Math 5334: Homework 3 Solutions

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1 Math 5334: Homework 3 Solutions Victoria E. Howle March 31, 28 Problem 4.3 a. Since I don t see any easy roots ( or ±1) of p(x) = 816x x 2 + 6x 3125 I will find them with MATLAB: >> format long >> roots([ ]) b. >> p = inline( 816*x.^3-3835*x.^2 + 6*x ) p = Inline function: p(x) = 816*x.^3-3835*x.^2 + 6*x >> x = 1.43:.1:1.71; >> y = p(x); >> plot(x,y) >> hold on >> plot( , p( ), o ) >> plot( , p( ), o ) >> plot( , p( ), o ) Texas Tech University, Department of Mathematics & Statistics 1

2 c. With x = 1.5, Newton s method finds the root in 11 iterations. >> pprime = inline( 3*816*x.^2-2*3835*x + 6 ) >> x = mynewton(p, pprime, 1.5) x = Here is my Newton code, mynewton.m: function x = mynewton(f,fprime,x) % function x = mynewton(f,fprime,x) % % define functions f and fprime as inlines or m-files % e.g., f x^2-2 % % finds a root of f near x 2

3 k = ; xprev = rand(1); x = x; while abs(x - xprev) > eps*abs(x) xprev = x; x = x - f(x)/fprime(x); % with m-file funcs, input func handles disp(sprintf( %5.16g, x)); k = k + 1; end d. With initial points x = 1 and x 1 = 2, the secant method finds the root in 12 iterations. >> x = mysecant(p,1,2) x = Here is my secant method code, mysecant.m: function x = mysecant(f,a,b) %function x = secant(f,a,b) % % define functions as inlines or m-files % e.g., f x^2-2 % % finds a root of f using initial guesses a,b k = ; while abs(b-a) > eps*abs(b) c = a; a = b; b = b + (b-c)/(f(c)/f(b)-1); 3

4 disp(sprintf( %5.16g, b)); k = k + 1; end e. Starting on the interval [1, 2], bisection finds the root after 52 iterations. >> x = mybisection(p,1,2)

5 x = Here is the bisection method, right out of the NCM book: function x = mybisection(f,a,b) % function x = mybisection(f,a,b) % see NCM p. 118 % % Does bisection method to find root of function f. % Bisects interval [a,b], so initial a and b must bracket a root. k = ; while b-a > eps % converge when bracket width is <= eps x = (a+b)/2; % midpt of interval [a,b] if sign(f(x)) == sign(f(b)) b = x; disp(sprintf( %5.16g,b)); else a = x; disp(sprintf( %5.16g,a)); end % now our new bracket is half as big k = k + 1; end e. fzerotx(p, [1,2]) finds the root The fzerotx code uses the secant method to take its first step. One step of secant gives , which is very near The following steps (IQI) zero in on this root. 5

6 Problem 4.8 Unlike Newton s method, which just bounced back and forth without making any progress on this function, the secant method does converge. Since it is only approximating the slope, the secant method doesn t get stuck in this case. Note that the requirements of the convergence theory for the secant method are the same as those for Newton. So like with Newton, they are violated by this function since it s first derivative is unbounded as x goes to a. But in this case we get lucky, and the method converges anyway. Depending on the given interval, the method converges in just a couple of iterations or after many iterations. In this example, I picked a = 3. f = Inline function: f(x) = sign(x - 3)*sqrt(abs(x - 3)) >> x = mysecant(f,2.5,4) x = 3 Problem 4.16 We ll start by rewriting the equation as a function T (x, t): x T (x, t) = erf( 2 αt )(T i T s ) + T s. Next we will write T (x, t) as an inline Matlab function of x only given α =.138e 6, T i = 2, T s = 15, and t = = 5184 seconds. >> T = inline( erf(x/(2*sqrt(.138e-6*5184)))*(2 + 15) - 15 ) T = Inline function: T(x) = erf(x/(2*sqrt(.138e-6*5184)))*(2 + 15) - 15 Finally, we find an interval on which T changes sign and call fzerotx to find a zero. >> T() -15 >> T(1) 2 >> fzerotx(t, [,1]).677 So we need to bury the water main at least.677 meters under ground. 6

7 Problem 5.4 a. We have defined ρ = 1/( σu k ) and want to show that ρ = 2/ u 2. First we will look at u 2 : u 2 = u ū Next we ll look at σu k : = (x + σe k ) ( x + σe k ) = x 1 x (x k + σ)( x k + σ) + + x n x n = x 2 + σ x k + σx k + σ σ = x 2 + x k x k x k x + x kx k x k x + x k x k x k 2 x 2 = 2 x σx k σu k = x k x k x u k = x k x k x (x k + σ) = x k x k x (x k + x k x k x ) = x kx k x k x + x k x k x k 2 x 2 = x 2 + σx k So we can see that u 2 = 2 σu k so that 1 = 2 σu k u 2. b. We ll show that H = H and H H = I with a Matlab example: >> x = rand(4,1) + i*rand(4,1) x = i i i i >> % we will take k = 2 >> sigma = sign(x(2))*norm(x); >> ek = [; 1; ; ] 7

8 ek = 1 >> u = x + sigma*ek u = i i i i >> rho = 2/norm(u)^2 rho =.291 >> H = I - rho*u*u H = i i i i i i i i i i i i.788 >> H i i i i i i i i i i i i.788 >> norm(h - H ) 1.284e-16 >> H *H i -. +.i -. +.i -. -.i i -. +.i -. -.i -. +.i i -. -.i -. -.i -. +.i 1. Note that H H above is equal to the identity up to a little round off error. 8

9 c. We ll use the same Matlab example as in (b). >> H*x. +.i i. +.i. -.i >> -sigma*ek i d. We can show this one directly: where τ = ρu y. Hy = (I ρuu )y = y ρuu y = y ρu(u y) = y u(ρu y) since ρ and u y are scalars = y τu Problem 5.5 a. For this problem, we just follow the steps for building the Householder reflector that zeros out all but the first element (k = 1) of the vector. >> x = [9 2 6] x = >> ek = [1; ; ] ek = 1 >> sigma = sign(x(1))*norm(x) sigma = 11 9

10 >> u = x + sigma*ek u = >> rho = 2/norm(u)^2 rho =.45 >> I = eye(3,3); >> H = I - rho*u*u H = >> H*x b. I am trying u from the construction of H (since the problem helpfully uses that letter...) >> u u = >> H*u That worked! For v, note that σ is defined with a plus or minus on p So we ll try v = x σ e 1. >> v = x - sigma*ek v =

11 >> H*v Problem 5.6 >> X = [1 2 3; 4 5 6; 7 8 9; ; ] X = a. X is rank deficient if its rank is < 3. >> rank(x) 2 b. >> Z = pinv(x) Z = >> B = X\eye(5,5) Warning: Rank deficient, rank = 2, tol = 2.471e-14. B = >> S = eye(3,3)/x Warning: Rank deficient, rank = 2, tol = e-14. S = >> norm(z, fro ).6777 >> norm(b, fro ).6791 >> norm(s, fro )

12 >> norm(x*z - I5, fro ) >> norm(x*b - I5, fro ) >> norm(x*s - I5, fro ) >> norm(z*x - I3, fro ) 1 >> norm(b*x - I3, fro ) >> norm(s*x - I3, fro ) 1 In all cases, the values involving Z are less than or equal to the values with B or S. c. >> norm(x*z - (X*Z) ) 1.942e-15 >> norm(z*x - (Z*X) ) 2.467e-15 >> norm(x*z*x - X) 2.466e-14 >> norm(z*x*z - Z) e-16 All are true for Z. >> norm(x*b - (X*B) ) e-15 >> norm(b*x - (B*X) ).771 >> norm(x*b*x - X) 2.254e-14 >> norm(b*x*b - B) 6.846e-16 Second property fails for B. >> norm(x*s - (X*S) ) >> norm(s*x - (S*X) ) e-16 >> norm(x*s*x - X) e-14 >> norm(s*x*s - S) 4.835e-16 First property fails for S. 12

13 Problem 5.12 a. >> x = [ ] ; >> y = [ ] ; >> M = [x.*x x.*y y.*y x y]; >> b = -1*ones(1,1); >> c = M\b; >> z = c(1)*x.^2 + c(2)*x.*y + c(3)*y.^2 + c(4)*x + c(5)*y + ones(1,1); >> [X,Y] = meshgrid(-.75:.5:1.25, :.5:1.25); >> Z = c(1)*x.^2 + c(2)*x.*y + c(3)*y.^2 + c(4)*x + c(5)*y + 1; >> contour(x,y,z,[ ]) >> hold on >> plot(x,y, bo ) >> title( orbit ) I made the meshgrid over a larger x and y range so we can see the full orbit. b. Perturbed system: >> rx = (.1)*rand(1,1); >> ry = (.1)*rand(1,1); >> x2 = x + rx; >> y2 = y + ry; >> M2 = [x2.*x2 x2.*y2 y2.*y2 x2 y2]; >> c2 = M2\b; >> % plot both orbits on same figure >> contour(x,y,z,[ ]) % from part a >> hold on 13

14 >> plot(x,y, bo ) >> plot(x2,y2, rx ) >> Z2 = c2(1)*x.^2 + c2(2)*x.*y + c2(3)*y.^2 + c2(4)*x + c2(5)*y + 1; >> contour(x,y,z2,[ ]) >> title( orbit and perturbed orbit ) 14

MATH 350: Introduction to Computational Mathematics

MATH 350: Introduction to Computational Mathematics Chapter IV: Locating Roots of Equations Greg Fasshauer Department of Applied Mathematics Illinois Institute of Technology Spring 2011 fasshauer@iit.edu