Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/August 2004

Solution 1a
The iteration can be represented in matrix formulation as follows:

$$ \begin{align} x_{n+1} &= x_n +y_n \\ y_{n+1} &= x_{n+1}+x_n = 2x_n + y_n \end{align} \!\,$$

which can be written as

$$ \underbrace{ \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} }_A \begin{bmatrix} x_n \\ y_n \end{bmatrix} = \begin{bmatrix} x_{n+1} \\ y_{n+1} \end{bmatrix} \!\,$$

Thus the iteration is just the power method since each step is represented by a multiplication by the matrix $$A \!\,$$.

The power method converges to the eigenvector of the largest eigenvalue.

The eigenvalues of $$A \!\,$$ are computed to be $$1 \pm \sqrt{2} \!\,$$. Hence the largest eigenvalue is $$1 + \sqrt{2} \!\,$$

The corresponding eigenvector is then

$$ \begin{bmatrix} \sqrt{2} \\ 2 \end{bmatrix} \!\,$$

Then $$y_n / x_n \rightarrow \sqrt{2} \!\,$$ as desired.

Solution 1b
Since convergence is linear, 7 steps is required to achieve the error bound.

Solution 2a
First notice that $$p_n - x p_{n-1} \in \Pi^{n-1} \!\,$$ and therefore we can express it as a linear combination of the monic polynomials of degree $$n-1 \!\,$$ or less i.e.

$$p_n - x p_{n-1}= -a_n p_{n-1}-b_n p_{n-2} + \sum_{i=0}^{n-3} \alpha_i p_i \qquad (*)\!\,$$

Taking the inner product of both side of $$(*) \!\,$$ with $$p_{n-1} \!\,$$ yields from the orthogonality of the polynomials:

$$-\langle x p_{n-1}, p_{n-1} \rangle = -a_n \langle p_{n-1}, p_{n-1} \rangle \!\,$$

Rearranging terms then yields

$$a_n = \frac{\langle xp_{n-1}, p_{n-1} \rangle}{ \langle p_{n-1}, p_{n-1} \rangle} \!\,$$

Similarly, taking the inner product of both side of $$(*) \!\,$$ with $$p_{n-2} \!\,$$ yields from the orthogonality of the polynomials:

$$b_n = \frac{\langle xp_{n-1}, p_{n-2} \rangle}{ \langle p_{n-2}, p_{n-2} \rangle} \!\,$$

Notice that

$$ \begin{align} \langle xp_{n-1}, p_{n-2} \rangle &= \langle p_{n-1}, x p_{n-2} \rangle \\ &= \langle p_{n-1}, p_{n-1}+ \sum_{i=0}^{n-2} \alpha_i p_i \rangle \\ &= \langle p_{n-1}, p_{n-1} \rangle \end{align} \!\,$$

Therefore,

$$b_n = \frac{\langle p_{n-1}, p_{n-1} \rangle}{ \langle p_{n-2}, p_{n-2} \rangle} \!\,$$

Finally, taking inner product of both side of $$(*) \!\,$$ with $$ p_k, k=0,\ldots,n-3\!\,$$ yields,

$$\alpha_k = - \frac{ \langle xp_{n-1}, p_k \rangle }{ \langle p_k, p_k \rangle} \!\,$$

Notice that

$$ \langle xp_{n-1},p_k \rangle = \langle p_{n-1},\underbrace{x p_k}_{\in \Pi^{n-2}} \rangle = 0 \!\,$$

which implies $$\alpha_k=0 \!\,$$ for $$k=0,1\ldots, n-3 \!\,$$

Solution 3a
Using Gram Schmidt with inner product defined as

$$\langle f,g \rangle = \int_0^\infty w(x) fg dx \!\,$$

and power basis $$1,x,x^2 \!\,$$ as starting vectors, we get

$$p_0=1 \!\,$$

$$p_1=x-1 \!\,$$

$$p_2=x^2-4x+2 \!\,$$

Solution 3b
Using test functions $$f(x)=1 \!\,$$ and $$f(x)=x \!\,$$ and using the roots of $$p_2 \!\,$$ as nodes we find

$$x_1 = 2 + \sqrt{2} \!\,$$

$$x_2 = 2 - \sqrt{2} \!\,$$

$$w_1 = \frac{2-\sqrt{2}}{4} \!\,$$

$$w_2 = \frac{2+\sqrt{2}}{4} \!\,$$