n}$$
Simplifying the powers of $y$ gives
$$xR^{n-2}|y|^n\cdot\frac{x|y|^2-R^2y}{|x|y|^2-R^2y|^n}$$
Presumably you have gotten this far by yourself. Here comes the fun part: what exactly is $|x|y|^2-R^2y|$?
It is, in terms of components, the square root of
$$\begin{align} \displaystyle{\sum_{i=1}^n}(|y|^2x_i-R^2y_i)^2 &=\sum_{i=1}^n|y|^4x_i^2-2|y|^2R^2\sum_{i=1}^nx_iy_i+R^4\sum_{i=1}^ny_i\\ &=|y|^4|x|^2-2|y|^2R^2 \langle x,y\rangle + R^4 |y|^2\\ \end{align} $$
Now we use the fact that $|x|^2=R^2$ to factorise the expression above as
$$|y|^2R^2(|y|^2-2\langle x,y \rangle + |x|^2)=|y|^2R^2|x-y|^2$$
Therefore the entire right hand side of the original equation can be written as
$$\begin{align} \frac{x(x-y)}{|x-y|^n}-\frac{xR^{n-2}|y|^n\cdot(x|y|^2-R^2y)}{(R|y|\cdot|x-y|)^n}=\frac{x(x-y)}{|x-y|^n}-\frac{\frac{x}{R^2}(x|y|^2-R^2y)}{|x-y|^n} \end{align}$$
Assuming $x(x-y)$ means the dot product of $x$ and $x-y$ and similarly for the second expression, the answer should come right out, using once again the fact that $|x|=R$