Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/January 2006

Solution 1a
Assume there exists $$q \in \Pi^n \!\,$$ such that

$$ E(q) < E(p) \!\,$$

Then for $$i=0,1,\ldots,n \!\,$$

$$|f(x_i)-q(x_i) | \leq |f(x_i) - p(x_i) | \!\,$$

Let $$ e(q) = f(x) - q(x) \!\,$$ and $$ e(p)=f(x)-p(x) \!\,$$.

Then $$ e \equiv e( q(x)) - e( p(x) )\!\,$$ takes on the sign of $$ e(p)\!\,$$ since

$$|e ( p(x_i) )| \geq |e (q(x_i) ) | \!\,$$

Since $$e(p) \!\,$$ changes signs $$n+1 \!\,$$ times (by hypothesis), $$e \!\,$$ has $$n+1 \!\,$$ zeros.

However $$e=p-q \!\,$$ and thus can only have at most $$n \!\,$$ zeros. Therefore $$e \equiv 0 \!\,$$ and $$p=q \!\,$$

Solution 1b
First we need to find the roots of $$T_2(x) \!\,$$ in [0,1], which are given by


 * $$x_k = \frac{1}{2} + \frac{1}{2}\cos(\frac{2k-1}{4}\pi) \!\,$$

So our points at which to interpolate are


 * $$x_1 = \frac{1}{2} + \frac{1}{2}\cos(\frac{\pi}{4}) \!\,$$


 * $$x_2 = \frac{1}{2} + \frac{1}{2}\cos(\frac{3\pi}{4}) = \frac{1}{2} - \frac{1}{2}\cos(\frac{\pi}{4}) \!\,$$

Our linear interpolant passes through the points $$(x_1,x_1^2) \!\,$$ and $$(x_2,x_2^2) \!\,$$, which using point-slope form gives the equation


 * $$y-x_1^2 = \frac{x_2^2-x_1^2}{x_2-x_1}(x - x_1) \!\,$$

or


 * $$y = x - \frac{1}{8} \!\,$$

The two problems are equivalent
First notice

$$ A = \begin{pmatrix} Q_1 Q_2 \end{pmatrix} \begin{pmatrix} R \\ 0 \end{pmatrix} = Q_1 R $$

Then we can write

$$ \begin{align} \|b - Ax \| &= \| b - Q_1Rx \| \\ &= \| Q_1^T b - Q_1^T Q_1 Rx \| \\ &= \|  Q_1^T b - Rx \| \end{align} $$

Note that multiplying by orthogonal matrices does not affect the norm.

Then solving $$\min_x \| b - Ax \|^2 \!\,$$ is equivalent to solving $$\min_x \| Q_1^T b - Rx \|^2 \!\,$$, which is equivalent to solving $$ Rx = Q_1^T b \!\,$$. Note that a solution exists and is unique since $$ R \!\,$$ is n-by-n and non-singular.

Show that $$ \rho(x) = \|Q_2^T b \| $$
Similarly

$$ \begin{align} \| b - Ax \| &=  \| b - Q_1Rx \| \\ &= \| Q_2^T b - \underbrace{Q_2^T Q_1}_{0} Rx \| \\ &= \| Q_2^T b \| \end{align} $$

Then

$$ \rho^2(x) = \| b - Ax \|^2 = \| Q_2^T b \|^2 $$, or simply $$ \rho(x) = \| Q_2^T b \| $$, as desired.

Solution 2b
$$ \begin{align} A^TAx &= (Q_1R)^TQ_1Rx \\ &= R^TQ_1^T Q_1Rx \\ &= R^T Rx \\ \\ A^Tb &= (Q_1R)^Tb \\ &= R^T\underbrace{Q_1^T b}_{Rx} \\ &= R^T Rx \end{align} \!\,$$

Solution 3a
Using Gram Schmidit, we have

$$ \begin{align} \alpha_k &= \frac{(Au_k,u_k)}{\|u_k\|^2} \\ \beta_k &= \frac{(Au_k,u_{k-1})}{\|u_{k-1}\|^2} \end{align} $$

Solution 3b
Since
 * $$\gamma_k=\frac{(Au_k,u_{k-2})}{\|u_{k-2}\|^2}$$

, if $$\gamma_k=0\!\,$$, then
 * $$ (Au_k,u_{k-2})=0 \!\,$$

Since $$A\!\,$$ is symmetric,


 * $$(Au_k,u_{k-2})=(u_k,A^Tu_{k-2})=(u_k,Au_{k-2})\!\,$$

From hypothesis, $$ (u_i,u_j) = \begin{cases} 1 & \mbox{if } i=j \\ 0 & \mbox{if } i\neq j \end{cases} $$

Also from hypothesis,

$$ \begin{align} Au_k  &=  u_{k+1}+\alpha_ku_k+\beta_ku_{k-1}+\gamma_ku_{k-2}+\ldots \\ Au_{k-2} &= u_{k-1}+\alpha_{k-2}u_{k-2}+\beta_{k-2}u_{k-3}+\gamma_{k-2}u_{k-4}+\ldots \end{align} $$

Using the above results we have,

$$ \begin{align} (Au_k,u_{k-2}) &= (u_k,Au_{k-2}) \\ &= (u_k,u_{k-1}+\alpha_{k-2}u_{k-2}+\beta_{k-2}u_{k-3}+\gamma_{k-2}u_{k-4}+\ldots)\\ &= (u_k,u_{k-1})+\alpha_{k-2}(u_k,u_{k-2})+\beta_{k-2}(u_k,u_{k-3})+\gamma_{k-2}(u_k,u_{k-4})+\ldots)\\               &= 0 \end{align} $$

Solution 3c
For $$k=0\!\,$$,
 * $$u_1  =Au_0-\alpha_0u_0 \!\,$$

If $$u_1=0\!\,$$, then
 * $$Au_0 =\alpha_0u_0 \!\,$$

Since $$\alpha_0\!\,$$ is a scalar, $$u_0\!\,$$ is an eigenvector of $$A\!\,$$.