## Description

## Question 1.

Part (a) – 3 pt

Consider the problem of finding the roots of the functions f1, f2, and f3. What is the (absolute) condition number of

each problem?

1. f1(x) = sin(

x

100 ), at x = 0

2. f2(x) = x

3 − 5x

2 + 8x − 4, at x = 2

3. f3(x) = x

3

, at x = 0

Include your solution in your PDF writeup. Show your work.

## Part (b) – 3 pt

What is the convergence rate of each of the following sequences? Your answer should be either “linear”, “superlinear

but not quadratic”, “quadratic”, “cubic”, or “something else”. Include your solution in your PDF writeup. Justify

your answer.

1. xn = 10−2n

2. xn = 2−n

2

3. xn = 2−n log n

## Question 2. Interval Bisection

## Part (a) – 3 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.

def bisect(f, a, b, n):

“””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,

1

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) – 2pt

Suppose you would like to use the interval bisection method to find the root of a function f(x), starting with an

interval (a, b).

What is the minimum number of interval bisection iterations necessary to guarantee that your estimate of a root is

accurate to 10 decimal places?

##
Question 3. Fixed-Point Iteration

Part (a) – 3 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)), …].

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) – 3 pt

To find a root of the equation

f(x) = x

2 − 5x + 4 = 0

we can consider finding the fixed-point of each of these functions:

1. g1(x) = x

2+4

5

2. g2(x) = √

5x − 4

3. g3(x) = 5 −

4

x

Suppose we use the fixed_point function to find the fixed point of each of these functions. If we choose a value x0

close to 1, would you expect the fixed point iteration to converge to 1?

In your PDF write up, explain why you would expect the iteration to converge (or not).

2

## Part (c) – 3 pt

Run the function that you wrote in part (a) to find the fixed points of g1, g2 and g3. Start at x0 = 3.

Does fixed-point iteration converge for each of the functions? Include the output of your call to fixed_point in your

PDF writeup.

If the iteration converges, what do you think is the approximate the convergence rate of convergence? (linear,

superliner, quadratic, etc).

Include your solution in your PDF writeup.

##
Question 4. Newton’s Method for Root Finding

Part (a) – 3 pts

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.

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]

“””

## Part (b) – 2 pts

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

f(x) = x

2 − 5x + 4 = 0

Start with x0 = 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.

def f(x):

return x ** 2 – 5 * x + 4

newton_root = None

## Part (c) – 6 pts

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

1. h1(x) = (x − 2)(x − 5)(x − 1)

3

2. h2(x) = x cos(πx) − x

3. h3(x) = e

−2x+4 + e

x−2 − x

Write out the statement for updating the iterate xk using Newton’s method for solving each of the equations hi(x) = 0.

For each problem, do you expect Newton’s method to converge if we start close to the root x = 2?

Include your solution in your PDF writeup.

##
Question 5. Secant Method

Part (a) [5 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.

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) [1 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. You can choose starting positions x0 and x1.

Save the result in the variable secant_root.

secant_root = None

## Part (c) [4 pt]

Show that the iterative method

xk+1 =

xk−1f(xk) − xkf(xk−1)

f(xk) − f(xk−1)

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

Include your solution in your PDF writeup.

## Part (d) [2 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:

xk+1 = xk − f(xk)

xk − xk−1

f(xk) − f(xk−1)

Include your solution in your PDF writeup.

4

##
Question 6. Golden Section Search

Part (a) – 6 pt

Write a function golden_section_search that uses the Golden Section search to find a minima of a function f on

the interval [a, b]. You may assume that the function f is unimodal on [a, b].

The function should evaluate f only as many times as necessary. There will be a penalty for calling f more times

than necessary.

Refer to algo 6.1 in the textbook, or slide 19 of the slides accompanying the textbook.

def golden_section_search(f, a, b, n=10):

“”” Return the Golden Section search intervals generated in

an attempt to find a minima of a function `f` on the interval

`[a, b]`. The returned list should have length `n+1`.

Do not evaluate the function `f` more times than necessary.

Example: (as always, these are for illustrative purposes only)

>>> golden_section_search(lambda x: x**2 + 2*x + 3, -2, 2, n=5)

[(-2, 2),

(-2, 0.4721359549995796),

(-2, -0.4721359549995796),

(-1.4164078649987384, -0.4721359549995796),

(-1.4164078649987384, -0.8328157299974766),

(-1.1934955049953735, -0.8328157299974766)]

“””

## Part (b) – 2 pt

Consider the following functions, both of which are unimodal on [0, 2]

f1(x) = 2x

2 − 2x + 3

f2(x) = −xe−x

2 CSC338 Assignment 3

Run the golden section search on the two functions for n = 10 iterations.

Save the returned lists in the variables golden_f1 and golden_f2

golden_f1 = None

golden_f2 = None

## Question 7.

Consider the function

f(x0, x1) = 2x

4

0 + 3x

4

1 − x

2

0x

2

1 − x

2

0 − 4x

2

1 + 7

## Part (a) – 2 pt

Derive the gradient. Include your solution in your PDF writeup.

## Part (b) – 2 pt

Derive the Hessian. Include your solution in your PDF writeup.

5 CSC338 Assignment 3

## Part (c) – 5 pt

Write a function steepest_descent_f that uses a variation of the steepest descent method to solve for a local

minimum of f(x0, x1) from part (a). Instead of performing a line search as in Algorithm 6.3, the parameter α will be

provided to you as a parameter. Likewise, the initial guess (x0, x1) will be provided.

Use (1, 1) as the initial value and perform 10 iterations of the steepest descent variant on the function f. Save the

result in the variable steepest. (The result steepest should be a list of length 11)

def steepest_descent_f(init_x0, init_x1, alpha, n=5):

“”” Return the $n$ steps of steepest descent on the function

f(x_0, x_1) given in part(a), starting at (init_x0, init_x1).

The returned value is a list of tuples (x_0, x_1) visited

by the steepest descent algorithm, and should be of length

n+1. The parameter alpha is used in place of performing

a line search.

Example:

>>> steepest_descent_f(0, 0, 0.5, n=0)

[(0, 0)]

“””

return []

steepest = []

6 CSC338 Assignment 3