Numerical Methods and Interpolation Questions and Answers

Answer: A

Interpolation is a method of constructing new data points within the range of a discrete set of known data points. Isolation means keeping something apart, integration finds the area under a curve or sums small parts, and a limit is the value that a function approaches, so none of these three estimates values between known points. An example is estimating the temperature at noon from readings taken at 11 am and 1 pm.

Answer: A

Both statements are true. Interpolation is a statistical method by which related known values are used to estimate an unknown value or set of values. In the Inverse Distance Weighted method the cell values are estimated by averaging the values of sample data points in the neighbourhood of each processing cell, with nearer points given more weight. So the first option is correct. In IDW a point close to the cell has more influence on its value than a point far away.

Answer: C

The condition is \(f(x_{1})f(x_{2}) \lt 0\).

A continuous monotonic function with exactly one root strictly between \(x_{1}\) and \(x_{2}\) crosses zero once there, so it has one sign at \(x_{1}\) and the opposite sign at \(x_{2}\). The product of two numbers of opposite sign is negative. This is the condition used to bracket a root in the bisection and false-position methods.

A positive product would mean the same sign at both ends, so a monotonic function would have no root between them. A zero product would put the root at an end point, not inside the interval. \(f(x_{1}) = f(x_{2})\) is impossible for a strictly monotonic function on distinct points.

Answer: C

The root is 0.658.

Test each option in \(f(x) = x^{4} + 2x^{3} - 4x^{2} + 3x - 1\). At \(x = 0.658\): \(x^{2} = 0.4330\), \(x^{3} = 0.2849\), \(x^{4} = 0.1875\), so

\[ \begin{aligned} f &= 0.1875 + 0.5698 - 1.7319 \\ &\quad + 1.974 - 1 \approx -0.0006 \end{aligned} \]

which is zero to the accuracy given; the root lies at about 0.6585, since \(f(0.66) \approx +0.002\). At \(x = 0.646\), \(f \approx -0.018\), thirty times further from zero, so it is not the root. At \(x = 0.333\), \(f \approx -0.36\), and at \(x = 1\), \(f = 1 + 2 - 4 + 3 - 1 = 1\), so neither of those is a solution either.

Answer: A

The diagonal elements are 2 and \(-1\).

The coefficient matrix has rows (2, 4) and (3, 5). Without pivoting the first row stays as the pivot row, and the second row has \(3/2\) times the first subtracted from it to clear the \(x\) term:

\[ \begin{aligned} m &= \frac{3}{2} \\ a_{22} &= 5 - \frac{3}{2}(4) = 5 - 6 = -1 \end{aligned} \]

So the upper triangular matrix has rows (2, 4) and (0, \(-1\)), and its diagonal is 2 and \(-1\).

The words 'without pivoting' matter: partial pivoting would swap the rows to put the larger 3 on top, and the diagonal would then be 3 and \(4 - \frac{2}{3}(5) = \frac{2}{3}\), which is not offered either. The pairs 3 and 2, 5 and \(-2\), and 3 and 4 do not arise from eliminating this system; 3 and 4 simply mixes entries of the original matrix.

Answer: A

The difference is 0.

The three points (0, 3), (0.5, 4) and (1, 5) lie on the straight line y = 3 + 2x. Both rules are exact for straight lines:

\[ \begin{aligned} \text{Trapezoidal} &= \tfrac{0.5}{2}[3 + 2(4) + 5] = 4 \\ \text{Simpson} &= \tfrac{0.5}{3}[3 + 4(4) + 5] = 4 \end{aligned} \]

Both give 4, the exact area, so they agree. Simpson's rule differs from the trapezoidal rule only when the data have curvature; 0.25, 0.5 and 1 would require a curve through the points.

Answer: D

The percentage error is 0.

Simpson's 1/3 rule fits a parabola through each pair of subintervals, and its error term involves the fourth derivative of the integrand. For a polynomial of degree three or less the fourth derivative is zero, so the rule is exact, whatever the number of subintervals.

The integrand \(4x^{2} + 2x + 6\) is a quadratic, so \(I_{s} = I_{e}\) and \(e = 0\). The exact value is

\[ \begin{aligned} I_{e} &= \left[\tfrac{4x^{3}}{3} + x^{2} + 6x\right]_{1}^{2} \\ &= \tfrac{28}{3} + 3 + 6 = \tfrac{55}{3} \end{aligned} \]

and Simpson's rule returns the same 55/3. The non-zero options would describe a rule that is not exact for quadratics, such as the trapezoidal rule.

Answer: D

The value of h is 2.

The scheme is the trapezoidal (Crank-Nicolson) rule. Solve it for \(y_{n+1}\):

\[ \begin{aligned} y_{n+1}\left(1 + \frac{\Delta t}{2}\right) &= y_{n}\left(1 - \frac{\Delta t}{2}\right) \\ y_{n+1} &= \frac{1 - \Delta t/2}{1 + \Delta t/2}\,y_{n} \end{aligned} \]

The exact solution \(e^{-t}\) decays without changing sign. The amplification factor is always smaller than 1 in size, so the scheme is stable, but it becomes negative when \(1 - \Delta t/2 \lt 0\), that is for \(\Delta t \gt 2\). The numerical solution then alternates in sign from step to step, a non-physical oscillation. So \(h = 2\); 1/2, 1 and 3/2 all still give a positive factor.

Answer: A

Newton-Raphson is based on the Taylor series.

Expand \(f\) about the current estimate and keep the linear term: \(f(x_{n} + h) \approx f(x_{n}) + hf'(x_{n}) = 0\), so \(h = -f(x_{n})/f'(x_{n})\) and \(x_{n+1} = x_{n} - f(x_{n})/f'(x_{n})\).

Fourier series represent periodic functions by sines and cosines, Laurent series expand complex functions about singularities, and 'power series' is a general class rather than the specific truncated expansion the method uses.

Answer: A

The difference \(m - n\) is 0.074.

Exactly, \(n = \ln 5 = 1.6094\). With four intervals, \(h = 1\) and the ordinates at \(x = 0, 1, 2, 3, 4\) are \(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\):

\[ \begin{aligned} m &= \tfrac{1}{2}\left(1 + \tfrac{1}{5}\right) + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} \\ &= 0.6 + 1.0833 = 1.6833 \end{aligned} \]

So \(m - n = 1.6833 - 1.6094 = 0.074\). The sign is positive because \(1/(1+x)\) is convex, and chords lie above a convex curve, so the trapezoidal rule overestimates; -0.074 has the wrong sign, and 0.003 or -0.003 is far smaller than the trapezoidal error with only four intervals.