University of Delaware Department of Mathematical Sciences Math 353 Engineering Mathematics III 06S C. Bacuta

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1 University of Delaware Department of Mathematical Sciences Math 353 Engineering Mathematics III 06S C. Bacuta Homework 4: Due Friday, 0/4/06, 10:00am Part I) (1) Section., Problems #3(a,d), 9, 11. () Apply the method of bisection 3 times by hand for #3(d) (Section.) for the interval [a 0, b 0 ] = [3, 4]. (3) Section.4, Problems #4, 5, 9. Note: (1) The answer in the back of the book is wrong for.4.5; it should be a cotangent function not a cosine. () It is fine to use one of the simple codes from the course web page or from your text and Matlab to solve the appropriate parts of these problems. However, I recommend strongly that you be able to do them by hand so that you can do so on an exam if so asked. II) For this part record a diary file showing your Matlab work. Include (paper) copies of script or function files written by you. (1) Download a bisection method code from the course web page and use it to solve the equation x = e x () Use newton.m function or another function or script file from the course web page to approximate the root r of f(x) = tan(x) x on the interval [7, 8]. Check that (N-R) iterations are not convergent if the initial guess is p 0 = 7 or p 0 = 8. Find two distinct values a, b (7, 8), such that b a 1/4 and (N-R) iterations converge for both choices of the initial guess p 0 = a and p 0 = b. Comment the result. (3) Modify newton.m function or write a new function called convorder.m to check that the Newton-Raphson iterations have quadratic convergence for simple roots and linear convergence for double roots. Consider the exact root as an input argument for your function. Use convorder.m to check the order of convergence for the Newton-Raphson iterations when solving for the root r = of the function f(x) = x 3 3x ( problem.4.4. ). Repeat the experiment with the root r = 1. Find estimates for the constant A (see Definition.5, page 75.) Solutions, hints and answers. Part I). Section., Problem 3(a): Plotting this function in Matlab with the commands x = linspace(-3,3); y = exp(x)--x; plot(x,y,[-3 3],[0 0]) shows that there are simple roots in the intervals [, 1] and [1, ]. The last two vectors draw a horizontal line at y = 0 to show where the function crosses y = 0. Then check (by hand or with MATLAB) that f( ) > 0, f( 1) < 0 and f(1) < 0, f() > 0. 1

2 Section., Problem 3(d): Plotting this function in Matlab with the commands x = linspace(0,10); y = x.^ - 10*x + 3; plot(x,y,[0 10],[0 0]) shows that there are simple roots in the intervals [3, 4] and [6, 7]. The last two vectors draw a horizontal line at y = 0 to show where the function crosses y = 0. Check that f(3) f(4) < 0. Section., Problem 9a: The method fails. The function does not change sign on the interval [3, 7]. Section., Problem 9b: The function does change sign on the interval [1, 7] but is not continuous on [1, 7]. Thus, Theorem.4 can not be applied. If the algorithm is used, then we still get that the sequences a n, b n converge to c =, but c = is a point of discontinuity for f and not a root for f. Section., Problem 11.: We can use equation (14), pg. 55, for this purpose. Let δ r c n be a given tolerance. Then from equation (14) we may write b 0 a 0 δ taking the natural log of both sides gives N+1 ; (N + 1) ln ln(b 0 a 0 ) ln δ. Solving for N and making it an integer gives N 1 + ceil((ln(b 0 a 0 ) ln δ)/ ln ), which is the same formula used in the code for the bisection method in the text. Using δ = , b 0 = 7 and a 0 =, one finds the samalest value N = 9. Here ceil converts its argument to the next largest integer. Problem I): (You can use Matlab to evaluate f). Starting with the interval [3,4], a 0 = 3, b 0 = 4, hence the first approximation is c 0 = a 0+b 0 = 3.5 f(a 0 ) =, f(c 0 ) = 0.5, f(b 0 ) = 1, hence a 1 = c 0 = 3.5, b 1 = b 0 = 4 and the second approximation is c 1 = a 1+b 1 = 3.75 f(a 1 ) =.5, f(c 1 ) =.4375, f(b 1 ) = 1, hence a = a 1 = 3.5, b = c 1 = 3.75 and the third approximation is c = a +b = 3.65 f(a ) =.5, f(c ) = , f(b ) = , hence a 3 = a = 3.5, b = c = 3.65 and c 3 = a 3+b 3 = is the fourth approximation of the root. Section.4, Problem 4.: f(x) = x 3 3x. (a) f (x) = 3x 3 and for Newton s method p k = p k 1 f(p k 1 )/f (p k 1 ) g(p k 1 ). For this problem, g(p k 1 ) = p k 1 (p 3 k 1 3p k 1 )/(3p k 1 3). (b) Some data including the iterates is given in the solution for II)3. (c) The sequence converges quadratically because f () 0 (r = is a simple root). Section.4, Problem 5.: f(x) = cos(x). (a) p k = p k 1 f(p k 1 )/f (p k 1 ) = p k 1 + cot(p k 1 ). (b) The initial guess p 0 = 3 is less than π and the slope of f(x) is negative there; the iterates will move towards the root at π/. This can be easily seen by sketching the tangent lines determined by Newton s method. (c) In this case, f (x) is positive and the iterates will decrease toward the root at 3π/. Again, this is easily seen graphically.

3 Section.4, Problem 9.: The iterates are determined by Part II). p k 1 p k p k = p k 1 f(p k 1 ) f(p k 1 ) f(p k ), and we find, with p 0 = 1.7 and p 1 = 1.67, that p = and p 3 = Problem II)1: We first need to write a.m file which gives the function f(x) = e x x called h4p1.m and stored in the working directory where the Matlab is opened. % function file for Homework 4, problem II)1 function y = h4p1(x); y=exp(-x)-x; Then, in the matlab command window, >>x=linspace(-1, 1); >> y=h4p1(x); >> plot(x,y,[-1 1], [0 0]) in order to see that the function has a root in the interval (0, 1). >> [r, er, y]=bisect( h4p1,0,1,10^-16) r = er = e-16 y = e-16 Here, r is the approximation to the root and y=f(r). Problem II): The files for f(x) = tan(x) x and the derivative f (x) = 1 cos (x) are: %Function file for Homework 4, Problem II: f(x) = tan(x) -x function y=h4p(x) y=tan(x)-x; %Function file for homework 4, Problem II: f (x) = 1/cos^(x) -1 function y=h4pd(x) y=1/(cos(x))^ -1; In the matlab command window: >>[r, er, k, y]=newton( h4p, h4pd,7, 10^-1, 10^-16, 100) r = e+66 er = e+65 k =100 y = e+66 proves that (N-R) iterations are not convergent if the initial guess is p0=7. The error er is huge, r is not in the interval (7, 8),

4 the maximum number of iterations is achieved and y=f(r) is also huge. Simply by slightly modifying the initial guess p0 we have: >> [r, er, k, y]=newton( h4p, h4pd,7.6, 10^-1, 10^-16, 100) r = er =0 k =14 y = e-14 which gives a good approximation to the root of f on (7, 8). After a few experiments (just changing the initial guess) we get [r, er, k, y]=newton( h4p, h4pd,7.851, 10^-1, 10^-16, 100) r = er =0 k =1 y = e-14 Thus, we can take a= 7.6 and b= Comment: N-R Method converges fast if the initial guess p0 is chosen close enough to the actual root. Problem II)3: My convorder function to check for the order of convergence for N-R Method. function convorder(f,df,p0,r, delta,epsilon,max1) % modified by CB % Checks if the Newton method has quadratic or linear convergence. %Input - f is the object function input as a string f % - df is the derivative of f input as a string df % - p0 is the initial approximation to a zero (root) of f % - r is the exact root % - delta is the tolerance for p0 % - epsilon is the tolerance for the function values y % - max1 is the maximum number of iterrations %Prints: k, pk, E(k)= p-pk, Q(k)= E(k)/E(k-1)^ and L(k) = E(k)/E(k-1) % to check for quadratic or linear convergence % Only the lines: 19, and 30 have been added. E0=abs(r-p0); disp(sprintf( k= 0 p=%7.8e f(p)=%7.8e E=%7.8e, p0, feval(f,p0), E0)) for k=1:max1 p1=p0-feval(f,p0)/feval(df,p0); err=abs(p1-p0); relerr=*err/(abs(p1)+delta); E1=abs(r-p1); Q=E1/E0^; L=E1/E0; disp(sprintf( k=%.0f p=%7.8e f(p)=%7.8e E=%7.8e Q=%7.8e L=%7.8e,k, p1, feval(f,p1), E1, Q, L))

5 p0=p1; E0=E1; y=feval(f,p0); if (err<delta) (relerr<delta) (abs(y)<epsilon),break,end end The files for $f(x)= x^3-3x -$ and the derivative $f (x)= 3 x^ -3$ are: %Function file for homework 4, Problem II)3: f(x) =x^3-3x-. function y=h4p3(x) y= x.^3-3*x -; %Function file for homework 4, Problem II)3: f (x) =3x^ -3 function y=h4p3d(x) y= 3*x.^ -3; To check that the convergence to the simple root r = is quadratic: >> convorder( h4p3, h4p3d,5,, 10^-8, 10^-10, 100) k= 0 p=5.0000e+00 f(p)=1.0800e+0 E= e+00 k= 1 p=3.5000e+00 f(p)=3.0375e+01 E= e+00 Q=1.6667e-01 L=5.0000e-01 k= p=.6000e+00 f(p)=7.7760e+00 E= e-01 Q=.6667e-01 L=4.0000e-01 k= 3 p=.1500e+00 f(p)=1.4884e+00 E= e-01 Q=4.1667e-01 L=.5000e-01 k= 4 p=.0130e+00 f(p)=1.1841e-01 E= e-0 Q=5.7971e-01 L=8.6957e-0 k= 5 p=.0001e+00 f(p)=1.0077e-03 E= e-04 Q=6.5808e-01 L=8.5837e-03 k= 6 p=.0000e+00 f(p)=7.503e-08 E= e-09 Q=6.6659e-01 L=7.463e-05 k= 7 p=.0000e+00 f(p)=0.0000e+00 E= e+00 Q=0.0000e+00 L=0.0000e+00 The sequence converges quadratically because the exponents of the error E k are roughly doubling with each iteration until k = 7, and the error there becomes zero because the error is too small to distinguish p 7 from p = in the number system we use. Note that the doubling of the error exponent is consistent with p p k A p p k 1, and that the iterates are quite close to having this behavior. The Q column is for Q k = E k /E k 1 (with E k = p p k ) and for what we can compute the ratio is quite close to a constant A /3; this is empirical evidence for quadratic convergence. The L column is for L k = E k /E k 1 and the ratio decreases fast as k increases. We should expect to be constant for the double root r = 1 (see next page).

6 To check that the convergence to the double root r = 1 is linear: convorder( h4p3, h4p3d,-,-1, 10^-8, 10^-10, 100) k= 0 p=-.0000e+00 f(p)= e+00 E= e+00 k= 1 p= e+00 f(p)= e+00 E= e-01 Q=5.5556e-01 L=5.5556e-01 k= p=-1.979e+00 f(p)=-.968e-01 E= e-01 Q=9.65e-01 L=5.363e-01 k= 3 p= e+00 f(p)= e-0 E= e-01 Q=1.7509e+00 L=5.161e-01 k= 4 p= e+00 f(p)= e-0 E=7.9561e-0 Q=3.950e+00 L=5.10e-01 k= 5 p= e+00 f(p)= e-03 E= e-0 Q=6.3645e+00 L=5.0638e-01 k= 6 p=-1.003e+00 f(p)=-1.418e-03 E=.07681e-0 Q=1.49e+01 L=5.039e-01 k= 7 p=-1.010e+00 f(p)= e-04 E=1.0173e-0 Q=.4741e+01 L=5.0167e-01 k= 8 p= e+00 f(p)= e-05 E= e-03 Q=4.936e+01 L=5.0084e-01 k= 9 p=-1.005e+00 f(p)= e-05 E=.54958e-03 Q=9.84e+01 L=5.004e-01 k=10 p= e+00 f(p)= e-06 E= e-03 Q=1.960e+0 L=5.001e-01 k=11 p= e+00 f(p)=-1.06e-06 E= e-04 Q=3.915e+0 L=5.0011e-01 k=1 p= e+00 f(p)= e-07 E= e-04 Q=7.8404e+0 L=5.0005e-01 k=13 p=-1.000e+00 f(p)=-7.698e-08 E= e-04 Q=1.5678e+03 L=5.0003e-01 k=14 p= e+00 f(p)= e-08 E= e-05 Q=3.1354e+03 L=5.0001e-01 k=15 p= e+00 f(p)= e-09 E= e-05 Q=6.706e+03 L=5.0001e-01 k=16 p= e+00 f(p)=-1.19e-09 E= e-05 Q=1.541e+04 L=5.0000e-01 k=17 p= e+00 f(p)=-.9806e-10 E= e-06 Q=.508e+04 L=5.0000e-01 k=18 p= e+00 f(p)= e-11 E= e-06 Q=5.0163e+04 L=5.0000e-01 Note that the L k = p p k p p k 1 1/, so A = 1/ in this case.