# Assignment 4¶

Due Date: March 25, 8:59pm

Please see the guidelines at https://www.cs.toronto.edu/~lczhang/338/homework.html

### What to Hand In¶

• Python file containing all your code, named csc338_a4.py. If you are using Jupyter Notebook to complete the work, your notebook can be exported as a .py file (File -> Download As -> Python). Your code will be auto-graded using Python 3.6, so please make sure that your code runs. There will be a 20% penalty if you need a remark due to small issues that render your code untestable.
• PDF file named csc338_a4_written.pdf containing your solutions to the written parts of the assignment. Your solutions can be hand-written, but must be legible. Graders may deduct marks for illegible or poorly presented solutions.

Submit the assignment on MarkUs by 9pm on the due date. See the syllabus for the course policy regarding late assignments. All assignments must be done individually.

In [30]:
import math
import numpy as np


## Q1. Interval Bisection [10 pt]¶

### Part (a) [5 pt]¶

Write a function bisect that returns a list of intervals where the root of the function f(x) lies. Each interval should be half the size of the previous, and should be obtained using the interval bisection method.

In [75]:
def bisect(f, a, b, n=20):
"""Returns a list of length n+1 of intervals
where f(x) = 0 lies, where each interval is half
the size of the previous, and is obtained using
the interval bisection method.

Precondition: f continuous,
a < b
f(a) and f(b) have opposite signs

Example:
>>> bisect(lambda x: x - 1, -0.5, 2, n=5)
[(-0.5, 2),
(0.75, 2),
(0.75, 1.375),
(0.75, 1.0625),
(0.90625, 1.0625),
(0.984375, 1.0625)]
"""



### Part (b) [2 pt]¶

In the bisect function, to obtain the midpoint between a and b, it is best to write

mid = a + (b - a) / 2

rather than

mid = (a + b) / 2

Why is the first computation better than the second? Provide an explanation in your PDF writeup.

### Part (c) [3 pt]¶

Use the interval bisection method to find the root of the function

$$f(x) = x^3 + x^2 + x - 4$$

accurate to 8 significant decimal digits. The root of $f(x)$ is between -1 and 4.

Show your work in your python file, and store the root you find in the variable bisection_root.

In [77]:
def f(x):
return x ** 3 + x ** 2 + x - 4

bisection_root = None


## Q2. Fixed-Point Iteration [20 pt]¶

### Part (a) [4 pt]¶

Write a function fixed_point to find the fixed-point of a function f by repeated application of f. The function should return a list of values [x, f(x), f(f(x)), ...].

In [114]:
def fixed_point(f, x, n=20):
""" Return a list lst = [x, f(x), f(f(x)), ...] with
lst[i+1] = f(lst[i]) and len(lst) == n + 1

>>> fixed_point(lambda x: math.sqrt(x + 1), 3, n=5)
[3,
2.0,
1.7320508075688772,
1.6528916502810695,
1.6287699807772333,
1.621348198499395]
"""


### Part (b) [8 pt]¶

To find a root of the equation

$$f(x) = x^2 - 3x + 2 = 0$$

we can consider fixed-point problems involving the following different functions :

1. $g_1(x) = \frac{x^2 + 2}{3}$
2. $g_2(x) = \sqrt{3x - 2}$
3. $g_3(x) = 3 - \frac{2}{x}$
4. $g_4(x) = \frac{x^2 - 2}{2x - 3}$

Analyze the convergence properties of each of the corresponding fixed-point iteration schemes for the root x = 2 by analyzing $g_i'(x)$. Do you expect the fix-point iteration to diverge or converge? What is the rate of convergence?

This question should be done by hand. Submit your solution in your PDF writeup.

### Part (c) [8 pt]¶

Confirm your analysis in Part (b) by using the fixed_point function to generate the fixed-point iterations. Verify the convergence of $g_i$, or lack thereof. Choose the starting value of $x$ to be $x_0 = 3$.

Analyze the numerical results (outputs of the fixed_point) function and approximate the observed convergence rate.

## Q3. Newton's Method [25 pt]¶

### Part (a) [5 pt]¶

Write a function newton to find a root of f(x) using Newton's Method. This Python function should take as argument both the mathematical function f and its derivative df, and return a list of successively better estimates of a root of f obtained from applying Newton's method.

In [51]:
def newton(f, df, x, n=5):
""" Return a list of successively better estimate of a root of f
obtained from applying Newton's method. The argument df is the
derivative of the function f. The argument x is the initial estimate
of the root. The length of the returned list is n + 1.

Precondition: f is continuous and differentiable
df is the derivative of f

>>> def f(x):
...     return x* x - 4 * np.sin(x)
>>> def df(x):
...     return 2 * x - 4 * np.cos(x)
>>> newton(f, df, 3, n=5)
[3,
2.1530576920133857,
1.9540386420058038,
1.9339715327520701,
1.933753788557627,
1.9337537628270216]
"""

In [52]:
def f(x):
return x* x - 4 * np.sin(x)
def df(x):
return 2 * x -4 * np.cos(x)
newton(f, df, 3)

Out[52]:
[3,
2.1530576920133857,
1.9540386420058038,
1.9339715327520701,
1.933753788557627,
1.9337537628270216]

### Part (b) [2 pt]¶

Use your function from part (a) to solve for a root of

$$f(x) = x^2 - 3x + 2 = 0$$

Start with $x_0$ = 3, and stop when the root is accurate to at least 8 significant decimal digits.

Show your work in your python file, and store the root you find in the variable newton_root.

In [1]:
def f(x):
return x ** 2 - 3 * x + 2

newton_root = None


### Part (c) [6 pt]¶

Consider the following non-linear equations $h_i(x) = 0$.

1. $h_1(x) = x^3 - 5x^2 + 8x - 4$
2. $h_2(x) = x cos(20x) - x$
3. $h_3(x) = e^{-2x} + e^{x} - x - 4$

Write out the statement for updating the iterate $x_k$ using Newtonâ€™s method for solving each of the equations $h_i(x) = 0$.

### Part (d) [3 pt]¶

Plot each of the three functions, showing the roots of each function (if any). Include your plots in your PDF writeup.

### Part (e) [3 pt]¶

Use the newton function to try and solve $h_i(x) = 0$, for n = 100 iterations, starting with x = 1.5.

Save the return values of calls to the function newton to the variables newton_h1, newton_h2, newton_h3,

In [ ]:
newton_h1 = None # newton(..., x=1.5, n=100)
newton_h2 = None # newton(..., x=1.5, n=100)
newton_h3 = None # newton(..., x=1.5, n=100)


### Part (f) [6 pt]¶

Explain why Newton's method either does not converge or converges slowly for each of the functions $h_1$, $h_2$ and $h_3$.

## Q4. Secant Method [17 pt]¶

### Part (a) [6 pt]¶

Write a function secant to find a root of f(x) using the secant method. The function should return a list of successively better estimates of a root of f obtained from applying the secant method.

In [1]:
def secant(f, x0, x1, n=5):
""" Return a list of successively better estimate of a root of f
obtained from applying secant method. The arguments x0 and x1 are
the two starting guesses. The length of the returned list is n + 2.

>>> secant(lambda x: x ** 2 + x - 4, 3, 2, n=6)
[3,
2,
1.6666666666666667,
1.5714285714285714,
1.5617977528089888,
1.5615533980582523,
1.561552812843596,
1.5615528128088303]
"""


### Part (b) [3 pt]¶

Use the secant function to find a root of $f(x) = x^3 + x^2 + x - 4$, accurate up to 8 significant decimal digits.

Show your work in your Python file. Save the result in the variable secant_root.

In [2]:
secant_root = None


### Part (c) [4 pt]¶

Show that the iterative method

$$x_{k+1} = \frac{x_{k-1} f(x_k) - x_k f(x_{k-1})}{f(x_k) - f(x_{k-1})}$$

is mathematically equivalent to the secant method for solving a scalar nonlinear equation $f(x) = 0$.

### Part (d) [4 pt]¶

When implemented in finite-precision floating-point arithmetic, what advantages or disadvantages does the formula given in part (c) have compared with the formula for the secant method given in lecture?

This is the formula given in lecture:

$$x_{k+1} = x_k - f(x_k)\frac{x_k - x_{k-1}}{f(x_k) - f(x_{k-1})}$$

## Q5. Conditioning [8 pt]¶

Consider the problem of finding the roots of the functions $f_1$, ... $f_4$. What is the condition number of each problem?

Save the condition numbers in the variables cond_f_1 to cond_f_4. If a root is not provided, you can find it using any method you wish, including by hand. Your condition numbers should be accurate up to 3 significant decimal digits.

1. $f_1(x) = x^3 + x^2 + x - 4$
2. $f_2(x) = x^3 - 5x^2 + 8x - 4$, at $x = 2$
3. $f_3(x) = x cos(20x) - x$, at $x = 0$
4. $f_4(x) = x^2 - 4 sin(x)$, at $x = 0$

If the condition number is infinite, store a very large number or a non-numerical value, but do not use the value None, so I know that you attempted the problem.

In [2]:
cond_f_1 = None
cond_f_2 = None
cond_f_3 = None
cond_f_4 = None