Fall 2004 Final View the exam on docuum

Under construction.

1Question 1¶

Newton's method

1.1Solution¶

1.1.1Part (a)¶

Use a Taylor expansion to derive Newton's method.

Let $x = x_n$ and $x^* = x_{n+1}$. Substituting these into the Taylor expansion (ignoring the error term), we get:

$$f(x_{n+1}) = f(x_n) + (x_{n+1} - x_n)f'(x_n) \approx 0 \tag{since f(x_{n+1}) \approx 0 as n \to \infty}$$

Solving for $x_{n+1}$, we get:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

which is indeed Newton's method.

1.1.2Part (b)¶

Write Newton's method as a fixed-point iteration

We need to find a $g$ such that we can write

$$x_{n+1} = g(x_n)$$

This is kind of a freebie ...

$$g(x) = x - \frac{f(x)}{f'(x)}$$

1.1.3Part (c)¶

Show that $g'(x^*) = 0$ provided $f'(x^*) \neq 0$.

Take the derivative of $g$:

$$g'(x) = \frac{f(x)f''(x)}{(f'(x))^2}$$

As long as $f'(x^*) \neq 0$, then the denominator is non-zero and so the expression is valid. Since $f(x^*) = 0$ if $x^*$ is the root, then $g(x^*) = 0$ (since the numerator is zero).

This tells us that the fixed-point iterations will converge quadratically provided our initial guess $x_0$ is close enough to $x^*$ (where "close enough" means $x_0$ is in some $\delta$-neighbourhood of $x^*$).

1.1.4Part (d)¶

Find some limit for $g'(x)$.

We're given that $f'(x^*) = 0$ but $f''(x^*) \neq 0$. Thus $f'(x)$ must be of the form $f'(x) = (x-x^*)h(x)$ for some function $h(x)$ (where $h(x^*) \neq 0)$. Then $f(x) = (x-x^*)^2i(x)$ where $i(x)$ is some function such that $i(x^*) \neq 0$. We can rewrite $f'(x)$ in terms of $i(x)$: $f'(x) = 2(x-x^*)i(x) + (x-x^*)^2i'(x) = (x-x^*)(2i(x) + (x-x^*)i'(x))$. Then, $f''(x) = 2(x-x^*)i'(x) + 2i(x) + 2(x-x^*)i'(x) + (x-x^*)^2i''(x) = 2i(x) + 4(x-x^*)i'(x) + (x-x^*)^2i''(x)$. Now let's find the limit for $g'(x)$:

$$\lim_{x \to x^*} g'(x) = \lim_{x \to x^*} \frac{f(x)f''(x)}{(f'(x))^2} = \text{some really long derivation} = \frac{1}{2}$$

This tells us that the fixed-point iteration will converge linearly to $x^*$ given $x_0$ is close enough to $x^*$.

1.1.5Part (e) (i)¶

Use Newton's method to compute $x_5$ for the root of some function.

$f(x) = x^3 - 9x^2 + 15x + 25$. Since we know that $x=5$ is one of the roots, we can rewrite this as $f(x) = (x-5)(x^2-4x-5) = (x-5)(x-5)(x+1)$. Also, $f'(x) = 3x^2 - 18x + 15$. Let $x_0 = 3$ be our initial approximation. We can calculate subsequent approximations using the formula from part (a):

\begin{align} x_1 & = x_0 - \frac{f(x_0)}{f'(x_0)} = 3 - \frac{f(3)}{f'(3)} = 3 - \frac{16}{-12} = \frac{13}{3} \\ x_2 & = x_1 - \frac{f(x_1)}{f'(x_1)} = \frac{211}{45} \\ x_3 & \approx 4.84881749219099 \tag{got lazy} \\ x_4 & \approx 4.9253984938432955 \\ x_5 & \approx 4.9629355449998265 \end{align}

1.1.6Part (e) (ii)¶

What's the apparent rate of convergence?

To find the apparent rate of convergence, we look at the following:

$$\lim_{k \to \infty} \frac{|x_{k+1} - 5|}{|x_{k} - 5|}$$

This seems to be around 0.5 so the order of convergence is 1.

1.2Accuracy and discussion¶

See q6 in the instructor-provided solutions to the final review exercises, hosted on docuum. Part (e) in the exercises doesn't have the "compute $x_5$" bit, though.

2Question 2¶

Fixed-point theorem & proving it. See HTSEFP for the proof.

3Question 3¶

Divided differences

3.1Solution¶

3.1.1Part (a)¶

Define the zeroth divided differences for all $j$.

Just $f[x_j] = f(x_j)$.

3.1.2Part (b)¶

Describe the interpolating polynomial in Newton form.

We are given the following formula for the interpolating polynomial in Newton form:

$$p_n(x) = \sum_{j=0}^n c_j w_j(x)$$

Our job is to define the $w$'s and $c$'s:

• $w_0(x) = 1$
• $\displaystyle w_j(x) = \prod_{i=0}^{j-1}(x-x_i) \quad j = 1, \ldots, n$
• $c_j = f[x_0, x_1, \ldots, x_j] \quad j=0, 1, \ldots, n$

3.1.3Part (c)¶

Assuming that $f$ is $n$-times diff'able, show there exists $\xi \in [x_0, x_n]$ such that

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

We define a new function $g$ as $g(x) = f(x) - p_n(x)$. By the definition of the interpolating polynomial, we know that $f(x_i) = p_n(x_i)$ for $i \in 0, \ldots, n$. So $g(x)$ has $n+1$ zeros. By the generalised version of Rolle's theorem, we know that $g^{(n)}(x)$ has a zero, which we'll call $\xi$. Thus there is a $\xi \in [x_0, x_n]$ such that $g^{(n)}(\xi) = f^{(n)}(\xi) - p_n^{(n)}(\xi) = 0$.

Let's look more closely at $p_n^{(n)}$. We know that the highest-order term in $p_n$ is $x^n$, which has a coefficient of $f[x_0, \ldots, x_n]$. Since all lower-order terms will become 0 in the $n$th derivative, we can ignore those. Then, if we take the $n$th of $p_n$, we get $p_n^{(n)}(x) = n!f[x_0, \ldots, x_n]$. If we substitute this into the formula for $g^{(n)}(\xi)$, we get:

$$g^{(n)}(\xi) = f^{(n)}(\xi) - n!f[x_0, \ldots, x_n] = 0$$

and if we rearrange this, we get

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

exactly as desired.

3.1.4Part (d)¶

Construct the table of centered differences given some values for $x_i$ and $f(x_i)$.

$f[x_i]$ $f[x_{i-1}, x_i]$ $f[x_{i-2}, x_{i-1}, x_i]$ $f[x_{i-3}, x_{i-2}, x_{i-1}, x]$
$x_0 = 0$ 3
$x_1 = 1$ 5 $(5-3)/(1-0) = 2$
$x_2 = 2$ 9 $(9-5)/(2-1) = 4$ $(4-2)/(2-0) = 1$
$x_3 = 3$ 17 $(17-9)/(3-2) = 8$ $(8-4)/(3-1) = 2$ $(2 - 1)/(3-0) = 1/3$

State the polynomial of degree 3 which interpolates at $x_0, x_1, x_2, x_3$.

Take the coefficients from the diagonal of the table:

$$p_3(x) = 3 + 2(x-0) + 1(x-0)(x-1) + 1/3(x-0)(x-1)(x-2) = 3+2x + x(x-1) + 1/3x(x-1)(x-2)$$

State the polynomial of degree 2 which interpolates at $x_0, x_1, x_2$.

Take the coefficients from one level below the diagonal:

$$p_2(x) = 5 + 4(x-1) + 2(x-1)(x-2)$$

3.2Accuracy and discussion¶

Same as question 2 for the Fall 2008 final. Solutions here. Note that part (d) has different numbers (but the method is identical).

4Question 4¶

Forward difference approximations

4.1Solution¶

4.1.1Part (a)¶

Use the fwd diff formula with $h=0.1$ and $h=0.2$ to get 2 approximations for $f'(0)$.

Free points basically ...

\begin{align} h = 0.1: & f'(0) \approx \frac{f(0+0.1) - f(0)}{0.1} = \frac{f(0.1) - f(0)}{0.1} = \frac{0.090483}{0.1} = 0.90483 \\ h = 0.2: & f'(0) \approx \frac{f(0 + 0.2) - f(0)}{0.1} = \frac{f(0.2) - f(0)}{0.2} = \frac{0.163746}{0.2} = 0.81873 \end{align}

4.1.2Part (b)¶

Using Taylor series or otherwise find the error term

Using Taylor series:

$$f(x_0 + h) = f(x_0) + hf'(x_0) + \frac{h^2}{2} f''(x_0) + \frac{h^3}{6} f'''(x_0) + \ldots$$

Rearranging to solve for $f'(x_0)$, we have:

$$f'(x_0) = \frac{f(x_0 + h) - f(x_0)}{h} - \underbrace{\left ( \frac{h^2}{2} f''(x_0) + \frac{h^3}{6} f'''(x_0) + \ldots \right )}_{\text{error terms}}$$

4.1.3Part (c)¶

Use Richardson extrapolation to get a better approximation

Replace $h$ by $h/2$, do 2(2) - (1), should get $\approx 0.99093$

4.1.4Part (d)¶

State a finite formula for approximating $f''(x_0)$, and use it to approximate $f''(0.1)$.

$$f''(x) \approx \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$

Let $h=0.1$. Then: $(f(0.2) - 2f(0.1) + f(0))/0.01 = 100*(0.163746 - 2*0.090483) = -1.722$.

4.2Accuracy and discussion¶

Similar to q3 in the exercises

5Question 5¶

5.1Solution¶

5.1.1Part (a)¶

Define degree of accuracy

see htsefp

Find deg of acc

see htsefp

5.1.3Part (c)¶

Find $k$

see htsefp

5.1.4Part (d)¶

Approximate $\ln(2)$

see htsefp

5.2Accuracy and discussion¶

Similar to q4 in the exercises (slightly different though). Also similar to q4 for fall 2008.

6Question 6¶

Simpson's rule

6.1Solution¶

6.1.1Part (a)¶

Derive composite rule

see htsefp

6.1.2Part (b)¶

Prove error formula for composite rule

see htsefp

6.1.3Part (c)¶

Show that $I(f) \leq I_h(f)$.

This whole question is so ridiculously easy. I can't believe I was actually worried for this exam1. To show that $I(f) \leq I_h(f)$, we can just use the formula from part (b) for $I(f) - I_h(f)$. As long as this difference is negative, $I(f) \leq I_h(f)$. Well, $b-a$ is obviously positive, and $h^4$ is also obviously positive. So to ensure that the difference is negative, we just need $f^{(4)}$ to be positive. Let's take a look at the fourth derivative:

$$f'(x) = 3x^2 + \pi\cos(\pi x) \quad f''(x) = 6x - \pi^2\sin(\pi x) \quad f'''(x) = 6 - \pi^3\cos(\pi x) \quad f^{(4)}(x) = \pi^4\sin(\pi x)$$

Consider the behaviour of $\sin(\pi x)$ over the interval $[0, 1]$. (This is equivalent to the behaviour of $\sin(x)$ over the interval $[0, \pi]$.) Notably, consider the fact that it's never negative at any point in this interval. So yeah, $I(f) \leq I_h(f)$. $\blacksquare$

Find an upper bound for the error when $h = 1/10$.

When $h=1/10$, the upper bound is given by

$$\max_{\xi \in [0, 1]} -\frac{(b-a)}{180}h^4 f^{(4)}(\xi) = -\frac{(1-0)}{180} \cdot \frac{1}{10^4} \cdot \max_{\xi \in [0, 1]} f^{(4)}(\xi) = -\frac{1}{1800000} \cdot \pi^4 \approx5.4109 \times 10^{-5} since the max of \sin(x) in the interval is obviously 1. What value of h is required to ensure that |I(f) - I_h(f)| \leq 10^{-4}?$$h \leq \sqrt{\frac{10^{-4} \cdot 180}{\pi^4}} \approx 0.116595$$6.2Accuracy and discussion¶ Identical to q5 for fall 2008. 7Question 7¶ Runge-Kutta 7.1Solution¶ 7.1.1Part (a)¶ Prove something same as exercises 7.1.2Part (b)¶ Prove something else ditto 7.1.3Part (c)¶ Conditions on h, etc same 7.1.4Part (d)¶ Define local truncation error, etc Slightly different from the one in the exercises. Define local truncation error as$$\tau_{i+1} = y(t_{i+1}) - w_{i+1}

and use a Taylor expansion on $y(t_{i+1}) = y(t_i + h)$. Recall that $y' = \lambda y,$y'' = \lambda^2 y\$, etc, and substitute those into the expansion. Should work out.

7.2Accuracy and discussion¶

Similar to q7 for the exercises and q6 for fall 2008.

8Question 8¶

Linear shooting method. Skipped cus that's not covered in fall 2013.

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