Matlab is needed...The image is just for visuals it doesnt contain any relevant information this is all the info I was given and all the info you need

This is the implementation that is referred to:

choice = menu('Choose a method','Bisection','False Position','Fixed-Point iteration','Newton','Secant','Modified Secant','MATLAB fzero','Muller','Bairstow','MATLAB roots');

%% Type in your function after @(x)

f=@(x)x^2+x^9-1;

%% Bisection Method

if(choice==1)

a=-10;

b=10; %% bisection interval

tol=1e-10;

fa=f(a);

fb=f(b);

while abs(b-a)>tol

c=(a+b)/2;

fc=f(c);

if fa*fc

b=c;

fb=fc;

elseif fb*fc

a=c;

fa=fc;

else

break

end

end

x=c;

end

%% False Position Method

if (choice==2)

x0 = -6 ;%initial guess

x1 = 5;

tolerance = 0.001;

for i=0:9223372036854775807

if ((f(x1)-f(x0))==0)

fprintf('Error in denominator equals 0! ');

break;

else

x2= x1 - (f(x1)* (x1-x0)/(f(x1)-f(x0))) ;

end

c = f(x2);

absolute_c= abs(c);

if absolute_c

break

end

if f(x0)*c

x1=x2;

continue;

else

x0=x2;

continue;

end

end

x2;

end

%% Fixed Point iteration

if (choice==3)

x = 0.5; %initial guess

Es = 0.1; %tolerance

Ea = 1000; %randomly large relative approximate error

xold = x;

n = 0; %iteration counter

while Ea > Es

x = -f(x)+x;

Ea = abs((x-xold)/x)*100;

xold = x;

n = n + 1;

end

x;

end

%% Newton-Raphson

if (choice==4)

x1 = 0.05; % Initial point

iter = 0;

tol=0.001;

while (1) %our loop handler is 10^-3 %error approximaiton

dif=(f(x1+0.001)-f(x1-0.001))/0.002;

x2=x1-f(x1)/dif;

if (abs(x2-x1)

break;

end

x1=x2;

end

x2;

end

%% Secant Method

if(choice==5)

x0=0;

x1=5; %% Initial values

tol=0.00001;

while (1)

x2=x1-(f(x1)*(x0-x1))/(f(x0)-f(x1));

if (abs(x2-x1)

break;

end

x0=x1;

x1=x2;

end

x2;

end

%% Modified secant method

if(choice==6)

delta=0.0001;

tol=0.000001;

while (1)

x2=x1-delta*f(x1)/(f(x1+delta)-f(x1));

if (abs(x2-x1)

break;

end

x1=x2;

end

x2;

end

%% fzero

if(choice==7)

x = fzero(f,0);

end

%% Muller

if (choice==8)

p0=0;

p1=5;

p2=10;

P(1) = p0;

P(2) = p1;

P(3) = p2; % starting points

delta=0.00001; % tolarance

epsilon=delta;

max1=10000;

y0 = feval(f,p0);

y1 = feval(f,p1);

y2 = feval(f,p2);

for k=1:max1,

h0 = p0 - p2;

h1 = p1 - p2;

c = y2;

e0 = y0 - c;

e1 = y1 - c;

det1 = h0*h1*(h0-h1);

a = (e0*h1 - h0*e1)/det1;

b = (h0^2*e1 - h1^2*e0)/det1;

if b^2 > 4*a*c,

disc = sqrt(b^2 - 4*a*c);

else

disc = 0;

end

if b

z = - 2*c/(b + disc);

p3 = p2 + z;

if abs(p3-p1)

u = p1;

p1 = p0;

p0 = u;

v = y1;

y1 = y0;

y0 = v;

end

if abs(p3-p2)

u = p2;

p2 = p1;

p1 = u;

v = y2;

y2 = y1;

y1 = v;

end

p2 = p3;

y2 = feval(f,p2);

P = [P,p2];

err = abs(z);

relerr = err/(abs(p3)+eps);

if (err

end

P(end);

end

%% Bairstow

if(choice==9)

tol=0.000001;

a=[1 0 1];

n=2;

it=1;

while n>2

%Initialise for this loop

u=1; v=1; st=1;

while st>tol

b(1)=a(1)-u; b(2)=a(2)-b(1)*u-v;

for k=3:n

b(k)=a(k)-b(k-1)*u-b(k-2)*v;

end;

c(1)=b(1)-u; c(2)=b(2)-c(1)*u-v;

for k=3:n-1

c(k)=b(k)-c(k-1)*u-c(k-2)*v;

end;

%calculate change in u and v

c1=c(n-1); b1=b(n); cb=c(n-1)*b(n-1);

c2=c(n-2)*c(n-2); bc=b(n-1)*c(n-2);

if n>3, c1=c1*c(n-3); b1=b1*c(n-3);end;

dn=c1-c2;

du=(b1-bc)/dn; dv=(cb-c(n-2)*b(n))/dn;

u=u+du; v=v+dv;

st=norm([du dv]); it=it+1;

end;

[r1,r2,im1,im2]=solveq(u,v,n,a);

rts(n,1:2)=[r1 im1]; rts(n-1,1:2)=[r2 im2];

n=n-2;

a(1:n)=b(1:n);

end;

%Solve last quadratic or linear equation

u=a(1); v=a(2);

[r1,r2,im1,im2]=solveq(u,v,n,a);

rts(n,1:2)=[r1 im1];

if n==2

rts(n-1,1:2)=[r2 im2];

end;

rts(1,1)

end

%% Matlab Roots

if (choice==10)

c=[1,0,-1]; %% Need to be changed for different equations

roots(c)

end

This is the function for solveq if it is not created you will get an error when running the code:

function [r1,r2,im1,im2]=solveq(u,v,n,a);

% Solves x^2 + ux + v = 0 (n 1) or x + a(1) = 0 (n = 1).

%

% Example call: [r1,r2,im1,im2]=solveq(u,v,n,a)

% r1, r2 are real parts of the roots,

% im1, im2 are the imaginary parts of the roots.

% Called by function bairstow.

%

if n==1

r1=-a(1);im1=0; r2=0; im2=0;

else

d=u*u-4*v;

if d

d=-d;

im1=sqrt(d)/2; r1=-u/2; r2=r1; im2=-im1;

elseif d>0

r1=(-u+sqrt(d))/2; im1=0; r2=(-u-sqrt(d))/2; im2=0;

else

r1=-u/2; im1=0; r2=-u/2; im2=0;

end;

end

end

Problem 2 hen two cylinders in contact, such as roller bearings, transmit a load F as shown in Fig. 1, the normal stress developed at a point along the z-axis is given by 1 (b) V max Where pmax is the maximum pressure along the contact line and 2b is the width of the deformed flat rectangular surface around the contact line. Rectangular contact area with semi-ellptical pressure Fig.1. Contact stress between Cylinder. a. Use the program implemented in problem 1 to estimate at which 0.5 pmax Use Es 0.01% a. Repeat part a using different initial guesses (3 different values where applicable). b. Plot a graph of the approximation percentage error for all the used algorithms in part a c. Which algorithm is the fastest? Problem 2 hen two cylinders in contact, such as roller bearings, transmit a load F as shown in Fig. 1, the normal stress developed at a point along the z-axis is given by 1 (b) V max Where pmax is the maximum pressure along the contact line and 2b is the width of the deformed flat rectangular surface around the contact line. Rectangular contact area with semi-ellptical pressure Fig.1. Contact stress between Cylinder. a. Use the program implemented in problem 1 to estimate at which 0.5 pmax Use Es 0.01% a. Repeat part a using different initial guesses (3 different values where applicable). b. Plot a graph of the approximation percentage error for all the used algorithms in part a c. Which algorithm is the fastest