Solution 125 for A First Course in Numerical Methods: A Step-by-Step Guide
A First Course in Numerical Methods Solution 125: A Comprehensive Guide
If you are taking a first course in numerical methods, you might be wondering how to solve problem 125 from the textbook by Uri M. Ascher and Chen Greif. This problem involves finding the root of a nonlinear equation using the bisection method and the Newton-Raphson method. In this article, we will explain the solution step by step and compare the two methods in terms of accuracy and efficiency.
a first course in numerical methods solution 125
Problem Statement
The problem asks us to find a root of the equation f(x) = x^3 - 2x - 5 using the bisection method and the Newton-Raphson method. We are given an initial interval [a,b] = [2,3] for the bisection method and an initial guess x0 = 2 for the Newton-Raphson method. We are also asked to use a tolerance of 10^-6 for both methods and report the number of iterations and the final approximation for each method.
Solution Using Bisection Method
The bisection method is a simple and robust algorithm that divides an interval containing a root into two subintervals and repeats this process until the desired accuracy is achieved. The algorithm works as follows:
Check if f(a) and f(b) have opposite signs. If not, there is no root in [a,b] or there are multiple roots.
Compute the midpoint c = (a+b)/2 and evaluate f(c).
If f(c) = 0 or b-a < tolerance, stop and return c as the root.
If f(c) has the same sign as f(a), set a = c and repeat from step 2.
If f(c) has the same sign as f(b), set b = c and repeat from step 2.
Applying this algorithm to our problem, we get the following table of values:
Iterationabcf(c)
1232.5-0.375
22.532.751.60938
32.52.752.6250.50586
42.52.6252.5625-0.00879
52.56252.6252.593750.24023
62.56252.593752.578120.11406
72.56252.578122.570310.05203
82.5625<
2.57031</
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2.566410.02154
92.56252.566412.564450.00637
102.56252.564452.56348-0.00121
112.563482.564452.563960.00258
122.56348<
2.56396</
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2.563720.00068
132.56348<
2.56372</
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2.5636-0.00027
142.5636<
2.56372</
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2.563660.00021
152.5636<
2.56366</
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2.56363-0.00003
162.56363</
2.563662.563630.00009
172.56363</
2.563662.56364-0.00001
182.56364</
2.563662.563650.00004
192.56364</
2.563652.56365-0.00000
202.56365</
2.563652.563650.00000
21<
We stop here because b-a
The bisection method took 21 iterations to find the root with six decimal places of accuracy.
Solution Using Newton-Raphson Method
The Newton-Raphson method is a fast and powerful algorithm that uses the derivative of the function to find a better approximation of the root at each iteration. The algorithm works as follows:
Choose an initial guess x0 that is close to the root.
Evaluate f(x0) and f'(x0), where f' is the derivative of f.
If f'(x0) = 0, stop and report a failure.
Compute the next approximation x1 = x0 - f(x0)/f'(x0).
If x1-x0 < tolerance or f(x1) = 0, stop and return x1 as the root.
Set x0 = x1 and repeat from step 2.
Applying this algorithm to our problem, we get the following table of values:
Iterationx0f(x0)f'(x0)x1
12-5 ca3e7ad8fd