== Basic Calculations *[[http://en.wikipedia.org/wiki/Vector_calculus_identities|Vector Calculus Identities]] **Including outer product about gradient operator and vectors *[[http://integral-table.com/|Integral Table]] *Integration by parts: Suppose $A$ is symmetric, we have $$\int_a^b (AU_x, U)dx = \frac{1}{2}\Big( (AU, U)|_a^b - \int_a^b (A_xU, U)dx\Big).$$ *Gaussian Functions: Suppose $A$ is $n\times n$ positive definite symmetric matrix, we have $$\int_{\mathbb{R}^n} e^{-x^t A x}\,dx = \frac{(\pi)^{n/2}}{\sqrt{det(A)}}.$$ *Gradient of quadratic form: let $f(x)=x^tAx$ with the convection $x=(x_1,\cdots,x_n)^t$, then the gradient $\grad f=(\f{\p f}{\p x_1},\cdots,\f{\p f}{\p x_n})^t$ is: $$\grad f=Ax+A^tx.$$ *Linear transformation in coordinates: let $x' = Ax$, then $dx' = Adx$, and we also have $$\begin{split} \inner{\f{\p}{\p x'}}{dx'} = I = \inner{ \f{\p}{\p x} }{ dx } ~\\ \inner{\f{\p}{\p x'}}{Adx} = \inner{ \f{\p}{\p x} }{ dx } ~\\ \inner{A^t\f{\p}{\p x'}}{dx} = \inner{ \f{\p}{\p x} }{ dx } ~\\ \end{split}$$ which shows that $$\f{\p}{\p x} = A^t \f{\p}{\p x'}.$$ In particular, if $A$ satisfies that $AA^t=I$ (i.e. $A$ is orthogonal matrix), then we will have $\Delta_{x'} = \Delta_{x}$.