# 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

## 2Question 2¶

### 2.1Solution¶

#### 2.1.1Part (a)¶

Define lagrange polys

see htsefp

Show uniqueness

see htsefp

#### 2.1.3Part (c)¶

Find specific poly

see htsefp

Find error bound

see htsefp

## 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

## 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:

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

## 5Question 5¶

### 5.1Solution¶

#### 5.1.1Part (a)¶

Define deg of acc for quad

see htsefp

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

## 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