ES 84 SAMPLE PROBLEM (Midterm)
Sample problem 1
Determine the positive real root of ln(x4) = 0.7 (a) using three iterations of the bisection method, with
initial guesses of xl = 0.5 and xu = 2, and (b) using three iterations of the false-point iteration, with the same
initial condition as in (a).
Solution:
(a) Using bisection, the first iteration is
25
.1
2
2
5
.0
rx
%
60
%
100
5
.0
2
5
.
0
2
a
66873
.0
)
19257
.0
(
47259
.3
)
25
.1
(
)
5
.
0
(
f
f
Therefore, the root is in the first interval and the upper guess is redefined as xu = 1.25. The second
iteration is
875
.0
2
25
.1
5
.0
rx
%
86
.
42
%
100
875
.0
25
.1
875
.0
a
28561
.4
)
23413
.1
(
47259
.3
)
875
.0
(
)
5
.
0
(
f
f
Consequently, the root is in the second interval and the lower guess is redefined as xl = 0.875. All the
iterations are displayed in the following table:
i
xl
xu
xr
f(xl)
f(xr)
f(xl) f(xr)
a
1
0.50000
2.00000
1.25000
-3.47259
0.19257
-0.66873
2
0.50000
1.25000
0.87500
-3.47259
-1.23413
4.28561
42.86%
3
0.87500
1.25000
1.06250
-1.23413
-0.4575
0.56461
17.65%
Thus, after three iterations, we obtain a root estimate of 1.0625 with an approximate error of 17.65%.
(b) Using false position, the first iteration is
43935
.1
07259
.2
47259
.3
)
2
5
.0
(
07259
.2
2
rx
62797
.2
)
75678
.0
(
47259
.3
)
43935
.1
(
)
5
.
0
(
f
f
Therefore, the root is in the first interval and the upper guess is redefined as xu = 1.43935. The second
iteration is
27127
.1
75678
.0
47259
.3
)
43935
.1
5
.0
(
75678
.0
43935
.1
rx
%
222
.
13
%
100
27127
.1
43935
.1
27127
.1
a
90312
.0
)
26007
.0
(
47259
.3
)
27127
.1
(
)
5
.
0
(
f
f
Consequently, the root is in the first interval and the upper guess is redefined as xu = 1.27127. All the
iterations are displayed in the following table:
iteration
xl
xu
f(xl)
f(xu)
xr
f(xr)
f(xl) f(xr)
a
1
0.5
2.00000
-3.47259 2.07259 1.43935 0.75678
-2.62797
2
0.5
1.43935
-3.47259 0.75678 1.27127 0.26007
-0.9031