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Homework Problems for Chapter 5
1. Bisection method
We consider the bisection method to find a root for f(x) = 0 where f is a continuous
function. If f(0) < 0 and f(1) > 0, then there is a root on the interval [0, 1], and we
take [0, 1] as the initial interval. How many steps of the bisection method are needed to
obtain an approximation to the root, if the absolute error is guaranteed to be ≤ 10−8
?
(Note that your answer does not depend on the actual function!)
2. Fixed point iteration
Given a function f(x) = e
−x − cos(x). We want to find a root r such that f(r) = 0.
(a). First show that there is a root inside the interval [1.1, 1.6].
(b). Next, we try to locate this root by a fixed point iteration. We observe that f(x) = 0
is equivalent to
x = g1(x), where g1(x) = f(x) + x.
Set up the corresponding fixed point iteration scheme. Then, choose the starting
point x0 = 1.6, and perform 4 iterations to compute the values x1, x2, x3, x4. What
do you observe? Does the method converge? Why?
(c). We now observe that f(x) = 0 is also equivalent to
x = g2(x), where g2(x) = x − f(x).
Based on this new function, set up the corresponding fixed point iteration. Choose
again x0 = 1.6 as your starting point , and perform 4 iterations to compute the
values x1, x2, x3, x4. What is your result? How different is it from part (b)? Does
the method converge? Why?
3. More on fixed point iteration.
Consider the iteration scheme:
xn+1 =
1
2
xn +
1
xn
, x0 = 1
(a). If the iteration converges, what is limn→∞ xn?
(b). Does the method converge? Why?
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4. Newton’s method
As you have seen in Example 5.7 in Section 5.4, the computation of √
R can be carried
out by finding the root of f(x) = x
2 − R using Newton’s method.
(a). Check that in this case Newton’s iteration scheme is
xn+1 =
1
2
xn +
R
xn
.
(b). Show that the sequence (xn)n≥1 satisfies
x
2
n+1 − R =
x
2
n − R
2xn
2
.
Interpret this equation in terms of quadratic convergence.
The quantity x
2
n − R = f(xn) is also called the residual. It gives a measure of the
error in the approximation xn.
(c). Now we test the iterations for R = 10. Choose x0 = 3, and compute x1, x2 x3 and
x4. Use 8 digits in your computation. Comment on your result.
5. When Newton’s Method Does not Work Well.
(i). Let m be a positive integer and consider the function
f(x) = (x − 1)m.
It is clear that r = 1 is a root (indeed, it is a root with multiplicity m).
We now try to find this root by Newton’s iteration, for m = 8.
(a). Set up the iteration scheme.
(b). Use x0 = 1.1 as your initial guess (note that this is very close to the root r = 1).
Complete 4 iterations to compute the values x1, x2, x3, x4.
(c). How does the method work for this problem? Do we still have quadratic convergence? Can you explain what is causing this?
(d). If m = 20 (or with any large value of m), can you predict the behavior of Newton’s
iteration? What type of convergence will you get? Explain in detail.
(ii). Use Newton’s method to solve the equation
0 =
1
2
+
1
4
x
2 − x sin(x) −
1
2
cos(2x), with x0 =
π
2
.
Iterate using Newton’s method until an accuracy of 10−3
is obtained. Explain why the
result seems unusual for Newton’s method.
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6. Newton’s Method in Matlab
Preparation: Use “help sprintf” and “help disp” in Matlab to understand how
to use “sprintf” and “disp” to display the data. Here is an example:
disp(sprintf(’I have n=%d and x=%g but f=%f.\n’,2,1.22, 1.22))
This will give the following result in Matlab:
I have n=2 and x=1.22 but f=1.220000.
The problem: Write a Matlab function for Newton’s method. Your file mynewton.m
should start with:
function x=mynewton(f,df,x0,tol,nmax)
Among the input variables, f,df are the function f and its derivative f
0
, x0 is the initial
guess, tol is the error tolerance, and nmax is the maximum number of iterations. The
output variable x is the result of the Newton iterations. Use sprintf and “disp” to
display your result in each iteration so you can check the convergence along the way.
First, test your function with Example 5.7 in Section 5.4, computing √
2.
Then, use your Matlab function to find a root of f(x) = e
−x − cos(x) on the interval
[1.1, 1.6]. Use tol=1e-12 and nmax=10. You should choose an initial guess x0 on the
interval [1.1, 1.6]. What is your answer?
What to hand in: Print out your files mynewton.m, files for functions f(x) and f
0
(x),
script file, test result for the example of √
2 and for the root of f(x) = e
−x − cos(x).
7. The secant method
a). Calculate an approximate value for 43/5 using the secant method with x0 = 3 and
x1 = 2. Make 3 steps, and compute the values x2, x3, x4. Comment on your result.
b). Use secant method to find the root for f(x) = x
3 −2x+ 1 with x0 = 4 and x1 = 2.
Compute the values x2, x3 and x4. Does the method converge? (You can easily
check that x = 1 is a root.)
c). Consider the iteration scheme
xn+1 = xn + (2 − e
xn
)(xn − xn−1)/(e
xn − e
xn−1
), x0 = 0, x1 = 1.
If the iteration converges, what is limn→∞ xn ?
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8. The Secant Method in Matlab
Write a Matlab function to locate a root by the secant method. Your function should
be put in the file mysecant.m, and should be called as
x=mysecant(f,x0,x1,tol,nmax)
Here f is the function f, x0,x1 are the initial guesses, tol is the error tolerance, nmax
is the maximum number of iterations, and x is the output of your function.
Test your function to find the value √
2, as the root for f(x) = x
2 − 2, to see if it works.
Use sprintf and “disp” to display your result in each iteration, so you can check the
convergence. Then, use it to find a root for the function f(x) = e
−x − cos(x) in the
interval [1.1, 1.6]. Use tol=1e-12 and nmax=10. Your two initial guesses should be on
the interval [1.1, 1.6]. How does the result compare to those with Newton’s method?
Comment in detail.
What to hand in: Your file mysecant.m, test result for f(x) = x
2 − 2, and the result
for f(x) = e
−x − cos(x).

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