FINAL EXAM SOLUTIONS

MATH 465

1. For precision three, we need the orthogonal polynomials of degree 0, 1, 2. They are \{1, x, x^{2} − \frac{2}{5}\}. Thus the nodes are ±\sqrt{\frac{2}{5}}. The weights are easily seen to be \frac{4}{3}. Thus the quadrature rule is

Q(f) = \frac{4}{3} (f(−\sqrt{\frac{2}{5}}) + f(\sqrt{\frac{2}{5}}))

2. Let the polynomials be {1, x − a, x^{2} + bx + c}. Then < 1, x − a >= 0 implies that 1(−a) + 2(1 − a) + 5(2 − a) = 0 and hence a = 3/2. Also < 1, x^{2} + bx + c >= 0, < x − 3/2, x^{2} + bx + c >= 0 imply that 12b + 8c = −22, 8b = −18, hence b = −9/4, c = 5/8. Thus \{1, x − \frac{3}{2}, x^{2} − \frac{9}{4}x + \frac{5}{8}\} are the orthogonal polynomials.

3. Let N = 2n and let F_{k} denote the discrete Fourier transform from C^{k} to

C^{k}.

(a) Let

f^{e} = [f_{0}, f_{2}, ⋯, f_{N−2}], f^{o} = [f_{1}, f_{3}, ⋯, f_{N−1}], ω = e^{\frac{2πi}{N}}

Then

(F_{N}f)_{k} = \frac{1}{2} ((F_{n}f^{e})_{k} + ω^{−k}(F_{n}f^{o})_{k}) , 0 ≤ k < n

and

(F_{N}f)_{k+n} = \frac{1}{2} ((F_{n}f^{e})_{k} − ω^{−k}(F_{n}f^{o})_{k}) , 0 ≤ k < n

(b)

F_{4} [\begin{matrix}1 \\ 0 \\ 1 \\ 3 \end{matrix}] = \frac{1}{4} [\begin{matrix}1 & 1 & 1 & 1 \\ 1 & −i & −1 & i \\ 1 & −1 & 1 & −1 \\ 1 & i & −1 & −i \end{matrix}] [\begin{matrix}1 \\ 0 \\ 1 \\ 3 \end{matrix}] = \frac{1}{4} [\begin{matrix}5 \\ 3i \\ −1 \\ −3i \end{matrix}]

(c)

p(x) = \frac{1}{4}(5 + 3ie^{ix} − e^{2ix} − 3e^{3ix})

4. Let

F (X) = [\begin{matrix}x^{2} + y^{2} − 2 \\ x + y \end{matrix}] , X_{0} = [\begin{matrix}1 \\ 0 \end{matrix}]

Then

J = DF (X_{0}) = [\begin{matrix}2 & 0 \\ 1 & 1 \end{matrix}]

and

X_{1} = X_{0} − J^{−1}F (X_{0}) = [\begin{matrix}1 \\ 0 \end{matrix}] − [\begin{matrix}−1/2 \\ 3/2 \end{matrix}] = [\begin{matrix}3/2 \\ −3/2 \end{matrix}]

5. u = [0 − 1 3]′. Then HAH, where H = I − \frac{2}{∥u∥^{2}} uu′ is

[\begin{matrix}1 & 5 & 0 \\ 5 & 48/25 & −14/25 \\ 0 & −14/25 & −48/25 \end{matrix}]

6. a) μ_{k} is the Rayleigh quotient for the power method for a symmetric matrix and hence converges to the eigenvalue of largest magnitude (if there is a unique such eigenvalue). In this case it is −6.

b) ν_{k} is the Rayleigh quotient for the inverse power method for the matrix A − 3I. It will converge to \frac{1}{λ_{j}−3}, where λ_{j} is the closest eigenvalue to 3. So in this case the answer is \frac{1}{−1} = −1.