Explanation The Matlab program run with the function replaced by the unstable two-step method. The resulting approximations follow: %Multistep method program % Inputs: time interval inters, ic = [y0] initial condition, number of steps n, % s=number of (multi)steps, e.g. 2 for 2-step method % Output: time steps t, solution y % Calls a multistep method such as ab2step.m % function [t,y]=exmultistep(inter,ic,n,s) h=(inter(2)-inter(1))/n; % Start-up phase y(1,:)=ic;t(1)=inter(1); for i=1:s-1 % start-up phase, using one-step method t(i+1)=t(i)+h; y(i+1,:)=trapstep(t(i),y(i,:),h); f(i,:)=ydot(t(i),y(i,:)); end for i=s:n % multistep method loop t(i+1)=t(i)+h; f(i,:)=ydot(t(i),y(i,:)); y(i+1,:)=ab2step(t(i),i,y,f,h); end plot(t,y) function y=trapstep(t,x,h) %one step of the Trapezoid Method z1=ydot(t,x); g=x+h*z1; z2=ydot(t+h,g); y=x+h*(z1+z2)/2; function z=ab2step(t,i,y,f,h) %one step of the Adams-Bashforth 2-step method z=y(i,:)+h*(3*f(i,:)/2-f(i-1,:)/2); function z=unstable2step(t,i,y,f,h) %one step of an unstable 2-step method z=-y(i,:)+2*y(i-1,:)+h*(5*f(i,:)/2+f(i-1,:)/2); function z=weaklystable2step(t,i,y,f,h) %one step of a weakly-stable 2-step method z=y(i-1,:)+h*2*f(i,:); function z = ydot(t,y) % IVP z = t*y+t^3; % initialize f = inline('y/x-y^2/x^2','x','y'); xrange=[1,2]; h=0.1; x=1:h:2; n = (xrange(2)-xrange(1))/h; y(1) = 1; % generate starting estimates using Runge-Kutta for i = 1 k1 = f(x(i), y(i)); k2 = f(x(i) + h/2, y(i) + h/2*k1); k3 = f(x(i) + h/2, y(i) + h/2*k2); k4 = f(x(i) + h, y(i) + h*k3); y(i+1) = y(i) + h/6*(k1 + 2*k2 + 2*k3 + k4); x(i+1) = x(i) + h; end % iterate for i = 3:n+1 % Adams-Moulton -- *correct* y(i) = y(i-1) + h/12*(5*f(x(i),y(i)) + 8*f(x(i-1),y(i-1))- f(x(i-2),y(i-2))); end [t,y]=exmultistep([0 1],1,20,2); (a) t i w i error 0.0 0.0000 0.0000 0.1 0.0050 0.0002 0.2 0.0213 0.0002 0.3 0.0493 0.0005 0.4 0.0916 0.0002 0.5 0.1474 0.0013 0.6 0.2222 0.0001 0.7 0.3105 0.0032 0.8 0.4276 0.0020 0.9 0.5510 0.0086 1.0 0.7283 0.0100 (b) t i w i error 0

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8.4.2 An article in Knee Surgery, Sports Traumatology, Arthroscopy [“Arthroscopic Meniscal Repair with an Absorbable Screw: Results and Surgical Technique” (2005, Vol. 13, pp. 273–279)] showed that only 25 out of 37 tears (67.6%) located between 3 and 6 mm from the meniscus rim were healed. a. Calculate a two-sided 95% confidence interval on the proportion of such tears that will heal. b. Calculate a 95% lower confidence bound on the proportion of such tears that will heal. 8.4.3 An article in the Journal of the American Statistical Association [“Illustration of Bayesian Inference in Normal Data Models Using Gibbs Sampling” (1990, Vol. 85, pp. 972–985)] measured the weight of 30 rats under experiment controls. Suppose that 12 were underweight rats. a. Calculate a 95% two-sided confidence interval on the true proportion of rats that would show underweight from the experiment. b. Using the point estimate of p obtained from the preliminary sample, what sample size is needed to be 95%…

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Numerical Analysis, Books A La Carte Edition (3rd Edition)

Ch. 6.1 - Show that the function y(t)=tsint is a solution of...Ch. 6.1 - Show that the function y(t)=esint is a solution of...Ch. 6.1 - Use separation of variables to find solutions of...Ch. 6.1 - Find the solutions of the IVP given by y(0)=0 and...Ch. 6.1 - Apply Eulers Method with step size h=1/4 to the...Ch. 6.1 - Apply Eulers Method with step size h=1/4 to the...Ch. 6.1 - (a) Show that y=tan(t+c) is a solution of the...Ch. 6.1 - (a) Show that y=tanh(t+c) is a solution of the...Ch. 6.1 - For which of these initial value problems on [0,...Ch. 6.1 - Sketch the slope field of the differential...

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