calling function with multi-variable functions

Question

I have the following function which finds the zero of a function using Newton-Raphson:

function [ x,i ] = nr_function( x0,f,fp )

N   = 15;
eps = 1e-5;

x=x0;
for i=0:N
    if( abs(f(x))<eps )
        return;
    end

    x=x-f(x)/fp(x);
end

I can call the function like this:

f = @(x) x.^3-1
fp = @(x) 3*x.^2
nr_function(3, f,fp)

However, say I instead define my function like this, i.e. taking 2 variables:

f = @(x, q) q*x.^3-1
fp = @(x, q) q*3*x.^2

Then how would I be able to call nr_function with f and fp? I tried nr_function(3, f,fp), but that doesn't work

q in this case is defined at the call of nr_function?

Matt
– Matt

2015-07-27 08:39:55 +00:00
Commented Jul 27, 2015 at 8:39 — Matt
– Matt, Commented Jul 27, 2015 at 8:39
@Matt Yes, it is defined when I call nr_function

BillyJean
– BillyJean

2015-07-27 08:41:12 +00:00
Commented Jul 27, 2015 at 8:41 — BillyJean
– BillyJean, Commented Jul 27, 2015 at 8:41

Matt · Accepted Answer · 2015-07-27 08:52:15Z

2

If q is defined when you call nr_function, you can use an anonymous function in the call. When you do this, then the argument you pass is a new function-handle with variable x and constant q.

f  = @(x, q) q*x.^3-1
fp = @(x, q) q*3*x.^2

q = 1;

nr_function(3, @(x)f(x,q), @(x)fp(x,q))

Note: It's not necessary that you use the variable x in the anonymous function. The only importance is to have a single argument in the end. So we can use for example y as the intermediate variable like this:

nr_function(3, @(y)f(y,q), @(y)fp(y,q))

If we expand it to multiple lines, it would look like this:

f  = @(x, q) q*x.^3-1
fp = @(x, q) q*3*x.^2

q = 1;

f2  = @(y) f(y,q)
fp2 = @(y) fp(y,q)

nr_function(3, f2, fp2)

edited Jul 27, 2015 at 8:52

answered Jul 27, 2015 at 8:43

Matt

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