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Unformatted text preview: 1 LECTURE 2.2 Dr. Vyas Overview: (Covers section 2.2) The need for learning numerical root finding methods Bisection Method RegulaFalsi Method Examples & Demos I. Why bother learning fundamentals of numerical root finding methods? 1. More often than not, engineers/scientists have to solve a nonlinear transcendental equation (or systems of such equations) as part of their professional practice. For example, roots of the equations like ( ) tan( ) f x x x = or even polynomials higher than fourth degree (except some special cases) cannot be solved explicitly. These have to be solved numerically. 2. At this juncture a reader may ask, cant I just use a command in the software like Matlab/Maple/Mathematica/MathCAD or equivalent, and not have to worry about learning fundamentals? The answer to this question follows. 3. Lecture 1.1 and 1.2 demonstrated limitations of a digital computer with regards to arithmetic operations. There are other software specific limitations with regards to the computational methods employed by the software. Typically, engineering/scientific software help menus contain brief information about the computational methods employed by software. If the user does not have adequate understanding of the fundamentals, then the help menu information is of no use, resulting in a reckless act of computation. 4. The occurrence (and numerical solution) of transcendental equations in engineering/scientific applications will be demonstrated as part of example problems at appropriate times. II. Bisection Method 5. Root/Zero : Assume that ( ) f x is a continuous function. Any number r for which ( ) f r = is called a root of the equation ( ) f x = . Also, we say that r is a zero of the function ( ) f x . 6. The basic idea / thought behind this method is that if a function value becomes zero somewhere, it might be located in an interval (say [ , ] a b ) such that on one end the function value is greater than zero ( ( ) f a is positive, say), while on the other end the function value is less than zero ( ( ) f b is negative, say). (Fig 2.6 ) 7. Logically, the next simplest step is to find whether the midpoint of the interval is the location where the function value goes to zero. To that end, compute the midpoint coordinate ( ) / 2 c a b = + . If the value of the function at c is indeed zero (that is ( ) f c = ), it was indeed very fortunate and the root is located. If the function value at the midpoint is not zero then two possibilities arise. 8. If ( ) f c < , then the root is in the interval, [ , ] a c . Else if, ( ) f c , the root lies in the interval [ , ] c b . Next, depending on the sign of ( ) f c , reassign either, 2 1 1 ; a a c b = = or 1 1 ; c a b b = = . Then, find a new midpoint 1 1 1 ( ) / 2 c a b = + and repeat the process of finding midpoints and function evaluations until the value of the function is sufficiently close to zero (concept of error and tolerance kicks in)....
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This note was uploaded on 04/14/2008 for the course MATH 353 taught by Professor Vyas during the Spring '08 term at University of Delaware.
 Spring '08
 Vyas
 Math

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