Chapter 5, Problem 2.
Determine the real root of :
(a) Graphically.
(b) Using bisection to locate the root. Employ initial guesses of and and iterate until the estimated error falls below a level of .
Answer:
(a) A plot indicates that a single real root occurs at about x = 0.42.
(b) First iteration:
Therefore, the new bracket is xl = 0 and xu = 0.5.
The process can be repeated until the approximate error falls below 10%. As summarized below, this occurs after 5 iterations yielding a root estimate of 0.40625.
iteration / xl / xu / xr / f(xl) / f(xr) / f(xl)´f(xr) / ea1 / 0 / 1 / 0.5 / -2 / 0.375 / -0.75
2 / 0 / 0.5 / 0.25 / -2 / -0.73438 / 1.46875 / 100.00%
3 / 0.25 / 0.5 / 0.375 / -0.73438 / -0.18945 / 0.13913 / 33.33%
4 / 0.375 / 0.5 / 0.4375 / -0.18945 / 0.08667 / -0.01642 / 14.29%
5 / 0.375 / 0.4375 / 0.40625 / -0.18945 / -0.05246 / 0.009939 / 7.69%
Chapter 5, Problem 8.
Find the positive square root of 18 using the false-position method to within . Employ initial guesses of and .
Answer:
The square root of 18 can be set up as a roots problem by determining the positive root of the function
Using false position, the first iteration is
Therefore, the root is in the second interval and the lower guess is redefined as xl = 4.22222. The second iteration is
Thus, the computation can be stopped after just two iterations because 0.442% < 0.5%. Note that the true value is 4.2426. The technique converges so quickly because the function is very close to being a straight line in the interval between the guesses as in the plot of the function shown below.