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HW6Solutions - Homework 6 Solutions 1 f(x =-x1/3 0.5x2-2=0...

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Unformatted text preview: Homework 6 Solutions 1. f(x) = -x1/3+0.5x2-2=0 x0 = 2 1/3 term g2(x) is for the 0.5x2 term g1(x) will be for the -x X” = 0.5x2- 2 a g1(x) = (0.5x2 - 2)3 0.5x2 = x“3 + 2 9 x2 = 2x“3 + 4 9 g2(x) = (2x1/3 + 4)“2 Earle) g1’(x) = 3(0.5x2 — 2)2(x) = 3x(o.5x2 - 2)z 81'(X) = 3*(-5*4 - 2)2(2) = 0 While this is less than 1, it doesn’t work. My guess is it is a singularity and something different happens at this point. The derivative is continuous. I don’t know why this doesn’t work. «wag ; «51X to new toolbar buttons: data brushing 8; linked glots g Q Play video t v, ..» a £5: d i .- , . g ‘ . . , A .,..;~.;M s . 1 .‘ an .mk‘uz g2'(x) = yzile/3 + 4)‘1/2(2/3*x'2/3) a (1/3*x'2/3)(2x1/3 + 4).”2 g2’(x) = (1/3*2'2/3)(2*21/3 + 4)“2 = 0.08224 ThereforeI we should use ggix) for our iterations. Pint—bl g1(x) = (0.5x2 — 2)3 ‘ 82(X) = (2x1/3 + 4?” Starting with gfl I gziz) = (2*21/3 + 4)“2 = 2.5534 g2(2.5534) = (2*255341/3 + 4)“2 = 2.5949 g2(2.5949) = (2*2.59491/3 + 4)“2 = 2.5978 g2(2.5978) = (2*259781/3 + 4)”2 = 2.59795 g2(2.59795) = (2*2597951/3 + 4)“2 = 2.59796 This is converged Now using gig); g1(2) = (0.5*22 — 2)3 = o g1(0) = (05"‘02—2)3 = -8 g1(—8) = (o.5*(-8)2 - 2)3 = 27000 g1(27000) = (o.5*270002 — 2)3 = 4.8428E25 g1(4.8428E25) = (O.5*4.8428E252 — 2)3 = 1.6124E153 This will not converge 2. xL = 2 xU = 3 f(x) = -x1/3+0.5x2—2=0 False Position Method: ck = [f(bk)ak — f(ak)bk]/[f(bk) - f(ak)] where ak = XL and bk = xu f(bk) = -(3)1/3 + 0.5*32 — 2 = 1.05775 f(ak) = -(2)“3 + 0.5*22 — 2 = -1.2599 ck = (1.05775*2 — (-1.2599*3))/(1.05775-(42599)) = 2.5436 f(ck) = -(2.5436)1/3 + 0.5*2.54362 .— 2 = -O.130067 Since this is the same sign as ak, ck becomes ak. cm = (1.05775*2.5436 — (-0.130067*3))/(1.05775-(-0.130067)) = 2.59359 f(ck+1) = -(2.59359)“3 + 0.5*2.593592 — 2 = 0010584 This is still the same sign as ak; therefore, cm becomes am. cm = (1.05775*2.59359 — (-0.010584*3))/(1.05775-(-0.010584)) = 2.59762 Secant Method: Xk+1 = Xk ' “kale " Xk-1)/(f(xk) — f(xk-1))l xk = XL xk.1 = xu flxk) = f(ak) = -1.2599 f(xk_1) = f(bk) = 1.05775 xk+1 = 2 — (-1.2599)[(2 - 3)/(-1.2599 — 105775)] = 2.5436 f(xk.1) = -(2.5436)“3 + o.5*2.54362 — 2 = 0.130067 x...2 = 2.5436 — (-0.130067)[(2.5436 — 2)/(-o.130067 — (-1.2599))] = 2.6062 f(xk+2) = —(2.6062)1/3 + O.5*2.60622 — 2 = 0.019967 Xk+3 = 2.6062 — (0.019967)[(2.6062 — 2.5436)/(0.019967 - (-0.130067))] = 2.59787 f(x) = -x1/3+0.5x2-2=0 f’ (x) = -1/3*x'2/3 + x x0 = 2 Xk+1 = Xk " f(xk)/f’(xk) f(2) = -(2)1/3 + 0.5*22 — 2 = -1.2599 f’(2) = -1/3*2"2/3 + 2 = 1.79001 xk.1 = 2 — (-1.2599/1.79oo1) = w % relative error = |2.70386 -— 2 [/2 *100 = w f(2.70386) = -(2.70386)1/3 + o.5*2.703862 — 2 = 0.262293 f’(2.70386) = -1/3*2.70386‘2/3 + 2.70386 = 2.53211 xk+2 = 2.70386 — (0.262293/2.53211) = mg; % relative error = [2.60027 — 2.70386 l/2.70386 = 3.83% f(2.60027) = —(2.60027)1/3 + 0.5*2.600272 — 2 = 0.005597 f’(2.60027) = 91/3*2.60027'2/3 + 2.60027 = 2.424 xk+3 = 2.60027 — (0.005597/2.424) = 2.59797 % relative error = [2.59797 — 2.60027 l/2.60027 = 0.0888% 10/17/08 1:52 PM C:\Documents and Settings\gtj223\Desktop\Grad\...\myrootfinder.m 1 of 3 function [xroot, froot] = myrootfinder(fun,x1,x2,method) % myrootfinder Root solver for any given function % % Synopsis: [xroot,froot] = myrootfinder(func,x1,x2,method) % Input: fun = (string) name of m~file that returns f(x) and f'(x) % x1 = An initial guess. Set as the lower bound. % x2 = A second initial guess. Set as the upper bound. % method 2 1 — bisection, 2—false position, 3—secant, 4—Newtons % Output: xroot = The x value where the function is zero % froot = The actual value of the function at xroot should be ~ 0 tol = 1E—8; %Tolerance value zeta = 2*tol; %Establish zeta as greater than the tolerance so that %you enter the while loop V % Bisection method if method == a = x1; b = x2; fa = feval(fun,a); %Calling a function that evaluates a given function. %dfdx isn't needed until Newton’s method so there fb = feval(fun,b); %is no need to reassign it. It will simply exist. %There are multiple ways to check here. You can do fa*fb and see if it %is negative (that implies that only one is negative). You can say (fa<0 %& fb<0)| (fa>0 & fb>0). I used the sign function in matlab. It is used %in the book as well. if sign(fa) == sign(fb) %Checking to make sure our signs are opposite for %the function values. Intermediate Value %Theorem check. fprintf(‘Root is not bracketed on interval [%f,%f]',x1,x2) xroot = 'No root found on interval‘; froot = 'No function value because no root“; else while zeta > tol %Just a randomly low value for our convergence xm = a + .5*(b—a); %Equation for bisection method fxm = feval(fun,xm); %Function call again but this one is %used to evaluate our new value for %xm. if sign(fxm) == sign(fa) %Sign check to make sure we have opposite signs. a = xm; %Reassign xm to a and then we can iterate again zeta = abs(a — b)/b; xroot = a; %We will claim a new root every time through. The %last time through will be the final xroot. froot = fxm; %Same for froot. else %Same as the if statement only it exists in case %fxm isn't the same sign as fa. b = xm; zeta = abs(b - a)/a; i l l i E . r i l i 10/17/08 1:52 PM C:\Documents and Settings\gtj223\Desktop\Grad\...\myrootfinder.m 2 of 3 W xroot = b; froot = fxm; end end end %False Position Method : elseif method == i xL=x1; xU=x2; g fo = feval(fun,xL); %Same function call. g fo = feval(fun,xU); E if sign(fo) == sign(fo) %Same check , fprintf('Root is not bracketed on interval [%f,%f]',x1,x2) xroot = 'No root on interval'; froot = 'No function value because no root'; else I while zeta > tol fo = feval(fun,xL); %We have to call this function again %because in this method we use different fo = feval(fun!xU); %function values where the bisection %only changes the root values. ck = (fo*xL - fo*xU)/(fo — fo); %False method equation fck = feval(fun,ck); if sign(fck) == sign(fo) xL = ck; zeta = abs(ck — xU)/xU; xroot = xL; froot = fck; elseif sign(fck) == sign(fo) xU = ck; zeta = abs(ck — xL)/xL; xroot = xU; froot = fck; end end end %Secant Method %Same as the false postition method but we have a slightly different ck %formula. Also, there is no need for sign checking since we are simply %incrementing not replacing if you will. elseif method == 3 xL = x1; xU = x2; %This while loop is different. Before, we would find an X value and %then we would plug it back in to solve for another x value but we %had to have two separate signs because of the IMV theorem. Now we %don't. We aren't trying to converge in the same manner. Sign is %now irrelevant as long as we know that a root exists in our %interval (that is why we do the check earlier). while zeta > tol fo: feval(fun,xL); 10/17/08 1:52 PM C:\Documents and Settings\gtj223\Desktop\Grad\...\myrootfinder.m 3 of 3 W“ fo= feval(fun,xU); xkl = xL — fo*(xL - xU)/(fo — fo); %Different iteration formula. ka1 = feval(fun,xkl); zeta = abs(xk1 — xU)/xU; XL = XU; xU = xkl; xroot = xkl; froot = ka1; end %Newton's Method elseif method == 4 XL = x1; xU = x2; %Note no Sign checking. This is similar to the secant method but it %uses a derivative for f as well. while zeta > tol [f,dfdx] = feval(fun,xL); fo = f; dfdxxL = dfdx; xkl = xL — fo/dfdxxL; kal = feval(fun,xkl); zeta = abs(xk1 — xL)/xL; xL xkl; xroot = xkl; froot = ka1; end %If someone typed in a method 5, we don’t have one that corresponds to it %so we add an error message in case. else fprintf(’Method chosen, %f, is not specified for this function.\n Chose 1 fort bisection,\n 2 for false position,\n 3 for secant,\n 4 for Newton.\n’,method) xroot = 'No xroot can be found with this method'; froot = ’No function value because no xroot'; end 10/17/08 1:52 PM C:\Documents and Settings\gtj223\Desktop\Gra...\mynonlineareqn.m 1 of 1 W function [f,dfdx] = mynonlineareqn(x) o z fx3n Evaluate f(x)= .5*x‘2 — X‘(1/3) — 2 and dfdx %We can manually change our function in this line and then the rootfinder %will use the new function. f = .5*(x)‘2 — x‘(1/3) — 2; dfdx = x — (1/3)*x‘(—2/3); vWWWMW._.M«—_—___mmmm.Mh‘wwmwwmw.»-_44« 4 10/17/08 1:54 PM MATLAB Command Window >> [xroot, froot] = myrootfinder('mynonlineareqn',2,3,1) xroot = 2.5980 froot = 7.6594e—009 >> [xroot, froot] = myrootfinder('mynonlineareqn',2,3,2) xroot = 2.5980 froot = 4.4409e—016 >> [xroot, froot] = myrootfinder('mynonlineareqn',2,3,3) xroot = 2.5980 froot = 4.4409e—016 >> [xroot, froot] = myrootfinder('mynonlineareqn',2,3,4) xroot = 2.5980 froot = 4.4409e—016 >> [xroot, froot] = myrootfinder('mynonlineareqn',2,3,5) Method chosen, 5.000000, is not specified for this function. Chose 1 for bisection, 2 for false position, 3 for secant, 4 for Newton. XI‘OOt = 1 of 2 WW“— ',‘1' 10/17/08 1:54 PM MATLAB Command Window 2 of 2 MW No xroot can be found with this method froot = No function value because no xroot >> [xroot, froot] = myrootfinder('mynonlineareqn',2.7,3.5,1) Root is not bracketed on interval [2.700000,3.500000] xroot = No root found on interval a E froot = t E No function value because no root 3 >> ...
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