Question

Only the matlab

nlinear equations x 0.75 Determine the roots of these equations using: a) The Fixed-point iteration method. b) The Newton Raphson method. Employ initial guesses of x y 1.2 and perform the iterations until E.<10%. Note: You can use to solve the problems, but you should sol at least two full iterations manually. AB bl Du Thursd 30/3/ 1. For the displacement in Q3 y 10 e cos at 0 St S 4. a) Plot the displacement y vs time tusing plot and fplot commands. Include in your plot all necessary annotations b) Find the first two roots using fzero built in function. Choose your initial guess based on the plot. 2. Solve Q5 using fsolve built in function. 3. Solve Q4 using roots built in function for polynomials. Expected MATLABoutputs a) Plot of y vs t with proper annotations upload your m files to schoology.

Answer 1

Answer: Note: Asked to solve matlab problem only. 1. Code to plot y vs t: plotmm %create t vector tI0 1 2 3 4; a. %create y vector y zerosl1,5); %find y fox kk-1:5 ỵikk) =10* (exp ( 1 )) ."( (-0.5)*t(kk))*cos (2+tikk)); end %plot y vs t plot (utine-)i titlel'y ys t) xlabel('y') ylabel't')

Sample output y vs t 4 3.5 2.5 1.5 0.5 -4 2 4 10

fplot of the function: Code: Afplotm . m %call fplot fplot(@ (t) 10* (exp (1)) (-0.5*t) *cos (2*t),[0 4]); 10 5 -5 0 0.5 1.5 2 2.5 3 4

b. Finding first 2 roots: %fzeron.m %create y y = (Lt) 10* (exp (1) ) ^ (-0.5*t) *cos (2*t) ; %find 1st root y1=fzerw (y, [O 2]); %find 2nd root y2-fzero (y3 4]) %display roots forintf("Two real roots are %f %f\n",y1,y2); Sample output: > fzerom Two real roots are 0.785398 3.926991

2. fsolve0 of the given function: Code to create equations: AfsolveFun.m %createing equations function FF fsolveFun ( p) end Code to solve the equations: f solvem %call fsolve() yp fsolve (fsoleu, 1.2,1.2]); %display root forintff 'The roots are x=%f y=%f\n' typ (1), .m (2) ) ;

Sample output: >solvem Equation solved. tasian completed because the vector of function vaiuss i e zero as measured by the default value of the function tolerance, and the problem appears regular as measured by the gradient. <stopping criteria details> The roots are x-1.372065 v 0.239502

3. roots0 of the given function: % roots m %create fx[1 2 4 8; %call roots() x-roots (fx) display root fprintf( 'The root is %f\nak) ; Sample output: rootsm -2.0000 2.00000.0000i 2.0000 0.0000i The root is -2.000000 The root is 2.000000 The root is 2.000000

copyable code:

a. Code to plot y vs t:

%create t vector

t=[0 1 2 3 4];

%create y vector

y=zeros(1,5);

%find y

for kk=1:5

    y(kk)=10*(exp(1)).^((-0.5)*t(kk))*cos(2*t(kk));

end

%plot y vs t

plot(y,t,'o-');

title('y vs t');

xlabel('y');

ylabel('t');

fplot of the function:

%fplotm.m

%call fplot

fplot(@(t) 10*(exp(1))^(-0.5*t)*cos(2*t),[0 4]);

b. Finding first 2 roots:

%fzerom.m

%create y

y = @(t) 10*(exp(1))^(-0.5*t)*cos(2*t);

%find 1st root

y1=fzero(y, [0 2]);

%find 2nd root

y2=fzero(y,[3 4]);

%display roots

fprintf('Two real roots are %f %f\n',y1,y2);

2.

Code to create equations:

%fsolveFun.m

%createing equations

function FF = fsolveFun( p )

FF(1)=p(2)+p(1)*p(1)-p(1)-0.75;

FF(2)=p(2)+5*p(1)*p(2)-p(1)*p(1);

end

Code to solve the equations:

%fsolvem.m

%call fsolve()

yp=fsolve(@fsolveFun,[1.2,1.2]);

%display root

fprintf('The roots are x=%f y=%f\n',yp(1),yp(2));

3.

%rootsm.m

%create fx

fx = [1 -2 -4 8];

%call roots()

x=roots(fx)

%display root

fprintf('The root is %f\n',x);