Index exceeds the number of array elements (1). Please help :(

%Método de la regla falsa
clear all;
close all;
clc;

syms x
prompt='Ingresa una función x: ';
y=input(prompt);
ai=input('Ingresa a: ');
bi=input('Ingresa b: ');
tole=input('Ingresa la tolerancia: ');
tol=double(tole);
ans1=subs(y,x,ai);
fa=double(ans1);
ans2=subs(y,x,bi);
fb=double(ans2);

if(fa*fb<1)
     a(1)=ai;
     ans1=subs(y,x,ai);
     fa=double(ans1);
     b(1)=bi;
     ans2=subs(y,x,bi);
     fb=double(ans2);
     xn(1)=b(1)-((fb*(a(1)-b(1)))/(fa-fb));
     f1=subs(y,x,xn(1));
     fx0=double(f1);
     i=1;
     eabs(i)=0;
     for i=0:1:eabs(i+1)<=tol
         if fx0<0
            a(i+1)=xn(i);
            ansa=subs(y,x,a(i+1));
            fa=double(ansa);
            b(i+1)=b(i);
            ansb=subs(y,x,b(i+1));
            fb=double(ansb);
         else
            a(i+1)=a(i);
            ansa=subs(y,x,a(i+1));
            fa=double(ansa);
            b(i+1)=xn(i);
            ansb=subs(y,x,b(i+1));
            fb=double(ansb);
         end
         xn(i+1)=b(i+1)-((fb*(a(i+1)-b(i+1)))/(fa-fb))
         eabs(i+1)=abs(xn(i+1)-xn(i))
         i=i+1
     end 
else
    fprintf('Ingresa un nuevo intervalo');
end

I have the error:

Ingresa una función x: sin(x)+1-x^2
Ingresa a: 1
Ingresa b: 2
Ingresa la tolerancia: 0.001
Index exceeds the number of array elements (1).

Error in mtl02 (line 30)
     for i=0:1:eabs(i+1)<=tol

And idk how to solve it, that's my condition, and if i put "eabs(i)<=tol" i can´t have more than two iterations :(

[G A]

before for loop

eabs(i)=0;

eabs(i+1) is not defined anywhere in your code

[dpb]

You'll need to arrange the test to work on the present and preceding (if needed) eabs value after the new one has been computed.

for i=0:1:eabs(i+1)<=tol
  ...
  eabs(i+1)=abs(xn(i+1)-xn(i))

the for loop is set up to run in integer increments by 1 to eabs(i+1)<=tol

But, even if eabs(i+1) were defined, the logical condition will be either F (0) or T (1) so the maximum the loop would run would be 0:1:1 or twice.

It looks like you need a while loop here instead of counted for

[Mariana Guzmán Ramírez]

I already used "while", but it's the same, I only have two iterations but with eabs(i)

%Método de la regla falsa
clear all;
close all;
clc;
syms x
prompt='Ingresa una función x: ';
y=input(prompt);
ai=input('Ingresa a: ');
bi=input('Ingresa b: ');
tole=input('Ingresa la tolerancia: ');
tol=double(tole);
ans1=subs(y,x,ai);
fa=double(ans1);
ans2=subs(y,x,bi);
fb=double(ans2);
i=1;
eabs(i)=0;
eabs(i+1)=0;
if(fa*fb<1)
     a(1)=ai;
     ans1=subs(y,x,ai);
     fa=double(ans1);
     b(1)=bi;
     ans2=subs(y,x,bi);
     fb=double(ans2);
     xn(1)=b(1)-((fb*(a(1)-b(1)))/(fa-fb));
     f1=subs(y,x,xn(1));
     fx0=double(f1);
     fprintf('It.           a           b            xn            Error abs \n');
     fprintf('%2d \t %11.7f \t %11.7f \t %11.7f \n',i,a(i),b(i),xn(i));
     while eabs(i+1)>tol
         if fa*fx0<0
            a(i+1)=xn(i);
            ansa=subs(y,x,a(i+1));
            fa=double(ansa);
            b(i+1)=b(i);
            ansb=subs(y,x,b(i+1));
            fb=double(ansb);
         else
            a(i+1)=a(i);
            ansa=subs(y,x,a(i+1));
            fa=double(ansa);
            b(i+1)=xn(i);
            ansb=subs(y,x,b(i+1));
            fb=double(ansb);
         end
         xn(i+1)=b(i+1)-((fb*(a(i+1)-b(i+1)))/(fa-fb));
         eabs(i+1)=abs(xn(i+1)-xn(i));
         fprintf('%2d \t %11.7f \t %11.7f \t %11.7f \t %7.3f \n',i+1,a(i+1),b(i+1),xn(i+1),eabs(i+1));
         i=i+1;
     end 
else
    fprintf('Ingresa un nuevo intervalo');
end

And now i have only one interation (the frist one):

Ingresa una función x: sin(x)+1-x^2
Ingresa a: 1
Ingresa b: 2
Ingresa la tolerancia: 0.001
It.           a           b            xn            Error aprox 
 1 	   1.0000000 	   2.0000000 	   1.2869786 

[dpb]

But you initialized to zero which would be perfect convergence for starters...