Using Differential Transform Method and Padé Approximant for Solving MHD Flow in a Laminar Liquid Film from a Horizontal Stretching Surface

Table 1

Original function	Transformed function
$f(t)=u(t)\pm v(t)$	$F(k)=U(k)\pm V(k)$
$f(t)=\lambda u(t)$	$F(k)=\lambda U(k)$
$f(t)=\frac{d^{n}u(t)}{d{t}^n}$	$F(k)=\frac{(k+n)!}{k!}U(k+n)$
$f(t)=t\frac{du(t)}{dt}$	$F(k)={\displaystyle \sum }_{r=0}^k\delta (r-1)(k-r+1)U(k-r+1))$
$f(t)=t\frac{d^{2}u(t)}{d{t}^2}$	$F(k)={\displaystyle \sum }_{r=0}^k\delta (r-1)(k-r+1)(k-r+2)U(k-r+2)$
$f(t)=\frac{du(t)}{dt}\frac{du(t)}{dt}$	$F(k)={\displaystyle \sum }_{r=0}^k(r+1)(k-r+1)U(r+1)U(k-r+1)$
$f(t)=\frac{d^{2}u(t)}{d{t}^2}\frac{d^{2}u(t)}{d{t}^2}$	$F(k)={\displaystyle \sum }_{r=0}^k\begin{array}{c}(r+1)(r+2)(k-r+2)\\ (k-r+1)U(r+2)U(k-r+2)\end{array}$
$f(t)=u(t)\frac{d^{2}u(t)}{d{t}^2}$	$F(k)={\displaystyle \sum }_{r=0}^k(k-r+2)(k-r+1)U(r)U(k-r+2)$