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Department of Mathematics
MTL107: Numerical Methods and Computations
Exercise Set 3: Bisection Method, Fixed-Point Iteration Method, Newton’s Method, Modified
Newton’s Method, Muller’s method for Polynomials
√
1. Use the Bisection Method to find the zero r for f(x) = x−cosx on [0,1].
3
2. Use the Bisection Method to find solution accurate within 10−2 for x3−7x2+14x−6 = 0
on the intervals:
a. [0,1]; b. [1,3.2]; c. [3.2,4].
3. Use the Bisection Method to find solution accurate within 10−5 for the following problems:
a. x −2−x = 0 on the interval [0,1],
b. ex −x2 +3x−2 = 0 on the interval [0,1],
c. 2xcos(2x)−(x+1)2 = 0 on the intervals [−3,−2] and [−1,0],
2
d. xcosx−2x +3x−1=0onthe intervals [0.2,0.3] and [1.2,1.3].
4. Sketch the graphs of y = x and y = 2sinx. Use the Bisection Method to find an approx-
imation to within 10−5 to the first positive value of x with x = 2sinx.
5. Sketch the graphs of y = ex − 2 and y = cos(ex − 2). Use the Bisection Method to find
−5 x x
an approximation to within 10 to a value in [0.5,1.5] with e − 2 = cos(e − 2).
6. To which zero of f(x) = (x+2)(x+1)2x(x−1)3(x−2) does the Bisection method converge
when applied on the following intervals ?
a. [−3,2.5] b. [−2.5,3] c. [−1.75,1.5] d. [−1.5,1.75].
√
7. Find an approximation to 3 25 correct to within 10−4 using the Bisection Algorithm.
8. Find a bound for the number of iterations needed to achieve an approximation with
−5 3
accuracy 10 to the solution of x − x − 1 = 0 lying in the interval [1,2]. Fina an
approximation to the root with this degree of accuracy.
9. Use a fixed-point iteration method to determine a solution accurate to within 10−2 for
x4 −3x2 −3 = 0 on [1,2]. Use the initial guess or starting guess r = 1.
0
10. Use a fixed-point iteration method to find an approximation to the fixed point that is
accurate to within 10−2 for g(x) = π + 0.5sin(x/2) on [0,2π]. Estimate the number of
iterations required to achieve 10−2 accuracy. Compare the theoretical estimate to the
number actually needed.
11. Use a fixed-point iteration method to find an approximation to √3 that is accurate to
within 10−4. Estimate the number of iterations required to achieve 10−4 accuracy. Com-
pare the theoretical estimate to the number actually needed.
12. Determine an interval [a,b] on which fixed-point iteration will converge. Estimate the
−5
number of iterations required to achieve 10 accuracy and perform the calculations:
2−ex+x2 5 x 1/2 −x −x
a. x = 3 b. x = x2 + 2 c. x = (e /3) d. x = 5 e. x = 6 f. x =
0.5(sinx+cosx).
13. Find all the zeros of f(x) = x2+10cosx by using the fixed-point iteration method for an
appropriate iteration function g. Find the zeros accurate to within 10−4.
14. Use a fixed-point iteration method to determine a solution accurate to within 10−2 for
2sinπx+x=0on[1,2]. Use the initial guess or starting guess r0 = 1.
15. Show that the sequence defined for n ≥ 1 by
x = 1x + 1 ,
n 2 n−1 xn−1
converge to √2 whenever x0 > √2.
√ 2 √ √
Now use the fact that o < (x0 − 2) whenever x0 6= 2 to show that if 0 < x0 < 2,
√
then x1 > 2. √
Using both the results show that the sequence converges to 2 whenever x0 > 0.
16. An object falling vertically through the air is subjected to viscous resistence as well as to
the force of gravity. Assume that an object with mass m is dropped from a height s and
0
that the height of the object after t seconds is
2
s(t) = s − mgt+ m g(1−e−kt/m),
0 k k2
2
whereg = 32.17ft/s andk represents the coefficient of air resistence in lb−s/ft. Suppose
s0 = 300ft, m = 0.25lb, and k = 0.1lb−s/ft. Find, to within 0.01s, the time it takes this
quarter-pounder to hit the ground.
17. Use Newton’s method to find r for f(x) = x2 −6 with an initial guess r = 1.
2 0
2
18. Use the Secant method and the method of False Position to find r3 for f(x) = x −6 with
√
an initial guesses r = 3 and r = 2. Which gives closer approximation to 6.
0 1
−4
19. Use Newton’s method to find solutions accurate within 10 for the following problems:
a. x3 −2x2 −5 = 0 on the interval [1,4],
b. x3 +3x2 −1 = 0 on the interval [−3,−2],
c. xcosx = 0 on the interval [0,π/2],
d. x−0.8−0.2sinx = 0 on the interval [0,π/2].
20. Repeat the above exercise using the Secant Method.
21. Repeat the above exercise using the Method of False Position.
−5
22. Use all three methods to find solutions to within 10 for the following problems:
a. 3xex = 0 on the interval [1,2],
b. 2x+3cosx−ex =0 on the interval [0,1].
−4
23. Use Newton’s method to approximate, to within 10 , the value of x that produces the
2 2
point on the graph of y = x that is closest to (1,0). [Hint: Minimize [d(x)] , where d(x)
2
represents the distance from (x,x ) to (1,0).]
24. Use all three methods for finding approximations of the two zeros, one in [−1,0] and the
−6 4 3 2
other in [0,1] to within 10 of f(x) = 230x +18x +9x −221x−9. Use the endpoints
of each interval as the initial guesses for the Secant Method and the Method of False
Position and the midpoints for Newton’s Method.
25. Use Newton’s method to approximate the two positive solutions x1 and x2 of f(x) =
2 x −x −5
4x −e −e =0towithin 10 with the following values of r :
0
a. r = −10, b. r = −5, c. r = −3, d. r = −1, e. r = 0, f. r = 1,
0 0 0 0 0 0
g. r = 3, h. r = 5, i. r = 10.
0 0 0
26. a. Determine, within 10−6, the only negative zero of f(x) = ln(x2 + 1) − e0.4x − cosπx.
−6 2 0.4x
b. Determine, within 10 , the four smallest positive zeros of f(x) = ln(x +1)−e −
cosπx.
c. Determine a reasonable initial approximation to find the nth smallest positive zero of
f(x). [Hint: Sketch an approximate graph of f(x).]
d. Determine, within 10−6, the 25th smallest positive zero of f(x) = ln(x2 + 1) − e0.4x −
cosπx.
27. Determine the two numbers x and y within 10−4, given
a. x +y = 20, and
√ √
b. (x + x)(y + y) = 155.55.
3x+1 2x
28. a. Plot f(x) = 3 −7·5 tofindinitial approximations to roots of f(x).
b. Use Newton’s method to find roots of f(x) to within 10−16.
29. Use Newton’s method and modified Newton’s method to find solutions accurate within
10−5 for the following problems:
a. x2 −2xe−x +e−2x = 0 on the interval [0,1],
√ √
b. cos(x + 2) +x(x/2+ x)=0 on the interval [−2,−1],
3 2 −x −x
c. x −3x 2 +3x4 =0onthe interval [0,1],
6x 2 2x 4x 3
d. e +3(ln2) e −(ln8)e −(ln2) =0 on the interval [−1,0].
Is there an improvement in speed or accuracy over Newton’s method ?
30. Use Newton’s method and modified Newton’s method to find solutions accurate within
10−5 for the problem: e6x + 1.441e2x − 2.079e4x − 0.3330 = 0 on the interval [−1,0].
This is the same problem as (d) in the above problem with the coefficients replaced by
their 4-digit approximations. Compare the solutions to the results obtained in the above
problems.
31. Use the Bisection method, Newton’s method, the Secant method, the method of False
Position and Muller’s method to find a solution in [0.1,1] accurate to within 10−4 for
f(x) = 600x4 −550x3 +200x2 −20x−1 = 0.
32. Use Newton’s method and Muller’s to find approximatins to all the real zeros accurate
within 10−4 for the following polynomials:
a. f(x) = x3 −2x2 −5,
b. f(x) = x3 +3x2 −1,
c. f(x) = x3 −x−1,
d. f(x) = x4 +2x2 −x−3,
e. f(x) = x3 +4.001x2 +4.002x+1.101,
f. f(x) = x5 − x4 + 2x3 − 3x2 + x − 4.
ANSWERS
1. p3 = 0.625.
2. The Bisection method gives:
a. p7 = 0.5859 b. p8 = 3.002 c. p7 = 3.419
3. The Bisection method gives:
a. p17 = 0.641182. b. p17 = 0.257530.
c. For the interval [-3,-2], we have p17 = −2.191307, and for the interval [-1,0], we
have p17 = −0.798164.
d. For the interval [0.2,0.3], we have p14 = 0.297528, and for the interval [1.2,1.3], we
have p14 = 1.256622.
4. b. Using [1.5,2] from graph of part (a) gives p16 = 1.89550018.
5. b. p17 = 1.00762177.
6. a. 2 b. -2 c. -1 d. 1
7. The third root of 25 is approximately p14 = 2.92401, using [2,3]
8. A bound is n ≥ 14, and p14 = 1.32477.
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