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ENGR 391 ASSESSMENTS ASSIGNMENT 2 - NONLINEAR EQUATIONS QUESTIONS AND ANSWERS CONCORDIA UNIVERSITY CA$13.90   Add to cart
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ENGR 391 ASSESSMENTS ASSIGNMENT 2 - NONLINEAR EQUATIONS QUESTIONS AND ANSWERS CONCORDIA UNIVERSITY
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ENGR 391 ASSESSMENTS ASSIGNMENT 2 - NONLINEAR EQUATIONS QUESTIONS AND ANSWERS CONCORDIA UNIVERSITY

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  • November 14, 2023
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  • 2023/2024
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ENGR 391 ASSESSMENTS ASSIGNMENT 2 - NONLINEAR EQUATIONS QUE
CONCORDIA UNIVERSITY




Make sure to understand well where you did mistakes and review the corresponding

Be careful not to fall behind by postponing the review of the topics.


Question 1 By checking the "yes" answer below I confirm
Complete
1. that I have neither given nor received unauthorized aid to answer the questions
Not graded
2. I agree to follow the rules in regard of online assignments as posted in the cours
3. I used only octave or Matlab to solve the questions (I am allowed to consult all c

Select one:
a. Yes I agree

b. No I do not agree



Your answer is correct.
The correct answer is: Yes I agree



Question 2
−5
You have to solve an equation f(x) = 0 with a precision below a relative error of 10 .If
Incorrect f(xr) ∣
< 10−5 then
Mark 0.00 out of ∣ xr ∣
1.00
Select one:
a. you have found an approximation xr to the specified precision

b. you have found an approximation xr to the specified precision as long as xr ≠0
c. you have found an approximation xr to the specified precision if xr is not a mult

d. you can't conclude anything



Your answer is incorrect.
The quantity |f (xr )|, which is the backward error, doesn't tell you anything about |r −
Similarly ∣ r ∣ doesn't tell you anything about ∣ r ∣ which is the relative error you wa
f(x ) r−x

∣ xr ∣ xr
tolerance.
Reference: lecture on backward and forward errors in Lesson 2 "Solving nonlinear equ

,Question 3
Find the absolute backward and forward error for the following functions, where the tru
Incorrect
approximated root is xr = 0.48.
Mark 0.00 out of
1.00 Fill out the following table to answer the question (use as many digits as possible)


f(x) Backward error Forward error

f(x) = 2x − 1 0.02 0.04


f(x) = (2x − 1)2 0.02 0.0016


f(x) = (2x − 1)3 0.02 0.000064




The absolute forward error is for all cases |r − xr| = 0.02.
The absolute backward error is different for each case and is computed as |f (xr )|.
The important element to understand here: even with a same forward error, one can h
This problem illustrates once more that computing the backward error doesn't tell us a
error.Reference: lecture on forward and backward errors in Lesson 2 "Solving nonlinear



Question 4 We want to apply the fixed point method to solve the equation x = g(x).
Partially correct Choose among the following examples of functions g(x) which will lead to a convergin
Mark 0.50 out of using the intial guess x0 specified.
1.00
Select one or more:
a. g(x) = ex − 7 and x0 = −2
b. g(x) = x2 − 3x cos x and x0 = 1
2
c. g(x) = and x0 = 1
√x
d. g(x) = sin x + ex + x − 4 and x0 = 0.5
e. g(x) = sin x + ex and x0 = 0.5


Your answer is partially correct.
You have correctly selected 1.

To decide which function g(x) leads to a converging algorithm you need to try out som
convergence plot.
Reference: lecture on fixed point method in Lesson 2 "Solving nonlinear equations".
2
The correct answers are: g(x) = ex − 7 and x = 0−2, g(x) = and x0 = 1
√x

,Question 5 The equation x3 − 4x2 − 11x + 30 = 0 has a solution in r = 2.
Correct You are using Newton's method to estimate this solution. If the absolute error in iterat
Mark 1.00 out of doyou expect to be the absolute error in iteration 5 ?
1.00

Select one:
a. 0.0002
b. We can not predict this error with the information given
c. 0.005
d. 0.03



Your answer is correct.
Newton's method convergences quadraticaly. The asymptotic error constant λ is given

∣ ≃ 0.13.
6r−8
This is λ = ∣
3 2 ∣r
2 2
11 5 ≃ λE 4≃ 0.13 ⋅ 0.04 ≃ 0.002
Consequently−8r−
E
Reference: Lecture on order of convergence form Lesson 2 "Nonlinear equations"
The correct answer is: 0.0002



Question 6 The root r = 0 of the equation ex − x − 1 has a multiplicity of
Correct

Mark 1.00 out of Answer:
2
1.00




Indeed r = 0 is a root as er − r − 1 = 0.
Define f(x) = ex − x − 1.
Then f ′(x) = ex − 1.
As f ′(r) = 0 the root is at least a double root.
Further f " (x) = ex and f " (r) ≠ 0. Consequently the root is a double root.


Reference: lecture on challenging problems in Lesson 2 "Solving nonlinear equations"
The correct answer is: 2



Question 7 We solve the equation x3 + 4x2 − 10 = 0 using the false-position method.
Incorrect If a0 = 1 and b0 = 2 then a1 and b1 are?
Mark 0.00 out of
1.00 Select one:
a. a1 = 1 and b1 = 1.5

, Question 8 You are implementing on a computer a numerical algorithm to solve an equation of th
Incorrect order to check your algorithm you decide to test it with the function f(x) = sin(x) − x.
Mark 0.00 out of
1.00 Select one:
a. This is not a good choice for the function f(x) to test the programmed algorithm
b. This is a valid choice for the function f(x) to test the programmed algorithm



Your answer is incorrect.
The function f(x) = sin(x) − x is flat around the root r = 0 as it is a root of multiplicity 3
Any numerical algorithm will face problems to find this root accurately. Some of them m
compared to what they should (see the examples with Newton's algorithm covered in
problemwith multiple roots becomes problematic).
It is not a good idea to check an implementation of an algorithm with a challenging pro
Reference: lecture on challenging problems in Lesson 2 "Solving nonlinear equations"

The correct answer is: This is not a good choice for the function f(x) to test the progra



Question 9 Solve the equation x3 + 4x2 − 10 = 0 using the bisection algorithm starting with a0
Correct Conduct three iterations (i.e. find ) and report the approximation x3 of the root com
Mark 1.00 out of x3
1.00
Use as many digits as possible.


Answer: 1.3125




First iteration: = 1 and b1 = 1.5
a1
Second iteration: a2 = 1.25 and b2 = 1.5
Third iteration: a3 = 1.25 and b3 = 1.375
Algorithm estimates the root in the third iteration as x3 = 1.125+1.375 = 1.3125
2
Reference: lecture on bisection method in lesson 2 "Solving nonlinear equations"
The correct answer is: 1.3125

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