Lecture 4: Numerical Methods False Position (Regula Falsi)

Review of Bisection Method At bisection method, the new position of the estimated root is determined at the middle of the given interval. The interval is divided into two parts and the parts are evaluated to determined where the root is located by checking the change of function signs. Only the interval containing the next root is considered and it is then divided into two to get the estimated root in the middle The outlined step is repeated until the interval is small enough and the consecutive results have no different more then the prescribed tolerance

y Bisection Method f(x) The estimated root + + - - + x

False Position Method The given interval is divided into two parts But the division point is calculated using congruent triangle formula, The new intervals may not be the same The calculation involves the lower and upper limits as well as the function value of the lower and upper limits The checking of the parts is still required to determine where the root exist

f(x) y f(xu) xl x xu f(xl) False Position Method It uses linear interpolation equation. f(xu) The new root position is estimated using the following equation. Note: A check for sign change is still required to determine where the interval that contain the root The estimated root (xr) xl x xu f(xl)

Example Problem A root of f(x) = 3x – 2 e 0.5 x is known to exist between x = 1 and x = 2. Calculate the guessed location of the root using false position method. Answer: Given xl = 1 and xu = 2, therefore: 1st Iteration f(xl ) = 3 – 2 e 0.5 = -0.2974 and f(xu ) = 3*2 – 2 e 0.5 *2= 0.5634 And and f(xr ) = 0.1173

Since f(xl). f(xr)< 0, the new interval is xl =1 and xu = 1 Since f(xl)*f(xr)< 0, the new interval is xl =1 and xu = 1.3455 2nd Iteration It is obtain from the 1st iteration that xl =1 and xu = 1.3455 f(xl ) = -0.2974 and f(xu ) = 0.1173, therefore: f(xr ) = 0.0110 Since f(xl)*f(xr)< 0, the new interval is xl =1 and xu = 1.2478 3rd Iteration It is obtain from the 1st iteration that xl =1 and xu = 1.2478 f(xl ) = -0.2974 and f(xu ) = 0.0110, therefore: and f(xr ) = 9.4635 x 10-4

The complete result for 10 iteration F(xl) F(xu) xroot f(xroot) 1 -0.29744 0.56344 1.34551 0.11727 2 1.247807 0.01102 3 1.238951 0.00095 4 1.238193 0.00008 5 1.238129 0.00001 6 1.238123 0.00000 7 8 9 10

Comparison with Bisection Iteration   Root Convergence False Position Bisection 1 1.345510323 1.5 0.11727 0.26600 2 1.247807131 1.25 0.01102 0.01351 3 1.238950941 1.125 0.00095 -0.13511 4 1.238193098 1.1875 0.00008 -0.05903 5 1.238128575 1.21875 0.00001 -0.02231 6 1.238123084 1.234375 0.00000 -0.00429 7 1.238122617 1.242188 0.00464 8 1.238122577 1.238281 0.00018 9 1.238122574 1.236328 -0.00205 10 1.238122573 1.237305 -0.00094

Comparison with Bisection

Assignment A root is known to exist between x = 0 and x = 1 for the function f(x) = x3 – 4.5 x2 + 5.75 x – 1.875 = 0. Find out the root using Bisection Method for the tolerance of 0. 001. A root is known to exist between x = 1 and x = 3 for the function f(x) = 3 x3 + 2 x2 - x – 30 = 0. Find out the root using False Position Method for the tolerance of 0. 001.

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