Interpolation Techniques Questions and Answers

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.

Answer: D

\(\Delta^{5}u_{0} =\) -61.

Expand \(\Delta^{5} = (E - 1)^{5}\) with binomial coefficients:

\[ \begin{aligned} \Delta^{5}u_{0} &= u_{5} - 5u_{4} + 10u_{3} \\ &\quad - 10u_{2} + 5u_{1} - u_{0} \\ &= 0 - 35 + 30 - 80 + 25 - 1 \\ &= -61 \end{aligned} \]

A difference table gives the same result. Option B, \(-60\), is what you get by dropping the final \(-u_{0}\), and 60 and 24 come from sign errors in the alternating sum.