User:Jefromi/Sandbox

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Let a,b \in S+T. Thus we know a = xa + ya and b = xb + yb for some x_a,x_b \in S and y_a, y_b \in T. Let 0 \le t\le 1. Since S and T are convex, we know that x_a + (x_b - x_a)t \in S and y_a + (y_b - y_a)t \in T. Then by definition of S+T, (x_a + (x_b - x_a)t) + y_a + (y_b - y_a)t \in S+T. By simple rearrangement, a + (b-a)t \in S+T.



\mbox{Let }u(\mathbf{x})=v(r)\mbox{, with }r=|x|.

u_{x_i} \equiv \frac{\partial u}{\partial x_i}
=v'(r)r_{x_i}

\begin{matrix} u_{x_i} &\equiv& \frac{\partial u}{\partial x_i} \\ &=& v'(r)r_{x_i} \end{matrix}

Now apply the product rule:

\begin{matrix} u_{x_i x_i} &=& v''(r)r_{x_i}^2+v'(r)r_{x_i x_i}\\ \end{matrix}

Now sum to get the laplacian:

\begin{matrix} \nabla^2 u &=& \sum\limits_{i=1}^n (v''(r)r_{x_i}^2 + v'(r)r_{x_i x_i})\\ &=& v''(r) \sum\limits_{i=1}^n r_{x_i}^2 + v'(r) \sum\limits_{i=1}^n r_{x_i x_i}\\ &=& v''(r) |\nabla r|^2 + v'(r) \nabla^2 r \end{matrix}

Now we examine further:

\begin{matrix} r_{x_i} &=& \frac{\partial}{\partial x_i} \sqrt{x_1^2+x_2^2+\ldots+x_n^2}) \\ &=& \frac{1}{2} (2x_i) (x_1^2+x_2^2+\ldots+x_n^2)^{-\frac{1}{2}} \\ &=& \frac{x_i}{r} \end{matrix}