Roots of Equations Definition Bisection Method FalsePosition Method

Roots of Equations • Definition • Bisection Method • False-Position Method

References: • Numerical Methods for Engineers “Steven C. Chapra, Raymond P. Canale” • http: //numericalmethods. eng. usf. edu “Autar Kaw, Luke Snyder”

Definitions • Consider an equation in one variable • The may be a polynomial or a transcendental function (function of polynomial, exponential, trigonometric, logarithmic functions) e. g.

Definitions(2) • We define following: • Root or zero: is a zero/root if • Simple root: is a simple root is , • e. g. : are the simple roots of

Definitions(3) • Root of multiplicity m • Ex:

Example • Remember that from Newton law, we got • And suppose that the problem is now • There is no way to re- arrange the equation so that c is isolated on one side of the equal sign ( it is impossible to write c=f(something) )

Example (2) • We can overcome the problem by constructing another function • The solution is the root of this equation e. g. c makes f(c)=0.

Roots of Equations • Why? • But 8

Nonlinear Equation Solvers Bracketing Graphical Bisection False Position (Regula-Falsi) Open Methods Newton Raphson Secant All Iterative 9

Bisection Method

Basis of Bisection Method Theorem An equation f(x)=0, where f(x) is a real continuous function, has at least one root between xl and xu if f(xl) f(xu) < 0. Figure 1 At least one root exists between the two points if the function is continuous, and changes sign. ethods. eng. usf. edu http: //numericalm real, 11

Algorithm for Bisection Methods. eng. usf. edu http: //numericalm 12

Step 1 Choose xl and xu as two guesses for the root such that f(xl) f(xu) < 0, or in other words, f(x) changes sign between xl and xu. This was demonstrated in Figure 1 ethods. eng. usf. edu http: //numericalm 13

Step 2 Estimate the root, xm of the equation f (x) = 0 as the mid point between xl and xu as Figure 5 Estimate of xm ethods. eng. usf. edu http: //numericalm 14

Step 3 Now check the following a) If , then the root lies between xl and xm; then xl = xl ; xu = xm. b) If , then the root lies between xm and xu; then xl = xm; xu = xu. c) If ; if this is true. then the root is xm. Stop the algorithm ethods. eng. usf. edu http: //numericalm 15

Step 4 Find the new estimate of the root Find the absolute relative approximate error where ethods. eng. usf. edu http: //numericalm 16

Step 5 Compare the absolute relative approximate error tolerance. Is with the pre-specified error Yes Go to Step 2 using new upper and lower guesses. No Stop the algorithm ? Note one should also check whether the number of iterations is more than the maximum number of iterations allowed. If so, one needs to terminate the algorithm and notify the user about it. ethods. eng. usf. edu http: //numericalm 17

Example 1 Cont. Solve a) Use the bisection method of finding roots of equations to find the depth x b) Find the absolute relative approximate error at the end of each iteration, and the number of significant digits at least correct at the end of each iteration. ethods. eng. usf. edu http: //numericalm 18

Example 1 Cont. Solution To aid in the understanding of how this method works to find the root of an equation, the graph of f(x) is shown to the right, where Figure 7 Graph of the function f(x) ethods. eng. usf. edu http: //numericalm 19

Example 1 Cont. Let us assume Check if the function changes sign between xl and xu. Hence So there is at least on root between xl and xu, that is between 0 and 0. 11 ethods. eng. usf. edu http: //numericalm 20

Example 1 Cont. Figure 8 Graph demonstrating sign change between initial limits ethods. eng. usf. edu http: //numericalm 21

Example 1 Cont. Iteration 1 The estimate of the root is Hence the root is bracketed between xm and xu, that is, between 0. 055 and 0. 11. So, the lower and upper limits of the new bracket are At this point, the absolute relative approximate error calculated as we do not have a previous approximation. ethods. eng. usf. edu http: //numericalm cannot be 22

Example 1 Cont. Figure 9 Estimate of the root for Iteration 1 ethods. eng. usf. edu http: //numericalm 23

Example 1 Cont. Iteration 2 The estimate of the root is Hence the root is bracketed between xl and xm, that is, between 0. 055 and 0. 0825. So, the lower and upper limits of the new bracket are ethods. eng. usf. edu http: //numericalm 24

Example 1 Cont. Figure 10 Estimate of the root for Iteration 2 ethods. eng. usf. edu http: //numericalm 25

Example 1 Cont. The absolute relative approximate error at the end of Iteration 2 is None of the significant digits are at least correct in the estimate root of x m = 0. 0825 because the absolute relative approximate error is greater than 5%. ethods. eng. usf. edu http: //numericalm 26

Example 1 Cont. Iteration 3 The estimate of the root is Hence the root is bracketed between xl and xm, that is, between 0. 055 and 0. 06875. So, the lower and upper limits of the new bracket are ethods. eng. usf. edu http: //numericalm 27

Example 1 Cont. Figure 11 Estimate of the root for Iteration 3 ethods. eng. usf. edu http: //numericalm 28

Example 1 Cont. The absolute relative approximate error at the end of Iteration 3 is Still none of the significant digits are at least correct in the estimated root of the equation as the absolute relative approximate error is greater than 5%. Seven more iterations were conducted and these iterations are shown in Table 1. ethods. eng. usf. edu http: //numericalm 29

Table 1 Cont. Table 1 Root of f(x)=0 as function of number of iterations for bisection method. ethods. eng. usf. edu http: //numericalm 30

Table 1 Cont. (Scarborough, 1966) Hence the number of significant digits at least correct is given by the largest value or m for which So The number of significant digits at least correct in the estimated root of 0. 06241 at the end of the 10 th iteration is 2. ethods. eng. usf. edu http: //numericalm 31

Advantages • Always convergent • The root bracket gets halved with each iteration - guaranteed. ethods. eng. usf. edu http: //numericalm 32

Drawbacks n n n Slow convergence If one of the initial guesses is close to the root, the convergence is slower If there are some roots in the given interval then only one root are to be found. ethods. eng. usf. edu http: //numericalm 33

Drawbacks (continued) • If a function f(x) is such that it just touches the x-axis it will be unable to find the lower and upper guesses. ethods. eng. usf. edu http: //numericalm 34

Drawbacks (continued) n Function changes sign but root does not exist ethods. eng. usf. edu http: //numericalm 35

False-Position Method 2/22/2021 http: //numericalmethods. eng. usf. edu 36

Introduction (1) In the Bisection method (2) (3) 1 37 Figure 1 False-Position Method lmethods. eng. usf. edu ht

False-Position Method Based on two similar triangles, shown in Figure 1, one gets: (4) The signs for both sides of Eq. (4) is consistent, since: ethods. eng. usf. edu http: //numericalm 38

From Eq. (4), one obtains The above equation can be solved to obtain the next predicted root , as (5) ethods. eng. usf. edu http: //numericalm 39

The above equation, (6) or (7) ethods. eng. usf. edu http: //numericalm 40

Step-By-Step False-Position Algorithms 1. Choose that and as two guesses for the root such 2. Estimate the root, 3. Now check the following , then the root lies between (a) If ; then and (b) If and , then the root lies between ; then and ethods. eng. usf. edu http: //numericalm 41

, then the root is (c) If Stop the algorithm if this is true. 4. Find the new estimate of the root Find the absolute relative approximate error as ethods. eng. usf. edu http: //numericalm 42

where = estimated root from present iteration = estimated root from previous iteration 5. If else stop the algorithm. , then go to step 3, Notes: The False-Position and Bisection algorithms are quite similar. The only difference is the formula used to calculate the new estimate of the root shown in steps #2 and 4! http: //numericalm ethods. eng. usf. edu 43

Example 2 Cont. Solve a) Use the false-position method of finding roots of equations to find the depth x b) Find the absolute relative approximate error at the end of each iteration, and the number of significant digits at least correct at the end of each iteration. ethods. eng. usf. edu http: //numericalm 44

Let us assume Hence, ethods. eng. usf. edu http: //numericalm 45

Graphical Intepretation

Iteration 1 ethods. eng. usf. edu http: //numericalm 47

Iteration 2 Hence, ethods. eng. usf. edu http: //numericalm 48

Iteration 3 ethods. eng. usf. edu http: //numericalm 49

Hence, ethods. eng. usf. edu http: //numericalm 50

Table 1: Root of for False-Position Method. Iteration 1 0. 0000 0. 1100 0. 0660 N/A -3. 1944 x 10 -5 2 0. 0000 0. 0660 0. 0611 8. 00 1. 1320 x 10 -5 3 0. 0611 0. 0660 0. 0624 2. 05 -1. 1313 x 10 -7 4 0. 0611 0. 0624 0. 0632377619 0. 02 -3. 3471 x 10 -10 ethods. eng. usf. edu http: //numericalm 51

The number of significant digits at least correct in the estimated root of 0. 062377619 at the end of 4 th iteration http: //numericalm is 3. ethods. eng. usf. edu 52

Exercise • Get the root of the following equation. x 3 – x – 2 = 0 1. Done by using bisection and false-position method until 14 iteration. (use a paper) 2. Make a script (m-file) for each method using Matlab software. Then, submit your assignment into edmodo. com. 3. Note: please be careful due to the deadline is 2 weeks.

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