transport to matlab language

STEP1: SET x=x0, SET i=0
STEP2: SET i=i+1
CALCULATE f(x)
CALCULATE f-­‐‑(x)= f(x-­‐‑h)
CALCULATE f+(x)= f(x+h)
CALCULATE f ’(x)=[f+(x)-­‐‑f-­‐‑(x)]/2h
STEP4: if f ’(x)=0 then {
if f(x) <= tol_root then goto 10 //success
else SET X=X0+h
}
STEP5: SET xnew=x-­‐‑f(x)/f ’(x)
STEP6: if i > imax then exit //failure
STEP7: if abs(xnew-­‐‑x) <= tol_x then goto STEP10 //success
else goto STEP8
STEP8: SET x=xnew
STEP9: goto STEP2
STEP10: exit

Can anyone help me with that? I have to transport this in matlab language in case to create a program

Answer 1

[Walter Roberson]

   while true 
      Do some things 
      if Condition
        break
      end
    end

[mar vouz]

So I created the following code :

% code
x=0;
h=0.001;
tol_root=1e-05;
tol_x=1e-05;
for i = 1:1:50
    f=@(x) x.^2 + 0.16*x -0.237;
    fpl=@(x) f(x+h);
    fout=@(x) f(x-h);
    df=@(x) (fpl(x)-fout(x))/(2*h);
    if df(x)==0 then
        if f(x)<=tol_root
            i=51;
        else
            x=x+h
        end
    end
    xnew=x-f(x)/df(x);
    if abs (xnew-x)<= tol_x
        e(i)=@(x) abs(xnew-x)
        i=51
    else
        x=xnew
    end
end
xroot=xnew

And it actually gives me seven roots.But also it gives me this message:

The following error occurred converting from function_handle to double:
Error using double
Conversion to double from function_handle is not possible.

why?

[Steven Lord]

e is a double precision array. When you try to assign a function handle into it MATLAB tries to convert that function handle into a double precision value. It can't, that conversion isn't defined.

From context you don't want e to contain a function handle that computes the residual, you want it to contain the residual itself. Get rid of the @(x) on the line where you assign to e(i).

But there are other problems or areas for improvement with this code. Two I see off the top of my head:

Assigning a value to the for loop variable will change the value of that variable for the rest of that iteration of the for loop. When MATLAB goes back to the top, your modified value is thrown away and replaced by the next iterate. This is described in the second item of the Tips section on the for documentation page. So your attempt to exit the loop with "i = 51" won't work.

To exit out of a for loop early, use the break keyword as suggested by the first item of the Tips section on the for documentation page.

Since your functions f, fpl, fout, and df don't depend on any variables that change inside the for loop, it will be more efficient to define them outside the loop. [Probably not much more efficient, but why define the same thing 50 times when you can define it once?]