2.4 - Convergence of the Newton Method and Modified Newton Method

Transcription

1 2.4 - Convergence of the Newton Method and Modified Newton Method Consider the problem of finding the solution of the equation: for in Assume that is continuous and for in 1. Newton's Method: Suppose that is a simple zero of Then we know where In Section 2.2, the Newton Method is introduced for solving Algorithm Newton Method: newton.m Recall Given in and for (1) compute (2) if or then the algorithm is terminated and otherwise goto Step (1). Convergence and Rate of Convergence: In this section, we study the convergence of Newton's Method and will answer the following questions: (i) Does the Newton Method always converge? (ii) If Newton Method converges, what is the rate of convergence and what is the order of convergence? (iii) Does the Newton Method still converge when has multiple zeros at From Section 2.3, we know that the sequence of iterations Algorithm with function generated by Newton's Method can also be considered as iterations generated by the Fixed-Point By the Fixed-Point Theorem, the sequence of iterations converges to if where in and the rate of convergence is We now investigate under what condition the function has the property that for. We know that If for all in and for all in then there exists an interval such that Notes: Let and be continuous in If is in and then there exists a such that generated by Newton's method converges to for any in Note that: 1. Since the convergence of Newton method deps on locally. 2. where for all in Determine if the iterations generated by Newton's Method for solving Consider converges.

2 (1) and (2) Approximate by finding the solution of for within 10 Let Then. By the Fixed-Point Theorem, we can find an upper bound for the number of iterations needed: Let Use the MatLab program newton.m to find an approximation. x.^2-2; 2*x; >>[xsol, flag, ct]=newton(fun, fpfun,1,10^(-8),100) flag = 1 ct = 5 Find an approximation of the solution of the equation for in within Compare the convergence with the Bisection method and the Fixed point Iteration with Newton's Method: >>fun=@(x) x.^3+x+1; >>fpfun=@(x) 3*x.^2+1; >>[xsol,flag,ct]=newton(fun,fpfun,-1,10^(-5),100) flag =1 ct =5 Bisection method: >>bisect the left point a = -1 the right point b = 0 epsilon = 10^(-5) the approximation to the solution - cn = function value at the solution - fcn = e-005 number of iterations - Nit =17 Fixed-point Algorithm: >>fixpt input initial point p0 = -1

3 input the tolerance for stopping criterion, ex:.0001,10^(-8) 10^(-5) input the maximum number of iterations 100 Algorithm converges with number of iterations = 25 fixed point p = Order of convergence: Theorem Modified Fixed Point Theorem: If and Then quadratically with an asymptotic error constant Assume that but By the Taylor Theorem, we have for Then So, the order of convergence is 2 and the asymptotic error constant Theorem Let be generated by the Newton Method. If is a simple zero of then quadratically with an asymptotic error constant By the Modified Fixed-Point Theorem, we have Then the asymptotic error constant is 2. Zero of Multiplicity : Definition A solution of is a zeros of multiplicity of if for can be expressed as where Theorem Proof Let be continuously differentiable on. has a simple zero at in if and only if but Let have a simple zero at in By the definition, where Since Now let Since exists on for by the Taylor Theorem Since So, where Theorem Let have continuous derivatives on has a zero of multiplicity at if and only if

4 but Let has a zero at Determine the multiplicity of this zero. Three ways to determine 1. Use Theorem 2.4.4: find such that but Check: So, 2. Use the definition: find such that where That is So, 3. Use the definition and Taylor series of So, Modified Newton's Methods: Let be a zero of multiplicity of on Then we know Though as So, for large In this case, does still converge to (theoretically)? If so, what is the order of convergence? Let To answer the first question, we need to find out if So, still converges to However, since linearly instead of quadratically. If but then

5 Consider solving the equation: using Newton's Method with. log(x+1)-x; 1./(x+1)-1; >>[xsol,flag,ct]=newton(fun,fpfun,1,10^(-5),100) flag =1 ct =9 >> orderconv(xsol,0,9,1) Clearly, linearly. Since when and when is a zero of multiplicity 2 of Can we improve the convergence of Newton's Method for the problems where 1. Modified Newton's Methods: newtonm.m 2. where Let be a solution of For each method, we will check if a. and b. quadratically. Let us just check out the case where By the definition, but 1. Let and Then So, is a fixed point of Since converges to and converges to quadratically. 2. This method can also be considered as solving using Newton's Method where has a simple zero at

6 Since has a simple zero at and quadratically. Consider solving the equation: using both modified Newton's Methods with >>fun=@(x) log(x+1)-x; >>fpfun=@(x) 1./(x+1)-1; >>[xsol,flag,ct]=newtonm(fun,fpfun,1,10^(-5),100,2) flag = 1 ct = 4 >>orderconv(xsol,0,4,2) and Define and >>fun=@(x) -(log(x+1)-x).*(x+1)./x; >>fpfun=@(x) 1+(log(x+1)-x).*(x+1)./x.^2; >>[xsol,flag,ct]=newton(fun,fpfun,1,10^(-5),100) flag =1 ct =4 >> orderconv(xsol,0,4,2) Use the Newton Method to solve for in Determine the multiplicity of the zero of >>fun=@(x) exp(3*x)-27*x.^6+27*x.^4.*exp(x)-9*x.^2.*exp(2*x); >>fpfun=@(x) 3*exp(3*x)-162*x.^5+27*x.^3*(4+x).*exp(x)-18*x.*(1+x).*exp(2*x); >>[xsol,flag,ct]=newton(fun,fpfun,3,10^(-8),200) ct=24 xsol(ct)=

7 r1(k)=abs(xsol(k+1)-xsol(ct))/abs(xsol(k)-xsol(ct)); >>plot(r1) >>title('\x_{n+1}-x*\/x_n-x*\') r2(k)=abs(xsol(k+1)-xsol(ct))/(abs(xsol(k)-xsol(ct)))^2; >>plot(r2) >>title('\x_{n+1}-x*\/x_n-x*\^2') So, the Newton Method converges linearly and Now let and use the Modified Newton Method: >> [xsol,flag,ct]=newtonm(fun,fpfun,3,10^(-8),200,2); ct=14 xsol(14)= r3(k)=abs(xsol(k+1)-xsol(ct))/abs(xsol(k)-xsol(ct)); >>subplot(1,2,1),plot(r3) >>title('\x_{n+1}-x*\/x_n-x*\') r4(k)=abs(xsol(k+1)-xsol(ct))/(abs(xsol(k)-xsol(ct)))^2; >>subplot(1,2,2),plot(r4) >>title('\x_{n+1}-x*\/x_n-x*\^2') So, the Modified Newton Method converges linearly with Now let and use the Modified Newton Method again: >> [xsol,flag,ct]=newtonm(fun,fpfun,3,10^(-8),200,3); ct=9 xsol(9)= r5(k)=abs(xsol(k+1)-xsol(ct))/abs(xsol(k)-xsol(ct)); >>subplot(1,2,1),plot(r5) >>title('\x_{n+1}-x*\/x_n-x*\') r6(k)=abs(xsol(k+1)-xsol(ct))/(abs(xsol(k)-xsol(ct)))^2; >>subplot(1,2,2),plot(r6) >>title('\x_{n+1}-x*\/x_n-x*\^2')

8 Hence, the Modified Newton Method convergence quadratically and Exercises: 1. Let 1. Show first that has a zero of multiplicity 3 at 2. Use Newton Method to solve this equation with within 3. Verify using the graphs of the sequences and that the iterations obtained in (2) converge linearly. Estimate the asymptotic error constant. 4. Apply one modified Newton Method to solve this equation again with within 5. Verify using the graphs of the sequences and that the iterations obtained in (4) converge quadratically. 2. Let 1. Show first that has a zero of multiplicity 3 at 2. Use Newton Method to solve this equation with within 3. Verify using the graphs of the sequences and that the iterations obtained in (2) converge linearly. Estimate the asymptotic error constant. 4. Apply one modified Newton Method to solve this equation again with within 5. Verify using the graphs of the sequences and that the iterations obtained in (4) converge quadratically. 3. Let 1. Verify that is a zero of Find the multiplicity of the zero 2. Use Newton Method to find the zero of within with Use a proper Modified Newton Method to find the zero (quadratically) within with. 3. Use Newton Method to find another zero of within with What is the zero? Estimate the order of convergence and asymptotic error constant Is this zero simple? If not, find a proper Modified Newton Method to find this zero (quadratically) within with What is the multiplicity of this zero? 4. The sum of two real numbers is 20. If each number is added to its square root, the product of the two sums is Determine the two numbers to within by the Newton Method. (Ref [1]) 5. Consider the population equation: Approximate within using Newton Method. Report your result including and (Ref [1])