Q1Advanced Engineering Mathematics I

We know the relationship between the forward difference ($\Delta$) and backward difference ($\nabla$) operators:

$\nabla =1-E^{-1}=E^{-1}(E-1)=E^{-1}\Delta$

Therefore, ${\nabla }^3=(E^{-1}\Delta {)}^3=E^{-3}{\Delta }^3$.

Applying this to $y_5$: ${\nabla }^3{y}_5=E^{-3}{\Delta }^3{y}_5$

Since $E^{-n}{y}_k=y_{k-n}$, we have: $E^{-3}({\Delta }^3{y}_5)={\Delta }^3{y}_{5-3}={\Delta }^3{y}_2$.

Hence, ${\Delta }^3{y}_2={\nabla }^3{y}_5$ is proven.