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Lecture 2 Typeset Notes

# Lecture 2 Typeset Notes - ME 218 ENGR COMPUTATIONAL METHODS...

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Unformatted text preview: ME 218: ENGR COMPUTATIONAL METHODS Lecture Notes (Set #2) Methods for solving nonlinear eguations l. Bisection Method Assume that the real function f(x) is continuous on the interval [a, b] and that f(a) and f(b) have different signs. Then, at least one root of f(x) lies in the interval [a, b]. Let xL = a and xR = b. Deﬁne xM to be the mid-point of [a, b]. Thus, xM = (xL+xR)/2. If f(xM) = 0, xM is the exact solution! If ﬂaw) is not equal to 0, either f(xL).f(xM)<0 or f(xM)f(xR)<0. In other words, a root lies either in the interval [xL, xM] or [xM, xR]. By using this step, a root of f(x) can be located in an interval (b — a)/2, which is half the original interval. Repeating this process leads to convergence towards a root of f(x). The process can be stopped when the length of the interval becomes less than a desired tolerance value, c. Algorithm: Given the above assumptions, Input xL, xR and stopping criterion/criteria n=0; REPEAT n=n+l; xM = (xL+xR)/2 IFf(xL).f(xM)<0, THEN xR = xM ELSE xL = xM UNTIL STOPPING CRITERION IS MET Note that after n iterations, your solution is enclosed within an interval of width (XL - xR)/2" (you reduce the width of the interval by one half with each section). Hence, aﬁer n iterations, you will be within (XL - xR)/2n of the actual solution!!! Possible stopping criteria 0 |xL — le < e for some small “e” (means we have narrowed down the interval so much that for all intents and purposes we do not need to continue. 0 n>Nnmit (means we have exceeded the maximal allowed number of iterations Nlimit) o | f(xM)| < e2 for some small e2 (this means that my middle point is almost the root of this function, so no need to continue ...
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Lecture 2 Typeset Notes - ME 218 ENGR COMPUTATIONAL METHODS...

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