Numerical Methods and Interpolation Questions and Answers

Answer: B

The relation is \(I_{E} = I_{S} \lt I_{T}\).

With h = 2 the nodes are 0, 2, 4, where \(x^{2} - 4\) is -4, 0 and 12.

\[ \begin{aligned} I_{E} &= \left[\tfrac{x^{3}}{3} - 4x\right]_{0}^{4} = \tfrac{64}{3} - 16 = \tfrac{16}{3} \\ I_{T} &= \tfrac{2}{2}\,[-4 + 2(0) + 12] = 8 \\ I_{S} &= \tfrac{2}{3}\,[-4 + 4(0) + 12] = \tfrac{16}{3} \end{aligned} \]

Simpson's rule is exact for polynomials up to degree three, so it reproduces the exact value. The trapezoidal rule overestimates the integral of a convex (upward-curving) function, so \(I_{T} = 8\) is larger. The other orderings either break the equality or reverse the trapezoidal error.

Answer: A

The relation that holds is \(c_{1} + \frac{c_{2}}{4} + c_{3} = 6\).

Exactness for degree up to 2 means the rule must integrate \(1\), \(x\) and \(x^{2}\) exactly over \([-1, 1]\), where the true integrals are 2, 0 and \(\frac{2}{3}\). Using \(f(x) = x^{2}\), with \(f(-1) = 1\), \(f(\frac{1}{2}) = \frac{1}{4}\), \(f(1) = 1\):

\[ \begin{aligned} \tfrac{1}{9}\left(c_{1} + \tfrac{c_{2}}{4} + c_{3}\right) &= \tfrac{2}{3} \\ c_{1} + \tfrac{c_{2}}{4} + c_{3} &= 6 \end{aligned} \]

The other two conditions fix the weights: \(f = 1\) gives \(c_{1} + c_{2} + c_{3} = 18\), and \(f = x\) gives \(-c_{1} + \frac{c_{2}}{2} + c_{3} = 0\). Together with the result above they give \(c_{1} = 5\), \(c_{2} = 16\), \(c_{3} = -3\).

Check the rest against these weights: \(c_{1} + \frac{c_{2}}{3} + c_{3} = \frac{22}{3}\), not 4; \(c_{1} + \frac{c_{2}}{2} + c_{3} = 10\), not 2; and \(c_{1} + c_{2} + c_{3} = 18\), not 5.

Answer: A

The correct statement is \(a = -2c,\ d = -2b\).

Substitute the four points into \(p(x) = a + bx + cx^{2} + dx^{3}\):

\[ \begin{aligned} p(0) &= a = 3 \\ p(-1) &= a-b+c-d = 1 \\ p(1) &= a+b+c+d = 2 \\ p(2) &= a+2b+4c+8d = 4 \end{aligned} \]

Adding the second and third equations gives 2a + 2c = 3, so c = -3/2, and indeed a = 3 = -2c. Then b + d = 1/2 and 2b + 8d = 7, giving d = 1 and b = -1/2, so d = -2b.

With a = 3, c = -1.5, b = -0.5 and d = 1, the other options fail: a = 2c would need c = 1.5, b = 3c gives -4.5, and b = 2c gives -3.