Root Finding Methods for Nonlinear Equations

Transcription

1 Root Finding Methods for Nonlinear Equations Dr. Sukanta Deb Department of Physics Cotton College State University College Hostel Road, Panbazar Guwahati, Pin Assam

2 1 Learning Objectives: Numerical Methods for the Root Finding

3 2 Solution of Algebraic and Transcendental Equations A variablexis said to be a root of the functionf(x) if f(x) = 0 (1) A root of this equation is also called a zero of the function. For example, the the two roots of a quadratic equation of the form are given by ax 2 +bx+c = 0 (2) x = b± b 2 4ac, (3) 2a where a,b,c are real numbers. The quatitity b 2 4ac is called the discriminant and is denoted by D. The value ofddetermines the nature of root of the equation (2). IfD = 0, then both the two roots are real and equal (repeated roots). IfD > 0, the roots are real and distinct. IfD < 0, then the roots are imaginary. Although the quadratic equation (2) can be solved exactly there exist many functions for which the exact solution does not exist. For an instance the function f(x) = e x x (4) cannot be solved analytically. For the solutions of those functions one has to resort to the approximate numerical methods (called iterative methods) which provides an efficient means to obtain the answer. Generically, an approximate solution of a function f(x) having any form can be obtained using the iterative method. Hence based on the above two cases, there are two methods that can be used to find the roots of the equation (1). 1. Direct methods: These methods provide the exact value of the roots without any round-off error) in a finite number of steps. All the roots can be determined at the same time using these methods. 2. Iterative methods: Iterative methods, also known as trial and error, are based on the idea of succesive approximations. These methods provide an approximate value of the roots. They start with one or more initial approximations and obtain a sequence of approximations by repeating a fixed sequence of steps till the result within a prescribed tolerance (or accuracy) is obtained. In this chapter, numerical methods for finding the roots nonlinear equations using the methods of iterations will be discussed.

4 3 Rate of convergence of an iterative method The rate of convergence of an iterative process is determined by the number of iterations taken to converge to the result. An iterative method is said to be of order p or has the rate of convergence p, if p is the largest positive number for which ǫ i+1 k ǫ i p, whereǫ i,ǫ i+1 are the absolute errors in thei th,(i+1) th iterations, respectively. The constant k is called the asymptotic error constant. The higher the value of p, the faster will be the rate of convergence. Ifp = 1, the convergence is linear. Ifp = 2, the convergence is quadratic and so on. The efficiency of an iterative procedure increases with smaller value of k. For the bisection method k = 0.5. Criteria for convergence One of the most important questions regarding the iterative methods is when the iteration should be stopped. The most obvious answer is that the procedure should be stopped when the error on the solution has been reduced to a prescribed tolerance. In other words, we say that the iteration should be stopped when the numerical algorithm has converged on a solution. Let ǫ be the prescribed tolerance, i.e., we would like to obtain the root with at most of ǫ. Given following are some of the commonly used criteria for convergence (a) f(x i ) < ǫ, whereǫis the tolerance limit close to zero. (b) x i+1 x i ǫ, whereǫis called the absolute tolerance. (c) x i+1 x i x i+1 ǫ, x i+1 0. where ǫ is called relative tolerance. The criteria (a) provides an way to determine if the convergence has reached by monitoring the value of the function for the solution at each and every iteration. if a zero of the function is sought, then the approach of the function towards zero can be used as the test for termination of the iterative method. On the other hand x i+1 >> 1, then condition (b) is more stringent. If x i+1 << 1, condition (c) gives more strict convergence crterion than condition (a). However, one has the option of applying one or multiple criteria for convergnce as deemed to be appropriate. Algebraic and Transcendental Equations 1. Albebraic Equations: An algebraic function of degree n in one variable is a function y = f(x) if it satisfies an equation of the form a 0 (x)+a 1 (x)y + +a n 1 (x)y n 1 +a n (x)y n = 0, (5) where a i s are ith order (i = 1,2,...,n) polynomials in x. Polynomials are a simple class of algebraic function that are represented genereally by f n (x) = a 0 +a 1 x+a 2 x 2 + +a n 1 x n 1 +a n x n, (6)

5 4 where n is the order of the polynomial and a i s are constants (i = 0,1,...,n) and are called the coefficients of the polynomial. 2. Transcendental Equation: A function which is nonalgebraic is called a transcendental function. These include trigonometric, exponential, logarithmic, and other less familiar functions. Examples are f(x) = logx 2 1 and f(x) = e 0.2x sin(3x 5). Although the roots of the algebraic and transendental equations may be either real or complex, we will be dealing only with the methods which give real roots in this course. Broadly, these two classes of equations fall into the category of nonlineear equations, since the functions belonging to either of these classes are nonlinear inx. This chapter focusses on finding the roots of nonlinear equations applying the rules of numerical techniques. Numerical techniques for solving nonlinear equations in one variable through iterative methods can be broadly classified into categories: (i) Bracketing methods. Examples include the bisection method and the regula-falsi method. (ii) open interval methods. Examples include Newton raphson method and the secant method. In bracketing methods, two initial guesses for the roots are required which must bracket the root, i.e, they should be on either side of the root. After each and every iteration, the width of the bracket which include the root is reduced until it approximates the correct answer. The bracketing method usually guarantees the convergence if the specified bracket contains the root. Brackting methods provide reliable soltions but converge slowly. But in numerical computation, faster methods of convergence are ususllay preferred. Open interval methods yield faster rate of converegence which explains their popularity. Nontheless, open interval methods do not always guarantee a solution. Fixed-Point Iteration Method A numberpis a fixed point of a functionφ(x) ifφ(p) = p. Let us suppose that the equationf(x) = 0 is written in the formx = φ(x), i.e.,f(x) = x φ(x) = 0. Then any fixed point p of φ(x) is a root of f(x) = 0, because f(p) = p φ(p) = 0. Thus a root of f(x) = 0 can be found by finding a fixed point ofx = φ(x), which corresponds tof(x) = 0. Finding a root off(x) = 0 by finding a fixed point ofx = φ(x) clearly suggests an iterative procedure of the following type. Let x 0 be an initial approximation to the root, and form a sequencex k defined by If the sequencex k converges, then x k+1 = φ(x k ),k = 0,1,2... lim x k = p k will be a root of off(x) = 0. Figure 3 shows the working of the fixed point iteration method.

6 5 Figure 1: Working of fixed point interation method Example 1. When does Fixed point iteration converge? Solution 1. The following theorem gives us a sufficient condition on φ(x) which ensures the convergence of the sequence x k. Theorem (1) (Fixed-Point Iteration Theorem). Let f(x) be written in the form x = φ(x). Let us assume that φ(x) has the following properties: 1. For allxin [a,b], φ(x) [a,b]; that isφ(x) takes every value between a and b. 2. φ (x) exists on(a,b) with the property that there exists a positive constantr < 1 such that φ (x) r, for all x in (a,b). Then (i) there is a unique fixed pointx = p ofφ(x) in [a,b]. (ii) For any pointx 0 in [a,b], the sequencex k defined by x k+1 = φ(x k ),k = 0,1,2... converges to the fixed point x = p; that is to the root p of f(x) = 0. The proof of this Theorem requires Mean Value Theorem of calculus: Let f(x) be a continuous function in [a,b], and be differentiable on (a,b). Then there is number c (a,b) such that f (c) = f(b) f(a). b a

7 6 Proof of the Fixed point Theorem: The proof comes in three parts: Existence, Uniqueness and Convergence. We first prove that there is a root of f(x) = 0 in [a,b] and it is unique. Since a fixed point of x = φ(x) is a root of f(x) = 0, this amounts to proving that there is a fixed point in[a,b] and it is unique. We then prove that the sequencex k+1 = φ(x) converges to the root. Existence: If a = φ(a), then a is a fixed point. If b = φ(b), then b is a fixed point. Since φ(x) [a,b] for all x [a,b], we have a φ(a) and φ(b) b. Thus, we have f(a) = φ(a) a 0 and f(b) = φ(b) b 0 Therefore, by the Intermediate Value Theorem to f(x) in [a,b], we conclude that there is a root of f(x) in[a, b]. This proves the existence. Uniqueness (by contradiction): To prove uniqueness, let us suppose thatp 1 andp 2 are two fixed points in[a,b], andp 1 p 2. Now, let us apply the MVT to φ(x). We can find a numbercin(p 1,p 2 ) such that Sinceφ(p 1 ) = p 1, andφ(p 2 ) = p 2, we have φ (c) = φ(p 2) φ(p 1 ) p 2 p 1. φ (c) = p 2 p 1 p 2 p 1. That is φ (c) = 1, which is a contradiction to our assumption 2. Thus, p 1 cannot be different from p 2. Therefore, p 1 = p 2. Convergence: Let p be the root off(x) = 0. Then the absolute error at step (k +1) is given by e k+1 = p x k+1. To prove that the sequence converges, we need to show that lim e k+1 = 0. k To show this, we apply the MVT to φ(x) in[x k,p]. Then φ (c) = φ(p) φ(x k) p x k,wherex k < c < p.

8 7 That is φ (c) = p x k+1 p x k (notethat φ(p) = p andx k+1 = φ(x)) Taking absolute values on both sides, we have φ (c) = p x k+1 p x k or φ (c) = e k+1 e k Sinceφ (c) r, we have e k+1 re k. Similarly, we have e k re k 1. Thus, e k+1 r 2 e k 1. Continuing in this way, we finally have e k+1 r k+1 e 0, where e 0 is the initial error (That is, e 0 = x 0 p ). Sincer < 1, we haver k+1 0 as k. Thus, lim e k+1 = lim x k+1 p k k lim k r k+1 e 0 =0. This proves that the sequencex k converges tox = p and x = p is the only fixed point ofφ(x). Algorithm 1. The algorithm for the fixed point iteration is given below Givenf(x) = 0,ǫ and an initial pointx 0 ; Givenmax=maximum number of iterations; Find a suitable equationx = φ(x) from f(x) = 0 where 1 φ (x) 1; Fori = 0 tomax Computex i+1 = φ(x i ); If x i+1 x i < ǫ; Solution=x i+1 ; Stop the iterations; Endif Endfor Example 2. Find the root off(x) = x 3 x 2 2 using the fixed-point iteration method, given the initial value ofx 0 = 1 with iterations until x i+1 x i < ǫ where, ǫ =

9 8 Table 1: Numerical results for the above problem i x i φ(x i ) φ (x i ) x i+1 x i Solution 2. Setting f(x) = 0 suggests several representations in the form of x = φ(x), including x = ± (x 3 2),x = 2 x 2 x and x = (x2 +2) 1/3. The first two functions are not suitable as they are not defined at x 0 = 1. Obviously φ(x) = (x 2 + 2) 1/3 is a good choice as φ 2x (x) = is a monotonically increasing 3(x 2 +2) 2/3 function with values between 0 and 1. Hencex i+1 = φ(x i ) = (x 2 i +2) 1/3 is a suitable choice here. For i = 0, x 1 = (x ) 1/3 = ( ) 1/3 = , with φ (x 0 ) = The error is x 1 x 0 = = > ǫ. For i = 1, x 2 = (x ) 1/3 = ( ) 1/3 = , with φ (x 1 ) = The error is x 2 x 1 = = > ǫ. Therefore, the process continues with the next iteration. The iterations stop ati = 5, where x 6 x 5 = < ǫ. The final solution is x = x 6 = Table 27 shows the complete results from this method. Example 3. Find a real root of the following equation correct to three decimal places using the fixed point iteration method: cosx = 3x 1. Solution 3. The given equation is cosx = 3x 1. Let us rewrite the given equation as x = 1 3 (cosx+1) x =φ(x),

10 9 Table 2: Numerical results for the above problem i x i φ(x i ) φ (x i ) x i+1 x i where φ(x) = 1 3 (cosx+1) φ (x) = sin(x) 3 φ (x) 1. Therefore, the fixed point interation method can be applied. Let us take the first approximation asx 0 = 1. Then the successive approximations of the root can be obtained as follows: x 1 =φ(x 0 ) = 1 [cos(1)+1] = x 2 =φ(x 1 ) = 1 [cos( )+1] = x 3 =φ(x 2 ) = 1 [cos( )+1] = x 4 =φ(x 3 ) = 1 [cos( )+1] = x 5 =φ(x 4 ) = 1 [cos( )+1] = x 6 =φ(x 5 ) = 1 [cos( )+1] = In the fifth and the sixth iterations, the successive approximations of the roots are obtained as and , respectively. Therefore, the required root of the equation correct to three decimal places is Table 2 shows the complete results from this method. Example 4. Find a real root of the following equation correct to two decimal places using the fixed point iteration method: xe x = 1.

11 10 Solution 4. The given equation is xe x = 1. Let us rewrite the given equation as x =e x x =φ(x), where φ(x) =e x φ (x) = e x φ (x) <1, x < 1. Therefore, the fixed point interation method can be applied. Let us take the first approximation as x 0 = 0.5. Then the successive approximations of the root can be obtained as follows: x 1 =φ(x 0 ) = exp( ) = x 2 =φ(x 1 ) = exp( ) = x 3 =φ(x 2 ) = exp( ) = x 4 =φ(x 3 ) = exp( ) = x 5 =φ(x 4 ) = exp( ) = x 6 =φ(x 5 ) = exp( ) = x 7 =φ(x 6 ) = exp( ) = In the sixth and the seventh iterations, the successive approximations of the roots are obtained as and , respectively. Therefore, the required root of the equation correct to three decimal places is Table 4 shows the complete results from this method. Example 5. Find a real root of the following equation correct to three decimal places using the fixed point iteration method: 2x log 10 x 7 = 0. Solution 5. The given equation is 2x log 10 x 7 = 0. Let us rewrite the given equation as x = 1 2 (log 10x+7) x =φ(x),

12 11 Table 3: Numerical results for the above problem i x i φ(x i ) φ (x i ) x i+1 x i where φ(x) = 1 2 (log 10x+7) φ (x) = log 10e 2x φ (x) <1, x < 1. Therefore, the fixed point interation method can be applied. Let us take the first approximation asx 0 = 1. Then the successive approximations of the root can be obtained as follows: x 1 =φ(x 0 ) = 1 2 [log 10( )+7] = x 2 =φ(x 1 ) = 1 2 [log 10( )+7] = x 3 =φ(x 2 ) = 1 2 [log 10( )+7] = x 4 =φ(x 3 ) = 1 2 [log 10( )+7] = x 5 =φ(x 4 ) = 1 2 [log 10( )+7] = In the fourth and the fifth iterations, the successive approximations of the roots are obtained as and , respectively. Therefore, the required root of the equation correct to three decimal places is Table 4 shows the complete results from this method. Example 6. Find a real root of the follwing equation correctto three decimaal places: sinx = 5x 2. Solution 6. Let us first express the given equation in the form x = φ(x) and then find out which of this form will give a convergent solution of the given equation.

13 12 Table 4: Numerical results for the above problem i x i φ(x i ) φ (x i ) x i+1 x i FormI : sinx =5x 2 x = sin 1 (5x 2) x = φ(x), where φ(x) = sin 1 (5x 2) φ 5 (x) =. 1 (5x 2) 2 Now, φ (x) > 1 x for which (5x 2) 2 < 1 x < 3 5 x < 0.6 Thus the method will not give a convergent solution. FormII : where sinx =5x 2 x = 1 5 (sinx+2) x =φ(x), φ(x) = 1 5 (sinx+2) φ (x) = 1 5 cosx φ (x) 1 < 1, x cosx 1. 5

14 13 Table 5: Numerical results for the above problem i x i φ(x i ) φ (x i ) x i+1 x i Therefore, we find that the φ(x) obtained from Form II will give a convergent solution. Therefore, we choose φ(x) = 1(sinx + 2). Let us take the first approximation as x 5 0 = 1. Then the successive approximations of the root can be obtained as follows: x 1 =φ(x 0 ) = 1 [sin( )+2] = x 2 =φ(x 1 ) = 1 [sin( )+2] = x 3 =φ(x 2 ) = 1 [sin( )+2] = x 4 =φ(x 3 ) = 1 [sin( )+2] = x 5 =φ(x 4 ) = 1 [sin( )+2] = In the fourth and the fifth iterations, the successive approximations of the roots are obtained as and , respectively. Therefore, the required root of the equation correct to three decimal places is Table 5 shows the complete results from this method. //C++ program to find root by fixed point iteration method using for loop //Written by Dr. Sukanta Deb, ANDC, University of Delhi #include < iostream > #include < cmath > #include < iomanip > #define iter 30 using namespace std; double f(double x) {double p, d; d = pow(x,2.0)+2; p = pow(d,1./3);

15 14 return p; } double g(double x) { double r,s; s = pow(x,2.0)+2; r = (2 x)/(3 pow(s,2./3)); return r; } int main() { int i,n; double t,x = 1.0; double eps = 0.005; cout.precision(6); cout.setf(ios :: fixed); for(i = 0;i < iter;i++) { t = x; x = f(x); if(fabs(x t) < eps) break; } cout << The solutionis : << endl; cout << setw(10) << x; cout << endl; return 0; } The same program is implemented below using the do-while loop. //C++ program to find root by fixed point iteration method using do-while loop //Written by Dr. Sukanta Deb, ANDC, University of Delhi #include < iostream > #include < cmath > #include < iomanip > using namespace std; double f(double x) { double p,d;

16 15 d = pow(x,2.0)+2; p = pow(d,1./3); return p; } int main() { double t,x = 1.0; double eps = 0.005; cout.precision(6); cout.setf(ios :: fixed); do { t = x; x = f(x); }while(fabs(t x) > eps); cout << x = << x; cout << endl; return 0; } Bisection Method (Bolzano Method) In this topic and in the other forthcoming topics of root finding, we will need to use one or a combination of the following theorems. These theorems find their importance in the root finding methods such as bisection in particular and other methods. Theorem (2) (Intermediate Mean-Value Theorem). Let f(x) be continuous on [a,b] and let k be any number between f(a) and f(b). Then there exists a number x [a, b] such that k = f(x). Theorem (3) (Mean-Value Theorem). Let f(x) be a continuous function defined on [a, b], such that f(a)f(b) < 0 (i.e., f(a) and f(b) are of opposite sign). Then there exists at least one root c [a,b] of f(x) = 0. Theorem (4) (Rolle s Theorem). If (i) f(x) is continuous in[a, b], (ii)f (x) exists in(a,b), and (iii)f(a) = f(b) = 0, then, there exists at least one value ofx, saycsuch that f (c) = 0,wherec (a,b).

17 16 Theorem (5) (Mean Value Theorem for Derivatives). If (i) f(x) is continuous in[a, b], (ii)f (x) exists in(a,b), and then, there exists at least one value ofx, saycsuch that f (c) = f(b) f(a),where c (a,b). b a Bisection method is one of the simplest iterative methods for nonlienar root finding. This method is based on the mean value theorem 3 which states that if a function f(x) is continuous in [x l,x u ] and f(x l )f(x u ) < 0 (i.e., f(x l ) and f(x u ) are of opposite sign). Then there exists at least one root x m [x l,x u ] off(x) = 0. As the title suggests, the method is based on repeated bisections of an interval containing the root. Let us suppose that f(x l ) and f(x u ) are such that f(x l ) f(x u ) < 0. This ensures that the root lies between x l and x u. Then bisect the interval [x l,x u ] and let x m = x l+x u 2 be the middle point of [x l,x u ]. Then, there are three possibilities that can aries (i) Iff(x m ) = 0, then x m is the root. (ii) Iff(x l ) f(x m ) < 0, then the root lies in the interval(x l,x m ). Thenx l = x l ;x u = x m. Therefore, the new interval now becomes[x l,x m ] which is half of the current interval is again bisected. (iii) Iff(x l ) f(x m ) > 0, then the root lies in the interval(x m,x u ). Thenx l = x m ;x u = x u. Therefore, the new interval now becomes[x m,x u ] which is again bisected. Therfore, by repeating this interval bisection procedure, we keep on enclosing the root in a new search interval, which is halved in each iteration. This iterative cycle is terminated when the search interval becomes smaller that the prescribed tolerance or the value of the function nearly vanishes at the new x value. Let ǫ be the prescribed tolerance in the required root. Then, the iterative cycle terminates when the absolute error becomes less than or equal to the prescribed tolerance, i.e., x u x l ǫ Example 7. How many iterations are needed in order that the length interval is less than ǫ in bisection method? Solution 7. Let L 0 = x u x l be the initial search interval that contains the root. After bisecting the intervalntimes, the search interval is reduced to the length given by L n = L 0 2 n.

18 Bisection method f(x u ) f(x) 10 0 x l x m -10 f(x m ) f(x l ) x x u Figure 2: Root finding using the bisection method. We require L n ǫ L 0 2 n ǫ (x u x l ) 2 n ǫ Taking logarithm on both sides of the inequality, we get [ ] (xu x l ) log e log 2 n e ǫ log e (x u x l ) log e ǫ nlog2 n log e [ ] (xu xl ) ǫ log e 2 Example 8. What is the minimum number of iterations needed in the bisection algorithm, givenx l = 2.5, x u = 3.5 and ǫ = Solution 8. We know that the minimum number of iterations needed to obtain a root of desired accuracy is given by n [ ] log (xu x l ) e ǫ log e 2 = 10

19 18 Table 6: Minimum number of iterations required ǫ n Let us assume that the root lies in the interval (1,2), then the minimum number of iterations required to find the root within the prescribed toleranceǫare listed in Table 6. Algorithm 1. The algorithm for the bisection method is given below Givenf(x) = 0,ǫ and the initial end points[x l,x u ], wheref(x l )f(x u ) < 0; Givenmax=maximum number of iterations; Fori = 0 tomax Computex m = (x l+x u) 2.0 ; Iff(x l )f(x m ) < 0 Updatex l = x l andx u = x m ; Iff(x l )f(x m ) > 0 Updatex l = x m and x u = x u ; If x u x l < ǫ Solution =x m ; Stop the iterations; Endfor Example 9. Find a real root of the following equation correct upto two decimal places: x 3 2x 5 = 0. Solution 9. Let f(x) = x 3 2x 5 = 0. Let us try to calculate f(x) for the values of x given in Table (16). It can be seen from the table that the interval [2,3] gives the accurate and precise location of the root among the other two intervals such as[0,3],[1,3]. Therfore, we choosex l = 2 andx u = 3 as the initial two guesses for bracketing the root. First Iteration : The interval [x l,x u ] = [2,3] and f(x l ) = f(2) = 1 < 0,f(x u ) = f(3) = 16 > 0. x u x l = 3 2 = 1. Therefore, x m = (x l +x u ) = (2+3) = f(x m ) = f(2.50) = > 0. f(2) f(2.5) < 0 Root lies between[2,2.50].

20 19 Table 7: This table shows how to select the bracketing interval [a,b] so as to include the root of the equation x 3 2x 5 = 0. It can be seen that the interval [2,3] gives the accurate and precise location of the root among the other two intervals such as [0,3],[1,3]. Therfore, we choosex l = 2 and x u = 3 as the initial two guesses for bracketing the root. x f(x) = x 3 2x 5 sign[f(x)] 0 5 < < < > 0 Second Iteration : The interval [x l,x u ] = [2,50] and f(x l ) = f(2) = 1 < 0,f(x u ) = f(2.50) = > 0. x u x l = = 0.5. Therefore, Now x m = (x l +x u ) = (2+2.5) = f(x m ) = f(2.25) = > 0. f(2) f(2.25) < 0 Root lies between[2,2.25]. Third Iteration : The interval [x l,x u ] = [2,2.25] and f(x l ) = f(2) = 1 < 0,f(x u ) = f(2.25) = > 0. x u x l = = 0.25 Now, x m = (x l +x u ) = (2+2.25) = f(x m ) = f(2.125) = > 0. f(2) f(2.125) < 0 Root lies between [2,2.1250]. FourthIteration : The interval [x l,x u ] = [2,2.1250] and f(x l ) = f(2) = 1 < 0,f(x u ) = f(2.1250) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(2.0625) = < 0. f(2.0625) f(2.125) < 0 Root lies between [2.0625,2.1250]. FifthIteration : The interval [x l,x u ] = [2.0625,2.1250] and f(x l ) = f(2) = 1 < 0,f(x u ) =

21 20 f(2.1250) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(2.0938) = < 0. f(2.0938) f(2.125) < 0 Root lies between [2.0938,2.1250]. Sixth Iteration : The interval [x l,x u ] = [2.0938,2.1250] and f(x l ) = f(2.0938) = < 0,f(x u ) = f(2.1250) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(2.1094) = > 0. f(2.0938) f(2.1094) < 0 Root lies between [2.0938,2.1094]. Seventh Iteration : The interval[x l,x u ] = [2.0938,2.1094] andf(x l ) = f(2.0938) = < 0,f(x u ) = f(2.1094) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(2.1016) = > 0. f(2.0938) f(2.1016) < 0 Root lies between [2.0938,2.1016]. Eighth Iteration : The interval [x l,x u ] = [2.0938,2.1016] and f(x l ) = f(2.0938) = < 0,f(x u ) = f(2.1016) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(2.0977) = > 0. f(2.0938) f(2.0977) < 0 Root lies between [2.0938,2.0977]. Ninth Iteration : The interval [x l,x u ] = [2.0938,2.0977] and f(x l ) = f(2.0938) = < 0,f(x u ) = f(2.0977) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(2.0957) = > 0. f(2.0938) f(2.0957) < 0 Root lies between [2.0938,2.0957].

22 21 Tenth Iteration : The interval [x l,x u ] = [2.0938,2.0957] and f(x l ) = f(2.0938) = < 0,f(x u ) = f(2.0957) = > 0. x u x l = = Now, x m = (x l +x u ) 2 = ( ) 2 = Iteration x l x u x m f(x l ) f(x u ) f(x m ) x u x l Example 10. Find a real root of the following equation correct using ǫ = if the root lies in the interval[0,1]: x = cosx. Solution 10. Let f(x) = x cosx. Heref(0) = and f(b) = First Iteration : The interval[x l,x u ] = [0,1] andf(x l ) = f(0) = < 0,f(x u ) = f(1) = > 0. x u x l = 1 0 = 1. Therefore, x m = (x l +x u ) = (0+1) = f(x m ) = f(0.5000) = < 0. f(0.5000) f(1.0000) < 0 Root lies between [0.5000,1.0000]. Second Iteration : The interval[x l,x u ] = [0.5000,1.0000] andf(x l ) = f(0.5000) = < 0,f(x u ) = f(1.0000) = > 0. x u x l = = Therefore, Now x m = (x l +x u ) = ( ) = f(x m ) = f(0.7500) = > 0. f(0.5000) f(0.7500) < 0 Root lies between [0.5000,0.7500].

23 22 Third Iteration : The interval [x l,x u ] = [0.5000,0.7500] and f(x l ) = f(0.5000) = < 0,f(x u ) = f(0.7500) = > 0. x u x l = = 0.25 Now, x m = (x l +x u ) = ( ) = f(x m ) = f(0.6250) = < 0. f(0.6250) f(0.7500) < 0 Root lies between [0.6250,0.7500]. FourthIteration : The interval [x l,x u ] = [0.6250,0.7500] and f(x l ) = f(0.6250) = < 0,f(x u ) = f(0.0183) => 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(0.6875) = < 0. f(0.6875) f(0.7500) < 0 Root lies between [0.6875,0.7500]. FifthIteration : The interval [x l,x u ] = [0.6875,0.7500] and f(x l ) = f(0.6875) = < 0,f(x u ) = f(0.7500) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(0.7188) = < 0. f(0.7188) f(0.7500) < 0 Root lies between [0.7188,0.7500]. Sixth Iteration : The interval [x l,x u ] = [0.7188,0.7500] and f(x l ) = f(0.7188) = < 0,f(x u ) = f(0.7500) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(0.7344) = > 0. f(0.7344) f(0.7500) < 0 Root lies between [0.7344,0.7500]. Seventh Iteration : The interval[x l,x u ] = [0.7344,0.7500] andf(x l ) = f(0.7344) = <

24 23 0,f(x u ) = f(0.7500) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(0.7422) = > 0. f(0.7344) f(0.7422) < 0 Root lies between [0.7344,0.7422]. Eighth Iteration : The interval [x l,x u ] = [0.7344,0.7422] and f(x l ) = f(0.7344) = < 0,f(x u ) = f(0.7422) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(0.7383) = < 0. f(0.7383) f(0.7422) < 0 Root lies between [0.7383,0.7422]. Ninth Iteration : The interval [x l,x u ] = [0.7383,0.7422] and f(x l ) = f(0.7383) = < 0,f(x u ) = f(0.7422) = > 0. x u x l = = Now, x m = (x l +x u ) = ( ) = f(x m ) = f(0.7402) = > 0. f(0.7383) f(0.7422) < 0 Root lies between [0.7383,0.7402]. Tenth Iteration : The interval [x l,x u ] = [0.7383,0.7402] and f(x l ) = f(0.7383) = < 0,f(x u ) = f(0.7402) = > 0. x u x l = = Now, x m = (x l +x u ) 2 = ( ) 2 = Newton-Raphson Method The Newton-Raphson method is one of the most powerful and well-known numerical methods for solving a root-finding problem. Let x 0 be an initial approximation of the root to f(x) = 0, whose real root is α(say). Then α = x 0 +h, where h is the small correction to be applied to x 0 to give the exact value of the root. Therefore, f(x 0 )+hf (x 0 )+ h2 2! f(α) = 0 f(x 0 +h) = = 0 (By Taylor series expansion)

25 24 Iteration x l x u x m f(x l ) f(x u ) f(x m ) x u x l Sincehis quite small, the higher powers ofh, i.e.,h 2,h 3,etc., can be neglected. Therefore f(x 0 )+hf (x 0 ) 0 h = f(x 0) f (x 0 ). Substituting this value ofhinα = x 0 +h, we get a better approximation to the rootαoff(x) = 0 as The successive approximations are x 1 = x 0 f(x 0) f (x 0 ). x 2 = x 1 f(x 1) f (x 1 ) x 3 = x 3 f(x 3) f (x 3 )... x n+1 = x n f(x n) f (x n ) This iterative formula is known as the Newton Raphson method. Newton Raphson method: Geometrical Significance The Newton Raphson method is an open interval method that requires only one initial guess. Let the initial guess at the root be x 0. A tangent line is drawn from the point (x 0,f(x 0 )) which intersects the x-axis at (x 1,0). The value x = x 1 is the new value at iteration i = 1. At i = 2, another tangent is drawn from (x 1,f(x 1 )) which intersects the x-axis at (x 2,0) to produce an improved value of x = 2.

26 25 The same step is repeated for i = 3,4... until the error e = x i+1 x i is smaller then a preset value ǫ. The final value ofxobtained at the last iteration wheree < ǫ becomes the final root of the equation. The computational steps in the Newton-raphson method are summarized as follows. The algorithm requires calculations of bothf(x) andf (x). Algorithm 1. The algorithm for the Newton-Raphson method is given below Givenf(x) = 0,ǫ and the initial pointsx 0 ; Givenmax=maximum number of iterations; Find f (x); Fori = 0 tomax Computex i+1 = x i f(x i) f (x i ) ; If x i+1 x i < ǫ Solution =x i+1 ; Stop the iterations; Endif Endfor 20 Newton-Raphson method 10 f(x) 0 (x 2,f(x 2 )) (x 1,f(x 1 )) -10 (x 0,f(x 0 )) x Figure 3: Working of Newton Raphson method

27 26 Example 11. Find a real root of the following equation using the Newton-Raphson method for ǫ = using the initial approximation of the root as x 0 = 2. 3x cosx 1 = 0 (7) Solution 11. Let f(x) = 3x cosx 1. Therefore, f (x) = 3+sinx. Givenx 0 = 2. First Iteration : x 0 = ,f(x 0 ) = ,f (x 0 ) = We know from the Newton- Raphson method that x =x 0 f(x 0) f (x 0 ) x = x = Second Iteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = Third Iteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = From the second and the third iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to four decimal places is The successive approximations of the root as computed above using the Newton-Raphson method are displayed in the following table: Example 12. Find a real root of the following equation correct upto five decimal places ln(x) = cos(x). Solution 12. Here f(x) = ln(x) cos(x). Therefore, f (x) = 1 +sin(x). x

28 27 Table 8: Numerical results for the above problem i x i f(x i ) f (x i ) h = f(x i) f (x i ) x i+1 x i+1 x i First Iteration : x 0 = 2.0, f(x 0 ) = , f (x 0 ) = We know from the Newton- Raphson method that x =x 0 f(x 0) f (x 0 ) x = x = Second Iteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = Third Iteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = FourthIteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = From the third and the fourth iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to four decimal places is

29 28 The successive approximations of the root as computed above using the Newton-Raphson method are displayed in the following table: Example 13. Find the cubic root of10, i.e., 3 10 correct to four decimal places. Solution 13. Let x = Thenx 3 10 = 0. Let f(x) = x 3 10 f (x) = 3x 2. First Iteration : x 0 = 2.0, f(x 0 ) = , f (x 0 ) = We know from the Newton- Raphson method that x =x 0 f(x 0) f (x 0 ) x = x = Second Iteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = Third Iteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = FourthIteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = From the third and the fourth iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to four decimal places is The successive approximations of the root as computed above using the Newton-Raphson method are displayed in the following table:

30 29 Table 9: Numerical results for the above problem i x i f(x i ) f (x i ) h = f(x i) f (x i ) x i+1 x i+1 x i Thusx = is a root off(x) = 0 correct upto five decimal places. Example 14. Find a real root of the following equation correct to three decimal places using the Newton- Rephson method taking the initial approximation to the root as x 0 = 2: x 3 2x 5 = 0. Solution 14. Givenf(x) = x 3 2.0x 5, x 0 = 2. Therefore, f (x) = 3x First Iteration : x 0 = 2.0, f(x 0 ) = x x 0 5 = = 1, f (x 0 ) = 3x = 10 We know from the Newton-Raphson method that x =x 0 f(x 0) f (x 0 ) x = x = Second Iteration : x 0 = x = , f(x 0 ) = x x 0 5 = ,f (x 0 ) = 3x = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = Third Iteration : x 0 = x = , f(x 0 ) = x 3 0 2x 0 5 = , f (x 0 ) = 3x = Therefore, x =x 0 f(x 0) f (x 0 ) x = x =

31 30 From the second and the third iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to three decimal places is The successive approximations of the root as computed above using the Newton-Raphson method are displayed in the following table: Table 10: Numerical results for the above problem i x i f(x i ) f (x i ) h = f(x i) f (x i ) x i+1 x i+1 x i Example 15. Find a real root of the following equation correct to three decimal places using the Newton- Rephson method taking the initial approximation to the root as x 0 = 2: x x +2x 6 = 0. Solution 15. Givenf(x) = x x +2x 6, x 0 = 2. Therefore, f (x) = x x (1+lnx)+2. First Iteration : x 0 = 2.0, f(x 0 ) = , f (x 0 ) = We know from the Newton- Raphson method that x =x 0 f(x 0) f (x 0 ) x = x = Second Iteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = Third Iteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x =

32 31 FourthIteration : x 0 = x = ,f(x 0 ) = ,f (x 0 ) = Therefore, x =x 0 f(x 0) f (x 0 ) x = x = From the third and the fourth iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to three decimal places is The successive approximations of the root as computed above using the Newton-Raphson method are displayed in the following table: Table 11: Numerical results for the above problem i x i f(x i ) f (x i ) h = f(x i) f (x i ) x i+1 x i+1 x i Secant Method One of the important difficulties with the Newton Raphson method is that it requires the computation of the derivative of f(x) at the iterated points. This requirement causes hardship to the computer by causing computational cost in solving the problem. The secant method helps eliminating the calculation of the derivative by replacing it with a secant line. The term secant denotes a straight line that intersects a curve in two or more parts. In secant method, two initial guesses say, x 1 and x 2 are provided, where f(x 1 ) f(x 2 ). These two points can be any points which are reliably close to the solution and they do not necessarily have to be on the opposite sides of thex-axis. A good choice for the two initial pointsx 1 andx 2 will bef(x 1 f(x 2 ) < 0 abiding by the mean value theorem. The first iteration with i = 1 produces an approximation at x = x 2 which is thex-intercept of the chord that passes through the points(x 1,f(x 1 )) and(x 2,f(x 2 )). The next approximated value x = x 3 is obtained from the x-intercept of the chord that passes through the points (x 2,f(x 2 )) and (x 3,f(x 3 )). The same step is repeated until the convergence to the solution is obtained through the stopping criteriae = x i+1 x i < ǫ. The secant method is derived as follows: The gradient of the line from(x 1,f(x 1 )) to (x 3,0) is m 1 = f(x 1 0) x 1 x 3.

33 32 which is equal to the the gradient of the line from (x 2,f(x 2 ) to(x 3,0) given by Therefore, m 2 = f(x 1 0) x 1 x 3 m 2 = f(x 2) 0 x 2 x 3. f(x 1 )(x 2 x 3 ) = f(x 2 )(x 1 x 3 ) x 3 = x 1f(x 2 ) x 2 f(x 1 ). f(x 2 ) f(x 1 ) Replacing x 1 with x i, x 2 with x i+1 and x 3 with x i+2 the iterative formula for the secant method can be written as: x i+2 = x if(x i+1 ) x i+1 f(x i ). f(x i+1 ) f(x i ) Algorithm 1. The algorithm for the secant method is given below Givenf(x) = 0,ǫ and the initial points[x 1,x 2 ]; Givenmax=maximum number of iterations; Fori = 0 tomax Computex i+2 = x if(x i+1 ) x i+1 f(x i ) f(x i+1 ) f(x i ) ; If x i+2 x i+1 < ǫ Solution =x i+2 ; Stop the iterations; Endif Endfor Example 16. Find the root of the following equation using the secant method with the initial values x 1 = 1 andx 2 and accuracy ǫ = x 3 x 2 2 = 0. Solution 16. Let f(x) = x 3 x 2 2. Givenx 1 = 1 and x 2 = 2. First Iteration : x 1 = 1, x 2 = 2. Therefore, f(x 1 ) = , f(x 2 ) = We know from the secant method that x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x 3 = ( ) ( ) x 3 =

34 33 Second Iteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x 3 = ( ) ( ) x 3 = Third Iteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = FourthIteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = FifthIteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = From the fourth and the fifth iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to three decimal places is The successive approximations of the root as computed above using the Newton-Raphson method are displayed in the following table:

35 34 Table 12: Numerical results for the above problem i x i x i+1 x i+2 f(x i ) f(x i+1 ) f(x i+2 ) x i+2 x i Secant method 2 f(x u ) 1 f(x) 0 x l x m x u -1 f(x m ) -2 f(x l ) x Figure 4: Root finding using the secant method. Example 17. Compute the root of the following equation in the interval [0,2] using the secant method correct upto three decimal places: x 2 e x 2 1 = 0. Solution 17. Let f(x) = x 2 e x 2 1. Givenx 1 = 0 andx 2 = 2. First Iteration : x 1 = 0, x 2 = 2. Therefore, f(x 1 ) = , f(x 2 ) = We know

36 35 from the secant method that x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x 3 = ( ) ( ) x 3 = Second Iteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x 3 = ( ) ( ) ( ) ( ) x 3 = Third Iteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = FourthIteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = From the third and the fourth iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to three decimal places is (after rounding off). The successive approximations of the root as computed above using the secant method are displayed in the following table: Example 18. Compute the root of the following equation in the interval [2,3] using the secant method correct upto three decimal places: x 3 2x 5 = 0.

37 36 Table 13: Numerical results for the above problem i x i x i+1 x i+2 f(x i ) f(x i+1 ) f(x i+2 ) x i+2 x i Solution 18. Let f(x) = x 3 2x 5. Givenx 1 = 2 andx 2 = 3. First Iteration : x 1 = 2, x 2 = 3. Therefore, f(x 1 ) = ,f(x 2 ) = We know from the secant method that x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x 3 = ( ) ( ) x 3 = Second Iteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x 3 = ( ) ( ) ( ) ( ) x 3 = Third Iteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = FourthIteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = ,

38 37 f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = From the third and the fourth iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to three decimal places is The successive approximations of the root as computed above using the secant method are displayed in the following table: Table 14: Numerical results for the above problem i x i x i+1 x i+2 f(x i ) f(x i+1 ) f(x i+2 ) x i+2 x i Example 19. Compute the root of the following equation in the interval [1,2] using the secant method correct upto three decimal places: x 4 x 10 = 0. Solution 19. Let f(x) = x 4 x 10. Givenx 1 = 1 andx 2 = 2. First Iteration : x 1 = 1, x 2 = 2. Therefore, f(x 1 ) = ,f(x 2 ) = We know from the secant method that x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x 3 = ( ) ( ) ( ) x 3 = Second Iteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = ,

39 38 f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x 3 = ( ) ( ) ( ) ( ) x 3 = Third Iteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = FourthIteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = FifthIteration : x 1 = x 2 = , x 2 = x 3 = Therefore, f(x 1 ) = , f(x 2 ) = Therefore, x 3 = x 1f(x 2 ) x 2 f(x 1 ) f(x 2 ) f(x 1 ) x = ( ) ( ) ( ) x 3 = From the fourth and the fifth iteration, we find that the successive approximations of the root are and Hence the required root of the given equation correct to three decimal places is The successive approximations of the root as computed above using the secant method are displayed in the following table: