Q2Advanced Engineering Mathematics I

Question

Prove that, $(\frac{Δ ^{2}}{E}) x^{3} = 6 x$ (if $h = 1$ ).

Answer 1

We need to prove: $(\frac{Δ ^{2}}{E}) x^{3} = 6 x$

We know the relationship between operators: $E = 1 + Δ$ , and $E^{- 1}$ is the backward shift operator. Thus, $\frac{Δ ^{2}}{E} f (x) = Δ^{2} E^{- 1} f (x)$ .

Let $f (x) = x^{3}$ . Then $E^{- 1} x^{3} = (x - 1)^{3}$ (since $h = 1$ ).

Now apply $Δ$ twice: $Δ (x - 1)^{3} = (x)^{3} - (x - 1)^{3} = x^{3} - (x^{3} - 3 x^{2} + 3 x - 1) = 3 x^{2} - 3 x + 1$

$Δ^{2} (x - 1)^{3} = Δ (3 x^{2} - 3 x + 1)$ $= [3 (x + 1)^{2} - 3 (x + 1) + 1] - [3 x^{2} - 3 x + 1]$ $= [3 (x^{2} + 2 x + 1) - 3 x - 3 + 1] - 3 x^{2} + 3 x - 1$ $= 3 x^{2} + 6 x + 3 - 3 x - 2 - 3 x^{2} + 3 x - 1$ $= 6 x$ . Hence Proved.

Q2Advanced Engineering Mathematics I

Question

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