Q11Advanced Engineering Mathematics

In numerical analysis, the forward difference operator is denoted by $\Delta$ and the backward difference operator by $\nabla$. A fundamental relationship connecting them is $\nabla y_n=\Delta y_{n-1}$. By applying this property successively to higher-order differences, we can prove the equivalence. Specifically, applying it three times yields: ${\nabla }^3{y}_5={\nabla }^2(\nabla y_5)={\nabla }^2(\Delta y_4)=\nabla (\nabla \Delta y_4)=\nabla ({\Delta }^2{y}_3)=\Delta ({\Delta }^2{y}_2)={\Delta }^3{y}_2$.