Fall 2009 Final

View the exam on docuum

Coming soon.

1Question 1¶

1.1Solution¶

1.1.1Part (a)¶

Exactly one root in interval

see htsefp

1.1.2Part (b)¶

Apply bisection method

just apply it

1.1.3Part (c)¶

number of steps req'd to ensure certain precision

$$n = \log_2\left ( \frac{b-a}{\text{max error}} \right )$$

1.1.4Part (d)¶

Rank iterative methods based on order of convergence

Compute $g'$ and $g''$ I guess.

(i) Both derivs are non-zero. Quadratic
(ii) $g'$ is 0. Linear
(iii) $g'$ is non-zero, $g''$ is non-zero. Quadratic

1.2Accuracy and discussion¶

2Question 2¶

2.1Solution¶

2.1.1Part (a)¶

Define lagrange polys

see htsefp

2.1.2Part (b)¶

Show uniqueness

see htsefp

2.1.3Part (c)¶

Find specific poly

see htsefp

2.1.4Part (d)¶

Find error bound

see htsefp

2.2Accuracy and discussion¶

3Question 3¶

3.1Solution¶

3.1.1Part (a)¶

Lagrange vs Hermite, clamped vs natural cubic spline

see htsefp

3.1.2Part (b)¶

Constants for natural cubic spline

see htsefp

3.1.3Part (c)¶

Bezier

I don't understand the question and I won't respond to it

3.2Accuracy and discussion¶

4Question 4¶

4.1Solution¶

4.1.1Part (a)¶

Show that some finite-diff formula holds for the second deriv

Use Taylor

4.1.2Part (b)¶

Apply the above formula to make some approximation

Apply the formula

4.1.3Part (c)¶

Find the error bound.

I think the error bound is given by

$$\frac{h^2}{12}M + \frac{4\delta}{h^2}$$

Determine the value of $h$ to minimise the error bound.

The function we want to minimise is

$$\frac{h^4M + 48\delta}{12h^2}$$

Plug in $\delta = 1\times 10^{-16}$ and $M = 3$:

$$\frac{3h^4 + 48\times 10^{-16}}{12h^2} = \frac{h^2}{4} + \frac{4 \times 10^{-16}}{h^2} \tag{not sure why I switched the format back}$$

Take the first derivative and set that equal to 0:

$$\frac{2h}{4} - \frac{8 \times 10^{-16}}{h^3} = \frac{2h^4 - 32\times 10^{-16}}{4h^3} = 0$$

Thus $2h^4 = 32\times 10^{-16}$. Solving for $h$, we get:

$$h = \sqrt[4]{\frac{32 \times 10^{-16}}{2}} = \sqrt[4]{16\times 10^{16}} = \sqrt{4 \times 10^{-8}} = 2\times 10^{-4} \; \blacksquare$$

4.2Accuracy and discussion¶

5Question 5¶

5.1Solution¶

Quadrature

5.1.1Part (a)¶

Define deg of acc for quad

see htsefp

5.1.2Part (b)¶

Find deg of acc

see htsefp

5.1.3Part (c)¶

Find some constant $k$

see htsefp

5.1.4Part (d)¶

Evaluate $I_h(f)$

see htsefp

5.2Accuracy and discussion¶

6Question 6¶

6.1Solution¶

Trapezoidal rule

6.1.1Part (a)¶

Apply composite trapezoidal rule

When $h=0.5$: the first trapezoid goes from $x=1$ to $x=1.5$. $f(1) = 1$, $f(1.5) = 2/3$. Area of the trapezoid is $0.5 * (5/6) = 5/12$. The second trapezoid goes from $x=1.5$ to $x=2$. $f(1.5 = 2/3$, $f(2) = 0.5$. Area of the trapezoid is $0.5 * 7/12 = 7/24$. The total area is $5/12 + 7/24 = 17/24 \approx $0.7083$. Not too bad.

When $h=0.25$: $f(1) = 1$, $f(1.25) = 4/5$. Area is $9/10 * 0.25 = 9/40$. Second: $f(1.25) = 4/5$, $f(1.5) = 2/3$. Area is $11/15 * 0.25 = 11/60$. Third: $f(1.5) = 2/3$, $f(1.75) = 4/7$. Area is $13/21 * 0.25 = 13/84$. Last: $f(1.75) = 4/7$, $f(2) = 1/2$. Area is $15/28 * 0.25 = 15/112$. Total area $\approx 0.69702$. Even better.

6.1.2Part (b)¶

Derive an error bound

Thought this was too hard but it's actually really easy. The error bound formula for the trapezoidal rule is just

$$\frac{(b-a)^3}{12} f''(\xi)$$

For each segment, replace $b$ by $a + h$, then just sum it up. Pretty cool.

Source

6.1.3Part (c)¶

Apply one step of Richardson extrap

whatever

6.1.4Part (d)¶

Compute relative error

ez

6.2Accuracy and discussion¶

7Question 7¶

Runge-Kutta

7.1Solution¶

7.1.1Part (a)¶

Define local truncation error, find order of method

basically the same as 2008 q6

7.1.2Part (b) (i)¶

Prove some identity for $w_{i+1}$

basically the same as 2008 q6

7.1.3Part (b) (ii)¶

Conditions on $h$ to get the limit of $w_i$ to be 0 at $\infty$?

basically the same as 2008 q6

7.2Accuracy and discussion¶