Roots of equation fx are the values of x which satisfy the above expression. Also referred to as the zeros of an equation

Transcription

1 LECTURE 20 SOLVING FOR ROOTS OF NONLINEAR EQUATIONS Consider the equation f = 0 Roots of equation f are the values of which satisfy the above epression. Also referred to as the zeros of an equation f() roots of f() Eample 1 Find the roots of f = Roots of this function are found by eamining the equation and solving for the values of which satisfy this equality = 0 p. 20.1

2 Eample 2 Solve tank = tank = tank = 0 f = tank Roots are found by eamining the equation tank = 0 Eample 3 Find 3 8 = = = 0 Find roots by eamining the equation f = 3 8 = 0. p. 20.2

3 Notes on root finding Roots of equations can be either real or comple. Recall = a is a real number; = a + ib is a comple number, where i = 1. A large variety of root finding algorithms eist, we will look at only a few. Each algorithm has advantages/disadvantages, possible restrictions, etc. Method Must Specify Interval Containing Root f 1 Continuous Features Bisection yes no Robust Newton-Raphson (Newton) no yes Fast and applies to comple roots p. 20.3

4 Bisection Method with One Root in a Specified Interval You know that the root lies in the interval a 1 b 1 = the root that we are looking for r f() a 1 c 1 b 1 r The midpoint of the starting interval is a 1 + b 1 c 1 = Evaluate fa 1, fc 1, fb 1. Then consider the product fa 1 fc 1 0 root r must lie in interval c 1 b 1 fa 1 fc 1 0 root r must lie in interval a 1 c 1 p. 20.4

5 Selection of the interval is based on the fact that the sign of f changes within the interval in which the root lies. Now reset the interval and repeat the process. Therefore for this case, the second iteration interval becomes a 2 b 2 = c 1 b 1. Now evaluate the midpoint of the second interval as a 2 + b 2 c 2 = f() a 2 c 2 b 2 Evaluate fa 2, fc 2, fb 2. Consider the product fa 2 fc 2 0 root r must lie in interval c 2 b 2 fa 2 fc 2 0 root r must lie in interval a 2 c 2 p. 20.5

6 Repeat until a certain level of convergence has been achieved. Interval size, I, after n steps I b n a n = = 2 b 1 a n I a n c n r b n The interval size at any given iteration also corresponds to the maimum error in r, therefore b 1 a 1 E n n p. 20.6

7 If you wish to limit the error to b 1 a n 2 n b 1 a ln b 1 a n ln2 b 1 a 1 n ln You must apply n iterations to ensure the level of convergence. p. 20.7

8 Eample 4 Find the root of f = 2 7 in the interval 14 n Iteration a n b n c n fa n fb n fc n E i The actual root to this equation is The actual error for our 6 th iteration estimate is p. 20.8

9 Problems with the Bisection Method Multiple roots in an interval a 1 b 1. If fa 1 fb 1 0, then the bisection method will find one of the roots. However it is not very useful to know only one root! Either use another method or provide better intervals. You can use graphical methods or tables to find intervals. Double roots a 1 b 1 f() 2 The bisection method will not work since the function does not change sign e.g. f = 2 2 p. 20.9

10 Singularities f() The bisection method will solve for a singularity as if it were a root. Therefore we must check the functional values to ensure convergence to see if it is indeed a root. p

11 Newton-Raphson Method (a.k.a. Newton Method) Finds the root if an initial estimate of the root is known Method may be applied to find comple roots Method uses a truncated Taylor Series epansion to find the root Basic Concept Slope is known at an estimate of the root. Compute the slope of f at the estimate of the root, o, and project this slope back to where it crosses the -ais to find a better estimate for the root. o f() f() p

12 Derivation of the Newton-Raphson Method We want to solve the equation f r = 0 Taylor series epand f about a point o where o is an estimate of the root f r f o f 1 o r o f 2 r o 2 = ! o r Eplicitly consider terms up to O r o. O r o 2 and higher order terms are not eplicitly accounted for but are represented by the last term in the series and are used to estimate the error. Set the series approimation to f r equal to zero and solve for r f o f 1 o r o f 2 r o = 0 2! f o r = o where f 1 + E E o = f 2 f 1 o r o ! p

13 The error can be estimated as: f 2 o E r o ! f 1 o We can iteratively update our estimate of with our most recently computed value o The error is equal to f n 1 n = n f 1 + E n n 1 E f 2 n 1 n n n 1 2 f ! n 1 Note that is not included to update but serves only as an estimate of the error. E n n p

14 Eample 5 Find the root of f = 2 7 using an initial estimate 0 = 4 f n 1 n = n f 1 E f 2 n 1 n n 1 2 n = f n 1 n 1 f 1 = 2 f 2 = 2 n n 1 f n 1 f 1 n 1 f 2 n 1 n E n E actual = (up to round-off accuracy of calculator). 4 2 p

15 Notes on Newton s Method Convergence rate for Newton s method is very high!! Error estimates are very good (however will be case dependent on the form of the function f ). Newton s method can find comple roots. Problems with Newton s Method If the local min/ma is selected as an initial guess f() 0 f (1) ( 0 ) = 0 The slope at does not intersect with -ais! o The formula for will lead to an infinite value. 1 p

16 If the initial guess is poor (what is poor depends on shape of f ) The solution may diverge: f() 0 1 The solution may converge to another root: f() 0 root you will find root you would like to find Therefore what is a good initial guess depends on the function and how it behaves. p

17 Secant Method For cases where it is difficult or epensive to evaluate the first derivative, we can apply the Newton Method using a difference approimation to evaluate the derivative Forward difference approimation difficult to apply since is included n f 1 n 1 = f n f n O n n n 1 n 1 Backward difference approimation readily applied f 1 n 1 f() = f n O n 2 f n 2 n 2 n 1 n 1 n n-1 n-2 Applying a backward approimation, the Secant algorithm to find the root becomes: f n 1 n = n f n 2 f n 1 n 2 n 1 p

18 SUMMARY OF LECTURE 20 There are many methods available to find roots of equations The Bisection method is a crude but simple method. It is based on the fact that the sign of a function changes in the vicinity of a root. Therefore given an interval within which the root lies, we can narrow down that interval, by eamining the sign of the function at the endpoints and midpoint of the interval (i.e. deciding if the sign change in the function occurs on the left half of the interval or the right half.) Newton-Raphson is based on using an initial guess for the root and finding the intersection with the ais of the straight line which represents the slope at the initial guess. It works very fast and converges assuming the initial guess was good. Very crude methods may have to be used to come up with intervals or initial guesses for roots. Brute force evaluations of functions noting sign changes (even with some iterative refinement in the vicinity of a root/roots) Graphics p