Stuck at a step in proving $\Delta + \nabla =\dfrac{\Delta}{\nabla} - \dfrac{\nabla}{\Delta}$

Question

$\Delta + \nabla =\dfrac{\Delta}{\nabla} - \dfrac{\nabla}{\Delta}$

L.H.S $= (E-1)+(1-E^{-1}) = E-E^{-1}$

R.H.S $= \dfrac{\Delta}{\nabla}-\dfrac{\nabla}{\Delta}=\dfrac{(E-1)}{(1-E^{-1})}-\dfrac{(1-E^{-1})}{(E-1)}=\dfrac{(E-1)^2-(1-E^{-1})^2}{(1-E^{-1})(E-1)}$

=$\dfrac{E^2+E^{-2}-2E+2E^{-1}}{(1-E^{-1})(E-1)}$

I am not able to simplify numerator.Someone please tell me an easy way to do this.

$\Delta =$ Forward difference operator

$\nabla =$ Backward difference operator

$E=$ Shift operator

Answer 1

Note that if you have $a^2-b^2$ it can be written as $(a+b)(a-b)$.

So we have:

$$\begin{align} \frac{(E-1)^2-(1-E^{-1})^2}{(1-E^{-1})(E-1)} &=\frac{(E-1+1-E^{-1})(E-1-1+E^{-1})}{E+E^{-1}-2} \\ \\ &=\frac{(E-E^{-1})(E+E^{-1}-2)}{E+E^{-1}-2} \\ \\ &=E-E^{-1} \end{align}$$

As was to be shown.