Original function

Transformed function

$q\left(x,t\right)=c\cdot u\left(x,t\right)$

$\tilde{q}\left(k_x,k_t\right)=c\cdot u\left(k_x,k_t\right)$

$q\left(x,t\right)={\partial }_{x}u\left(x,t\right)$

$\tilde{q}\left(k_x,k_t\right)=\left(k_x+1\right)\cdot \tilde{u}\left(k_x+1,k_t\right)$

$q\left(x,t\right)={\partial }_{t}u\left(x,t\right)$

$\tilde{q}\left(k_x,k_t\right)=\left(k_t+1\right)\cdot \tilde{u}\left(k_x,k_t+1\right)$

$q\left(x,t\right)=u\left(x,t\right)\cdot v\left(x,t\right)$

$\tilde{q}\left(k_x,k_t\right)={\displaystyle \sum _{r=0}^{k_x}{\displaystyle \sum _{s=0}^{k_t}\tilde{u}\left(r,s\right)\cdot \tilde{v}\left(k_x-r,k_t-s\right)}}$

$q\left(x,t\right)=1/u\left(x,t\right)$

$\begin{array}{l}\tilde{q}\left(0,0\right)=1/\tilde{u}\left(0,0\right)\\ \tilde{q}\left(k_x,k_t\right)=-\frac{1}{\tilde{u}\left(0,0\right)}{\displaystyle \sum _{r=0}^{k_x}{\displaystyle \sum _{\begin{array}{c}s=0\\ s+r>0\end{array}}^{k_t}\tilde{u}\left(r,s\right)\cdot \tilde{q}\left(k_x-r,k_t-s\right)}}\end{array}$