Consider finding the root of f(x) = x2 - 3. Let εstep = 0.01, εabs = 0.01 and start with the interval [1, 2].
Table 1. False-position method applied to f(x) = x2 - 3.
|1.0||2.0||-2.00||1.00||1.6667||-0.2221||a = c||0.6667|
|1.6667||2.0||-0.2221||1.0||1.7273||-0.0164||a = c||0.0606|
|1.7273||2.0||-0.0164||1.0||1.7317||0.0012||a = c||0.0044|
Thus, with the third iteration, we note that the last step 1.7273 → 1.7317 is less than 0.01 and |f(1.7317)| < 0.01, and therefore we chose b = 1.7317 to be our approximation of the root.
Note that after three iterations of the false-position method, we have an acceptable answer (1.7317 where f(1.7317) = -0.0044) whereas with the bisection method, it took seven iterations to find a (notable less accurate) acceptable answer (1.71344 where f(1.73144) = 0.0082)
Consider finding the root of f(x) = e-x(3.2 sin(x) - 0.5 cos(x)) on the interval [3, 4], this time with εstep = 0.001, εabs = 0.001.
Table 2. False-position method applied to f(x) = e-x(3.2 sin(x) - 0.5 cos(x)).
|3.0||4.0||0.047127||-0.038372||3.5513||-0.023411||b = c||0.4487|
|3.0||3.5513||0.047127||-0.023411||3.3683||-0.0079940||b = c||0.1830|
|3.0||3.3683||0.047127||-0.0079940||3.3149||-0.0021548||b = c||0.0534|
|3.0||3.3149||0.047127||-0.0021548||3.3010||-0.00052616||b = c||0.0139|
|3.0||3.3010||0.047127||-0.00052616||3.2978||-0.00014453||b = c||0.0032|
|3.0||3.2978||0.047127||-0.00014453||3.2969||-0.000036998||b = c||0.0009|
Thus, after the sixth iteration, we note that the final step, 3.2978 → 3.2969 has a size less than 0.001 and |f(3.2969)| < 0.001 and therefore we chose b = 3.2969 to be our approximation of the root.
In this case, the solution we found was not as good as the solution we found using the bisection method (f(3.2963) = 0.000034799) however, we only used six instead of eleven iterations.
Copyright ©2005 by Douglas Wilhelm Harder. All rights reserved.