Question

For 01-3, use f (x)E-3.3x3 +2.5x -0.6, and be sure to modify f m and f and fprime.m accordingly.

% Bisection.m Lines of code 17-26 and 43-47 are bold

% This code finds the root of a function f(x) in the interval [a, b] using the Bisection method

%

% It uses f.m to define f(x), and assumes f(x) is continuous

% It requires specification of a, b and the maximum error

% It defines error using |f(xnew)|

% Define inputs for problem

a=0; %Defines lower limit of initial bracketing interval

b=1; %Defines upper limit of initial bracketing interval

errmax = 1E-4; %Defines maximum error

%Get function values associated with initial interval endpoints

fa = f(a); %get function value at a

fb = f(b); %get function value at b

% Let MATLAB figure out which end point is "plus" and which is "minus"

if fa>0 & fb<0

    xplus = a;

    xminus = b;

elseif fa <0 & fb>0

    xplus = b;

    xminus = a;

else

    'Not a valid bracketing interval'

    return

end

iter = 0; %Initialize iteration counter (not required, but for interest)

err = 1; %Initial value of err is set large to ensure enter loop.

while err > errmax %Iteration loop

   

    %Increase iteration counter

    iter=iter+1;

   

    %Get xnew using Bisection

    xnew = (xplus + xminus)/2;

    %Get f(xnew)

    fnew = f(xnew);

    %Adjust bracketing interval based on sign of f(xnew)

    if fnew > 0

        xplus = xnew;

    else

        xminus = xnew;

    end

    err = abs(fnew); %This is the error metric for this example.

   

end

'Root is'

xnew

iter

......................................................................................NEW CODE...............................................................

%Textbooks bisect.m lines of code 19 – 24 are Bold

function bisect(f,a,b,tol,n)

% Bisection method for solving the nonlinear

%equation f(x)=0.

a0=a;

b0=b;

iter=0;

u=feval(f,a);

v=feval(f,b);

c=(a+b)*0.5;

err=abs(b-a)*0.5;

disp('_____________________________________________________________________')

disp(' iter    a            b           c           f(c)      |b-a|/2 ')

disp('_____________________________________________________________________')

fprintf('\n')

if (u*v<=0)

   while (err>tol)&(iter<=n)

      w=feval(f,c);

      fprintf('%2.0f %10.4f %10.4f %12.6f %10.6f %10.6f\n',iter,a,b,c,w,err)

      if (w*u<0)

         b=c;v=w;

      end

      if (w*u>0)

         a=c;u=w;

      end

      iter=iter+1;

      c=(a+b)*0.5;

      err=abs(b-a)*0.5;

   end

   if (iter>n)

      disp(' Method failed to converge')

   end  

else

   disp('   The method cannot be applied f(a)f(b)>0')

end

% Plot f(x) in the interval [a,b].

fplot(f, [a0 b0])

xlabel('x');ylabel('f(x)');

grid

     

        

  

Answer 1

PART A)

MATLAB CODE call bisect.m function

f = @(x) -3.3*x.^3+2.5*x-0.6;
a =0;
b = 0.5;
tol = 1E-4;
n=21;
bisect(f,b,a,tol,n);

OUTPUT

_____________________________________________________________________
iter    a            b           c           f(c)      |b-a|/2
_____________________________________________________________________

0      0.5000      0.0000      0.250000   -0.026562    0.250000
1      0.5000      0.2500      0.375000    0.163477    0.125000
2      0.3750      0.2500      0.312500    0.080542    0.062500
3      0.3125      0.2500      0.281250    0.029709    0.031250
4      0.2812      0.2500      0.265625    0.002215    0.015625
5      0.2656      0.2500      0.257812   -0.012018    0.007812
6      0.2656      0.2578      0.261719   -0.004862    0.003906
7      0.2656      0.2617      0.263672   -0.001313    0.001953
8      0.2656      0.2637      0.264648    0.000453    0.000977
9      0.2646      0.2637      0.264160   -0.000429    0.000488
10      0.2646      0.2642      0.264404    0.000012    0.000244
11      0.2644      0.2642      0.264282   -0.000209    0.000122

0.3 0.2 0.1 0 0.1 0.2 0.3 -0.4 0.5 0.4 0.5 -0.6 0.2 0.3 0.1

ANSWER

Initial xminus = 0 and initial xplus =0.5

f(x) is calculated 14 times

PART B)

Modified Bisection.m function

%Textbooks bisect.m lines of code 19 – 24 are Bold
function bisect(f,a,b,tol,n)
% Bisection method for solving the nonlinear
%equation f(x)=0.
fa = feval(f,a);
fb = feval(f,b);
if(fa>0 && fb <0)
    xplus = a;
    xminus = b;
elseif(fa<0&&fb>0)
    xplus = b;
    xminus = a;
else
    fprintf('Not a valid bracketing interval\n');
    return
end
iter=0;
xnew=(xplus+xminus)*0.5;
err=abs(xplus-xminus)*0.5;
disp('_____________________________________________________________________')
disp(' iter xminus          xplus        xnew         f(xnew)     |xplus-xminus|/2 ')
disp('_____________________________________________________________________')
fprintf('\n')
while ((err>tol)&&(iter<=n))
   fnew=feval(f,xnew);
   fprintf('%2.0f %10.4f %10.4f %12.6f %10.6f %10.6f\n',iter,xminus,xplus,xnew,fnew,err)
    if(fnew>0)
        xplus=xnew;
    else
        xminus=xnew;
    end
   iter=iter+1;
   xnew=(xplus+xminus)*0.5;
   err=abs(xplus-xminus)*0.5;
end
if (iter>n)
   disp(' Method failed to converge')
end
end

FUNCTION CALL

f = @(x) -3.3*x.^3+2.5*x-0.6;
a =0;
b = 0.5;
tol = 1E-4;
n=21;
bisect(f,b,a,tol,n);

OUTPUT

_____________________________________________________________________
iter xminus          xplus        xnew         f(xnew)     |xplus-xminus|/2
_____________________________________________________________________

0      0.0000      0.5000      0.250000   -0.026562    0.250000
1      0.2500      0.5000      0.375000    0.163477    0.125000
2      0.2500      0.3750      0.312500    0.080542    0.062500
3      0.2500      0.3125      0.281250    0.029709    0.031250
4      0.2500      0.2812      0.265625    0.002215    0.015625
5      0.2500      0.2656      0.257812   -0.012018    0.007812
6      0.2578      0.2656      0.261719   -0.004862    0.003906
7      0.2617      0.2656      0.263672   -0.001313    0.001953
8      0.2637      0.2656      0.264648    0.000453    0.000977
9      0.2637      0.2646      0.264160   -0.000429    0.000488
10      0.2642      0.2646      0.264404    0.000012    0.000244
11      0.2642      0.2644      0.264282   -0.000209    0.000122

Part C)

function regularfalsi(f,a,b,tol,n)
% The secant method for solving the nonlinear
% equation f(x)=0.
iter=0;
u=feval(f,a);
v=feval(f,b);
err=abs(b-a);
disp('______________________________________________________________')
disp('iter     xn          f(xn)      f(xn+1)-f(xn)    |xn+1-xn|')
disp('______________________________________________________________')
fprintf('%2.0f %12.6f %12.6f\n',iter,a,u)
fprintf('%2.0f %12.6f %12.6f %12.6f %12.6f\n',iter,b,v,v-u,err)
while (err>tol)&(iter<=n)&((v-u)~=0)
      x=(v*a-u*b)/(v-u);
      a=b;
      u=v;
      b=x;
      v=feval(f,b);
      err=abs(b-a);
      iter=iter+1;
      fprintf('%2.0f %12.6f %12.6f %12.6f %12.6f\n',iter,b,v,v-u,err)
end
if ((v-u)==0)
      disp(' Division by zero')
end
if (iter>n)
      disp(' Method failed to converge')
end

Part D)

Bisection.m modified to regulafalsi.m

f = @(x) -3.3*x.^3+2.5*x-0.6;
a =0;
b = 0.5;
tol = 1E-4;
n=21;
regulafalsi(f,b,a,tol,n);

OUTPUT

______________________________________________________________
iter     xn          f(xn)      f(xn+1)-f(xn)    |xn+1-xn|
______________________________________________________________
0      0.500000      0.237500
0      0.000000     -0.600000     -0.837500      0.500000
1      0.358209      0.143844      0.743844      0.358209
2      0.288939      0.042743     -0.101101      0.069270
3      0.259653     -0.008637     -0.051381      0.029286
4      0.264576      0.000322      0.008959      0.004923
5      0.264399      0.000002     -0.000320      0.000177
6      0.264398     -0.000000     -0.000002      0.000001

ANSWER

f(x) is calculated 8 times

In general Bisection method take more time to converge to the root, but it have less number of poerations in the loop as compared to ragula falsi method