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Numerical methods Assignment 3

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CSC338. Assignment 3
Due Date: April 9, 10pm
What to Hand In
• Python File containing all your code, named a3.py.
• PDF file named a3_written.pdf containing your solutions to the written parts of the assignment. Your solution
can be hand-written, but must be legible. Graders may deduct marks for illegible or poorly presented solutions.
If you are using Jupyter Notebook to complete the work, your notebook can be exported as a .py file (File ->
There will be a 15% penalty if you need a remark due to small issues that renders your code untestable. (Please note
the penalty is higher than in A1!)
Make sure to remove or comment out all matplotlib or other expensive code before submitting your
code! Expensive code can render your code untestable, and you will incur the 15% penalty for remark.
Submit the assignment on MarkUs by 10pm on the due date. See the syllabus for the course policy regarding late
assignments. All assignments must be done individually.
import math
import numpy as np
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
Part (b) – 3 pt
What is the convergence rate of each of the following sequences? Your answer should be either “linear”, “superlinear
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,
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?
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.
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)
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
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
Part (b) – 2 pt
5
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 = []
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Numerical methods Assignment 3
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