[R] Help in Bissection algorithm

[William Dunlap]

And have the function return a value.  All values
set in the function will be destroyed when the
function is done - you must return something and
the caller must assign the return value to a variable
if you want to keep it.

E.g.,
raiz
function (f, a, b, e) 
{
    i <- 0
    repeat {
        if (i > 50) {
            break
        }
        if (abs(a - b) < e) {
            m = (a + b)/2
            raiz1 = m
        }
        if (abs(a - b) > e) {
            m = (a + b)/2
            af = f(a)
            if ((af * f(m)) > 0) {
                a = m
            }
            if ((af * f(m)) < 0) {
                b = m
            }
        }
        i = i + 1
    }
    c(a, b) # bounds on estimate of root
}
> z <- raiz(function(x)x^4 - 3, -2, -1, 1e-10)
> z
[1] -1.316074 -1.316074
> z^4 - 3
[1]  7.566836e-12 -5.231717e-10

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com

> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf
> Of Berend Hasselman
> Sent: Thursday, November 29, 2012 12:13 PM
> To: finehko
> Cc: r-help at r-project.org
> Subject: Re: [R] Help in Bissection algorithm
> 
> 
> On 29-11-2012, at 16:18, finehko wrote:
> 
> > I'm trying to make a function witch recieves a function "f", "a" and "b", and
> > the error "e". When I run the algorithm without labeling it a function and
> > typing the values it works, but when I just try to save it in a function It
> > doesn't ' bug but don't do anything.
> 
> I don't understand what you mean with this: "doesn't ' bug but don't do anything"
> 
> > Here's the code, anyone know whats
> > happening?
> > raiz=function(f,a,b,e){
> >  repeat{
> >    if(i>50){break}
> >    if(abs(a-b)<e){m=(a+b)/2
> >                  raiz1=m}
> >    if(abs(a-b)>e){
> >          m=(a+b)/2
> >          af=f(a)
> >           if((af*f(m))>0){a=m}
> >           if((af*f(m))<0){b=m}}
> >        i=i+1}}
> >
> 
> Initialize i before starting the repeat loop.
> i <- 1
> 
> Berend
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