CODEWITHSUNDEEP
matlab
ttl
ttl

Reputation: 19

Matlab - user defined function as an argument of a function

So I'm trying to code a function that will implement the secant method to a function(f) that will be input by the user and add the result of each iteration to the vector. I get an error stating that x is undefined which is understandable, but I'm unsure of how to define it. I'm trying to write it in a way where the argument f can be a function in terms of x, eg. x^3 +12, as opposed to the name of a seperate function file.

function [xans, xi, iter] = secant( f, x0, x1, tol )

    k = 1;
    a = x0;
    b = x1;
    c = f(b)*((b-a)/(f(b)-f(a)));

    while abs(c) >= tol
        xi(k) = b-c;
        a = b
        b = xi(k)
        k = k+1;

    end
disp(x)
disp(xi)
disp(iter)

Upvotes: 1

Views: 203

Answers (3)

evaristor
evaristor

Reputation: 1

Execute your function with >>secant(@(x) x^3 +12, x0, x1, tol)

Upvotes: 0

Rody Oldenhuis
Rody Oldenhuis

Reputation: 38032

Use a function_handle: 

[xans,xi,k] = secant(@myFunc,1,2,0.0001);

Also, call the funciton inside the loop, otherwise, it doesn't update:

function [xans, xi, k] = secant(f, x0, x1, tol)

    %// etc.  

    while abs(c) >= tol
        c = f(b) * (b-a)/(f(b)-f(a));
        %// etc.
    end
end

Upvotes: 0

Chris Taylor
Chris Taylor

Reputation: 47392

You seem a little confused about loops and functions in Matlab. The function you have written doesn't do any updating of the variables inside the loop. When you write

c = f(b) * (b-a) / (f(b) - f(a));

that stores one value in c, but it doesn't automatically update c every time you go through the loop. Instead, I'd write something like this

function x1 = secant(f, x0, x1, tol)

y1 = f(x1);
y0 = f(x0);
while abs(y1) > tol
    tmp = x1;                              %// Store the old value of x1
    x1  = x1 - y1 * (x1 - x0) / (y1 - y0); %// Use the secant method to update x1
    x0  = tmp;                             %// x0 gets the old value of x1
    y0  = y1;                              %// We already know what f(x0) is
    y1  = f(x1);                           %// Need to re-compute f(x1)
end

which you can then call as follows. The first argument is known as a function handle.

>> secant( @(x)x^2-2, 0, 1, 1e-6)
ans = 
   1.414213562057320

Upvotes: 2

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