Numerical Methods Qualification Exam Problems and Solutions (University of Maryland)/January 2000

Proof of Hint
Claim: There exists $$\eta \in (x_0, x_2)\!\,$$ such that $$f^{\prime\prime}(\eta)-q^{\prime\prime}(\eta)=0 \!\,$$

Proof: The interpolation polynomial may be expressed using dividing difference coefficients i.e.

$$ q(x)=f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)\!\,$$

which implies

$$q^{\prime\prime}(x)=2f[x_0,x_1,x_2] \!\,$$

In general, the divided difference coefficients may be expressed as a factorial weighted point of a derivative of $$f \!\,$$ i.e.

$$(\triangle) \qquad f[x_0, \ldots,x_n]=\frac{f^{(n)}(\xi)}{(n)!} \qquad \xi \in I[x_0,\ldots,x_n]\!\,$$

Hence,

$$f[x_0,x_1,x_2]=\frac{f^{\prime\prime}(\eta)}{2!} \qquad \eta \in [x_0, x_2] \!\,$$

which implies

$$q^{\prime\prime}(\eta)-f^{\prime\prime}(\eta)=0 \!\,$$

Application of Hint
From the hint we know that

$$f^{\prime\prime}(\eta)-q^{\prime\prime}(\eta)=0 \!\,$$

which implies

$$f^{\prime\prime}(x)-q^{\prime\prime}(x)= f^{\prime\prime}(x)-q^{\prime\prime}(x) -f^{\prime\prime}(\eta)+q^{\prime\prime}(\eta)\!\,$$

Since $$ q(x)\!\,$$ is quadratic, $$q^{\prime\prime}(x) \!\,$$ is constant i.e. $$ q^{\prime\prime}(x)=K\!\,$$ for all $$ x\!\,$$

Therefore,

$$f^{\prime\prime}(x)-q^{\prime\prime}(x)= f^{\prime\prime}(x)-q^{\prime\prime}(x) -f^{\prime\prime}(\eta)+q^{\prime\prime}(\eta)=f^{\prime\prime}(x)-f^{\prime\prime}(\eta)\!\,$$

By the fundamental theorem of calculus,

$$ \begin{align} f^{\prime\prime}(x)-f^{\prime\prime}(\eta) &=\int_\eta^x f^{\prime\prime\prime}(t) dt \\ &\leq \|f^{\prime\prime\prime}(x)\|_\infty |x-\eta| \\ &\leq 2h \|f^{\prime\prime\prime}\|_\infty \end{align} $$

Therefore,

$$\max_{x\in [x_0,x_2]} |f^{\prime\prime}(x)-q^{\prime\prime}(x)| \leq \underbrace{ 2\|f^{\prime\prime\prime}(x)\|_\infty}_C h^1 \!\,$$

Third Derivative of f has Zero
We know that $$ f[x_0,x_1,x_2,x]=0 \!\,$$, because $$ p \in P_2 \!\,$$. Now, by $$ (\triangle) \!\,$$ we can conclude that there exists $$ \eta^* \in I[x_0,x_1,x_2,x] \!\,$$ such that $$ f'''(\eta^*)=0 \!\,$$.

Application of Fundamental Theorem of Calculus (Twice)


\begin{align} f(x_1)-f(\eta) &=\int_{\eta}^{x_1}f^{(3)}(x)dx\\ &=\int_{\eta}^{x_1}f^{(3)}(x)-\underbrace{f^{(3)}(\eta^*)}_0dx\\ &=\int_{\eta}^{x_1} \int_{\eta*}^x f^{(4)}(t)dtdx\\ &\leq \| f^{(4)} \|_{\infty}|x-\eta^*||x_1-\eta| \\ &\leq \underbrace{2\| f^{(4)} \|_{\infty}}_{C'}h^2 , \end{align} $$

Apply Gram Schmidt
To find orthogonal $$\{ p_0, p_1, p_2 \} \!\,$$ use the Gram Schmidt method.

Let $$ \{v_0=1,v_1=x,v_2=x^2 \} \!\,$$ be a basis of $$ P_2 \!\,$$.

Calculate p_0
Choose $$ p_0=v_0=1 \!\,$$.

Calculate p_1
From Gram Schmidt, we have

$$ p_1=\underbrace{x}_{v_1}- \frac{(v_1, p_0)}{(p_0\cdot, p_0)} \cdot \underbrace{1}_{p_0} \!\,$$, where

$$ (v_1, p_0)= \int_{0}^{\infty} e^{-x}\cdot 1\cdot x dx= 1 \!\,$$

$$ (p_0, p_0)= \int_{0}^{\infty} e^{-x}\cdot 1\cdot 1 dx= 1 \!\,$$

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

Calculate p_2
Proceeding with Gram Schmidt, we have

$$ p_2=\underbrace{x^2}_{v_2}- \frac{(v_2, p_1)}{(p_1, p_1)} \cdot \underbrace{(x-1)}_{p_1} - \frac{(v_2,p_0)}{(p_0,p_0)} \cdot \underbrace{1}_{p_0} \!\,$$  where

$$ \begin{align} (v_2, p_1) &= \int_{0}^{\infty} e^{-x}\cdot x^2 \cdot (x-1) dx \\ &= \underbrace{\int_{0}^{\infty} e^{-x}\cdot x^3 dx}_{3!} - \underbrace{\int_{0}^{\infty} e^{-x}\cdot x^2 dx}_{2!} \\ &=4 \end{align} $$

$$ \begin{align} (p_1, p_1) &= \int_{0}^{\infty} e^{-x}\cdot (x-1) \cdot (x-1) dx \\ &= \int_{0}^{\infty} e^{-x}\cdot (x^2-2x+1) dx \\ &= \underbrace{\int_{0}^{\infty} e^{-x} x^2 dx}_{2!}-2 \underbrace{\int_{0}^{\infty} e^{-x}\cdot x dx}_{1!}+\underbrace{\int_{0}^{\infty} e^{-x}\cdot 1 dx}_{0!} \\ &= 1 \end{align} $$

$$ (v_2, p_0)= \int_{0}^{\infty} e^{-x}\cdot x^2 \cdot 1 dx= 2!=2 \!\,$$

$$ (p_0, p_0)= \int_{0}^{\infty} e^{-x}\cdot 1\cdot 1 dx= 1 \!\,$$

Therefore

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

Find the Nodes
The nodes $$x_1 \!\,$$ and $$x_2 \!\,$$ are the roots of the $$n\!\,$$th orthogonal polynomial i.e. $$ p_2(x)=x^2-4x+2 \!\,$$

Applying the quadratic formula yields the roots:

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

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

Find the Weights
The approximation is exact for polynomials at most of degree $$2n-1=3\!\,$$. Hence, we have the following system of equations

$$\int_0^\infty e^{-x}dx = 1 = w_1\cdot \underbrace{ 1}_{f(x_1)} + w_2 \cdot \underbrace{1}_{f(x_2)}\!\,$$

$$\int_0^\infty x\cdot e^{-x}dx=1 = w_1\cdot\underbrace{(2-\sqrt{2})}_{f(x_1)}+w_2\cdot \underbrace{(2+\sqrt{2})}_{f(x_2)} \!\,$$

Solving the solving the system of equation by substitution yields the weights:

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

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

Solution 3a
Choose $$x_0 \!\,$$

for $$k=0,1,2,\dots \!\,$$
 * $$x^{(i+1)} = D^{-1}(b-(L+U)x^{(i)}) \!\,$$

end

Where $$A=D+L+U \!\,$$, $$D \!\,$$ is diagonal, $$L \mbox{ and }U \!\,$$ are lower and upper triangular, respectively.

Solution 3b
The $$ n \!\,$$ diagonal entries of $$A \!\,$$ are non-zero since otherwise $$D^{-1} \!\,$$ would not exist.

Therefore $$A \!\,$$ contains $$m-n \!\,$$ off-diagonal non-zero entries.

The computation during each iteration is given by

$$x^{(i+1)} = \underbrace{D^{-1}(b-\underbrace{(L+U)x^{(i)}}_{m-n\mbox{ multiplies}})}_{n \mbox{ multiplies}} \!\,$$

Therefore there are $$(m-n)+n =m \!\,$$ multiplies in each iteration.

Solution 3c
Let $$ E:= D^{-1}(L+U) \,\!$$.

Theorem 8.2.1 [SB] states that $$ \rho( E ) < 1 \,\!$$ if and only if the Jacobi iteration converges.

Matrix multiplication and the definitions of $$D^{-1},L, U \!\,$$ gives the explicit entrywise value of $$ E\!\,$$

$$ e_{ij} = \frac{a_{ij}}{a_{ii}} \,\!$$ for $$ j \ne i \!\, $$ and $$ e_{ii} = 0 \,\!$$

Then, using Gerschgorin's Theorem and diagonal dominance, we have the result.


 * $$ | \lambda - \underbrace{e_{ii}}_0 | < \sum_{\underset{j \ne i}{j = 1}}^n |e_{ij}| = \sum_{\underset{j \ne i}{j = 1}}^n \frac{|a_{ij}|}{|a_{ii}|} < 1 \!\, $$