is there any mistake !!! method bisection to find root of equation

function ys = fs(vp)
% pour les modes symetriques, pour plaque d'acier : 
% vitesse longitudinale= 6144 m/s ,vitesse transversale =3095 m/s, 
% module de Young E= 2e11Pa, coefficient de poisson  0.33, 
% densité = 7850kg/m3,

f = 4000000;
d = 0.001;
w = 2*pi*f;
%   d =  1.0000e-03; % epaisseur de la palaque 1 mm
vt = 3095;
vl = 6144;
k = w/vp; 
kt = w/vt; 
kl = w/vl; 
s = sqrt ( k^2 - kt^2 );
q = sqrt ( k^2 - kl^2 );
ys = ((k^2 + s^2)^2)*cosh(q*d)*sinh(s*d)+ 4*(k^2)*q*s*sinh(q*d)*cosh(s*d) ;

function  vp = bisection (f,a,b)

if  f(a)*f(b) < 0
     vp =  (a+b)/2;
     err = abs (f(vp));    
     while err > 1.0e-5
            if  f(a)*f(vp) <0
                b = vp;
            end
            if  f(a)*f(vp) >0
                a = vp;
            end
            if f (vp) == 0
                break
            end
            vp =  (a+b)/2;
            err = abs (f(vp));         
     end
else     
     fprintf ( ' pas de solution ');  
end

[John D'Errico]

There is no question here. What is your problem? Haave you tried to run it and it did not work? Or are we expected to completely test and debug your code, while figuring out what it is supposed to do?

[Geoff Hayes]

salah - your implementation of the Bisection method seems reasonable though it isn't clear to me why the initial condition of

 f(a)*f(b) < 0

is necessary. I can see you wanting to ensure that

 b > a

so that the interval is correctly set but not the other. Also, when using while loops, it is good practice to implement a strategy to ensure that your code doesn't become stuck in an infinite loop. You do this by counting the number of iterations for the loop and once the maximum allowed has been reached, then you exit the loop. For example, your method could become

       err      = realmax;
       iters    = 0;
       TOL      = 1.0e-5;
       MAXITERS = 1000;
       while err > TOL && iters > MAXITERS
           iters = iters + 1;
           vp    =  (a+b)/2;
           err   = abs(f(vp)); 
           if  f(a)*f(vp) <0
               b = vp;
           elseif  f(a)*f(vp) >0
               a = vp;
           elseif abs(f(vp)) < eps
               break;
           end          
       end

Note the use of the elseif in the above. Also, when using floating point precision, it a good idea to avoid comparing the float (or double) to zero as

 f(vp) == 0

since this not likely to succeed (this is valid for integer comparisons but not when using doubles). Instead, we compare against some small number, eps, and assume that if this condition is met then the number is close enough to being zero.