Discussion
The value of \(\left(\dfrac{\Delta^{2}}{E}\right) x^{2}\) (taking \(h = 1\)) is :
where :
\(\Delta\) - forward difference operator
E - shift operator.
Answer: D
The value is 2.
The shift operator acts by \(E f(x) = f(x + h)\), so \(E^{-1}f(x) = f(x - 1)\) with \(h = 1\), and \(\Delta\) commutes with E. Hence
\[ \frac{\Delta^{2}}{E}x^{2} = \Delta^{2}(x - 1)^{2} \]
The second forward difference of any quadratic \(ax^{2} + bx + c\) with step 1 is the constant \(2a\). Here \(a = 1\), so the result is 2, whatever the shift. Directly: \(\Delta x^{2} = 2x + 1\) and \(\Delta(2x + 1) = 2\).
0 would be the third difference of a quadratic, and \(\pm 1\) come from forgetting the factor 2.
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