How to store each iteration value and plot it ?

Hi everyone I would like to store each L2 values after each iteration and plot it vs iteration number. How can I do it with the attached code?

clc;
clear all;
n=input('Enter the no of iterations : \n');
N=n+1;
a=input('\n Enter lower limit : \n'); 
b=input('\n\nEnter upper limit : \n');
fold=1;
fnew=1;
func = @(x)(.65-[.75/(1+x^2)]-.65*x*atan(1/x));
for i=1:N
  if i==1 || i==2
    f(i)=1;
   continue;
  end
  f(i)=fold+fnew;
  fold=fnew;
  fnew=f(i);
end
L2=(b-a)*f(N-2)/f(N);
j=2;
while j<N
  L1=(b-a);
  if L2>L1/2
    anew=b-L2;
    bnew=a+L2;
  else 
    if L2<=L1/2
      anew=a+L2;
      bnew=b-L2;
    end
  end
  k1=func(anew);
  k2=func(bnew);
  if k2>k1
    b=bnew;
    L2=f(N-j)*L1/f(N-j+2);
  else 
    if k2<k1
      a=anew;
      L2=f(N-j)*L1/f(N-(j-2));
    else 
      if k2==k1
        b=bnew;
        L2=f(N-j)*[b-a]/f(N-(j-2));
        j=j+1;
      end
    end
  end
  j=j+1;
end
disp(a);
disp(b);

[engineer]

when I run , it only shows the last iteration result.

Answer 1

[dpb]

Preallocate for L2 and keep and increment an index inside the loop and store into an array

L2=zeros(N,1);                % preallocate
L2(1)=(b-a)*f(N-2)/f(N);      % save initial value
j=2;
while j<N
  L1=(b-a);
  if L2>L1/2
    ...

every store is then

L2(j)=...;

Or, change to counted for loop instead of while since there's no conditional exit.

L2(1)=...;
for j=2:N-1
  ...
  L2(j)=...;
  ...

BTW, Matlab "elseif" is one string, I had to reformat code to try to read it for indention levels being all fouled up where you wrote

...
else if L2<=L1/2
  ...

Rewrite

if L2>L1/2
  anew=b-L2;
  bnew=a+L2;
else 
  if L2<=L1/2
    anew=a+L2;
    bnew=b-L2;
  end
end

as

if L2>L1/2
  anew=b-L2;
  bnew=a+L2;
elseif L2<=L1/2
  anew=a+L2;
  bnew=b-L2;
end

which since the 'elseif' is the negation of the 'if' condition, this is really

if L2>L1/2
  anew=b-L2;
  bnew=a+L2;
else 
  anew=a+L2;
  bnew=b-L2;
end

there's no need for the conditional if on the else clause here.

[engineer]

Sir, I tried but it gave me the same answer for all iteration result..Would it be possible to send one single code for this.

[dpb]

Show the code as you've modified it...

[engineer]

sure.

if true
  clc;
clear all;
n=input('Enter the no of iterations : \n');
N=n+1;
a=input('\n Enter lower limit : \n'); 
b=input('\n\nEnter upper limit : \n');
fold=1;
fnew=1;
func = @(x)0.3039*exp(-((x-0.1524)/0.3671).^2) + 6.593e+13*exp(-((x+495.5)/86.41).^2) + 0.245*exp(-((x-0.1098)/0.1452).^2) + 0.06194*exp(-((x-0.3992)/0.2167).^2) + 0.09388*exp(-((x--0.3931)/0.5412).^2) + 0.143 *exp(-((x-1.001)/1.158).^2);
for i=1:N
if i==1 || i==2
  f(i)=1;
 continue;
end
f(i)=fold+fnew;
fold=fnew;
fnew=f(i);
end
L2=(b-a)*f(N-2)/f(N);
L2=zeros(N,1);                % preallocate
L2(1)=(b-a)*f(N-2)/f(N);      % save initial value
j=2;
while j<N
L1=(b-a);
if L2>L1/2
  anew=b-L2;
  bnew=a+L2;
else 
  if L2<=L1/2
    anew=a+L2;
    bnew=b-L2;
  end
end
k1=func(anew);
k2=func(bnew);
if k2>k1
  b=bnew;
  L2(j)=f(N-j)*L1/f(N-j+2);
else 
  if k2<k1
    a=anew;
    L2=f(N-j)*L1/f(N-(j-2));
  else 
    if k2==k1
      b=bnew;
      L2(j)=f(N-j)*(b-a)/f(N-(j-2));
      j=j+1;
    end
  end
end
j=j+1;
end
disp(a);
disp(b);
end

[dpb]

1. Don't need the L2= line before preallocating
2. You missed at least one path in the loop of the (j) subscript on L2 so that will write the one value into the whole array

Make sure you got each and every "L2=..." line corrected.