## Topic 10.2: False-Position Method (Examples)

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# Example 1

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.

a b f(a) f(b) c f(c) Update Step Size
1.02.0-2.001.001.6667-0.2221a = c0.6667
1.66672.0-0.22211.01.7273-0.0164a = c0.0606
1.72732.0-0.01641.01.73170.0012a = c0.0044

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)

# Example 2

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)).

a b f(a) f(b) c f(c) Update Step Size
3.04.00.047127-0.0383723.5513-0.023411b = c0.4487
3.03.55130.047127-0.0234113.3683-0.0079940b = c0.1830
3.03.36830.047127-0.00799403.3149-0.0021548b = c0.0534
3.03.31490.047127-0.00215483.3010-0.00052616b = c0.0139
3.03.30100.047127-0.000526163.2978-0.00014453b = c0.0032
3.03.29780.047127-0.000144533.2969-0.000036998b = c0.0009

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.