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*1*Question 1¶

*1.1*Solution¶

*1.1.1*Part (a)¶

Exactly one root in interval

see htsefp

*1.1.2*Part (b)¶

Apply bisection method

just apply it

*1.1.3*Part (c)¶

number of steps req'd to ensure certain precision

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

*1.1.4*Part (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

*1.2*Accuracy and discussion¶

*2*Question 2¶

*2.1*Solution¶

*2.1.1*Part (a)¶

Define lagrange polys

see htsefp

*2.1.2*Part (b)¶

Show uniqueness

see htsefp

*2.1.3*Part (c)¶

Find specific poly

see htsefp

*2.1.4*Part (d)¶

Find error bound

see htsefp

*2.2*Accuracy and discussion¶

*3*Question 3¶

*3.1*Solution¶

*3.1.1*Part (a)¶

Lagrange vs Hermite, clamped vs natural cubic spline

see htsefp

*3.1.2*Part (b)¶

Constants for natural cubic spline

see htsefp

*3.1.3*Part (c)¶

Bezier

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

*3.2*Accuracy and discussion¶

*4*Question 4¶

*4.1*Solution¶

*4.1.1*Part (a)¶

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

Use Taylor

*4.1.2*Part (b)¶

Apply the above formula to make some approximation

Apply the formula

*4.1.3*Part (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:

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

*4.2*Accuracy and discussion¶

*5*Question 5¶

*5.1*Solution¶

Quadrature

*5.1.1*Part (a)¶

Define deg of acc for quad

see htsefp

*5.1.2*Part (b)¶

Find deg of acc

see htsefp

*5.1.3*Part (c)¶

Find some constant $k$

see htsefp

*5.1.4*Part (d)¶

Evaluate $I_h(f)$

see htsefp

*5.2*Accuracy and discussion¶

*6*Question 6¶

*6.1*Solution¶

Trapezoidal rule

*6.1.1*Part (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.

*6.1.2*Part (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.

*6.1.3*Part (c)¶

Apply one step of Richardson extrap

whatever

*6.1.4*Part (d)¶

Compute relative error

ez

*6.2*Accuracy and discussion¶

*7*Question 7¶

Runge-Kutta

*7.1*Solution¶

*7.1.1*Part (a)¶

Define local truncation error, find order of method

basically the same as 2008 q6

*7.1.2*Part (b) (i)¶

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

basically the same as 2008 q6

*7.1.3*Part (b) (ii)¶

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

basically the same as 2008 q6