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# lecture 10 - CE 30125 Lecture 10 LECTURE 10 SOLVING FOR...

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CE 30125 - Lecture 10 p. 10.1 LECTURE 10 SOLVING FOR ROOTS OF NONLINEAR EQUATIONS • Consider the equation • Roots of equation are the values of which satisfy the above expression. Also referred to as the zeros of an equation Example 1 • Find the roots of • Roots of this function are found by examining the equation and solving for the values of which satisfy this equality. fx  0 = x x 1 x 2 x 3 x 4 x f(x) roots of f(x) 3 x 5 2 x 2 x 10 ++ = 3 x 5 2 x 2 x 10 0 = x

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CE 30125 - Lecture 10 p. 10.2 Example 2 •So lv e • Roots are found by examining the equation Example 3 • Find • Find roots by examining the equation . kx tan x = tan x = x tan 0 = fx  x tan = x tan 0 = 8 3 x 8 3 = x 3 8 = x 3 8 –0 = x 3 8 ==
CE 30125 - Lecture 10 p. 10.3 • Notes on root finding • Roots of equations can be either real or complex. • Recall is a real number; is a complex number, where . • A large variety of root finding algorithms exist, we will look at only a few. • Each algorithm has advantages/disadvantages, possible restrictions, etc. Method Must Specify Interval Containing Root Continuous Features Bisection yes no Robust Newton-Raphson (Newton) no yes Fast and applies to complex roots xa = i b + = i 1 = f 1  x

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CE 30125 - Lecture 10 p. 10.4 Bisection Method with One Root in a Specified Interval • You know that the root lies in the interval = the root that we are looking for • The midpoint of the starting interval is • Evaluate , , . Then consider the product root must lie in interval root must lie in interval a 1 b 1  x r a 1 b 1 c 1 x x r f(x) c 1 a 1 b 1 + 2 ---------------- = fa 1  fc 1 fb 1 1 1 0 x r c 1 b 1 1 1 0 x r a 1 c 1
CE 30125 - Lecture 10 p. 10.5 • Selection of the interval is based on the fact that the sign of changes within the interval in which the root lies. • Now reset the interval and repeat the process. Therefore for this case, the second itera- tion interval becomes . • Now evaluate the midpoint of the second interval as • Evaluate , , . Consider the product root must lie in interval root must lie in interval fx  a 2 b 2  c 1 b 1 = c 2 a 2 b 2 + 2 ---------------- = x f(x) b 2 c 2 a 2 fa 2 fc 2 fb 2 2 2 0 x r c 2 b 2 2 2 0 x r a 2 c 2

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CE 30125 - Lecture 10 p. 10.6 • Repeat until a certain level of convergence has been achieved.
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lecture 10 - CE 30125 Lecture 10 LECTURE 10 SOLVING FOR...

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