R help - Jul 2002 - help with adapt function

Dear People,

I'm trying to use the function adapt, from the adapt library package,
which does multidimensional numerical integration. I think I must be using
the wrong syntax or something, because even a simple example does not
work. Consider

 foo <- function(x){x[1]*x[2]}

and

adapt(2, lo = c(-1,-1), up = c(1,1), functn = foo)

This simply hangs. A more complicated example crashes R, but never
mind that for the moment. Perhaps some kind person could explain what
I am doing wrong here?

The help page says

"functn an R function which should take a single vector argument and
        possibly some parameters and return the function value at that
        point. functn must return a single numeric value."

which is what I think I am doing here, but I expect I am missing
something as usual.

By the way, the help page does not say this explicitly, but I assume
that the first component in the vectors "lower" and "upper"
correspond
respectively to the lower and upper limits of the integral for the
first component of the vector argument in the function, and so on?

                                       Sincerely, Faheem Mitha.

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On Sat, 13 Jul 2002, Faheem Mitha wrote:
>
> Dear People,
>
> I'm trying to use the function adapt, from the adapt library package,
> which does multidimensional numerical integration. I think I must be using
> the wrong syntax or something, because even a simple example does not
> work. Consider
>
>  foo <- function(x){x[1]*x[2]}
>
> and
>
> adapt(2, lo = c(-1,-1), up = c(1,1), functn = foo)
>
> This simply hangs. A more complicated example crashes R, but never
> mind that for the moment. Perhaps some kind person could explain what
> I am doing wrong here?

You're trying to compute an integral that's exactly zero and not
specifying a maximum number of evaluations. The Fortran code
in adapt() uses an estimate of the relative error in deciding if the
answer is accurate, and if you don't specify the maxpts argument it will
keep increasing the number of evaluations until the relative error
criterion is satisfied.

If the true integral is zero the relative error will never be less than
the specified level and the function will never stop.  With your example I
get
>  adapt(2, lo = c(-1,-1), up = c(1,1), functn = foo,maxpts=10000)
                                                    value
                                             8.673617e-19
                                                   relerr
                                                        1
                                                   minpts
                                                     9999
                                                   lenwrk
                                                     1403
                                                    ifail
                                                        1
                                                     warn
Ifail=1, maxpts was too small. Check the returned relerr!
Warning message:
Ifail=1, maxpts was too small. Check the returned relerr! in: adapt(2, lo
= c(-1, -1), up = c(1, 1), functn = foo, maxpts = 10000)

or integrating over just one quadrant so the answer isn't zero
>  adapt(2, lo = c(-1,-1), up = c(0,0), functn = foo)
      value      relerr      minpts      lenwrk       ifail
       0.25 7.38275e-08         165          73           0

or changing the function slightly
> bar<-function(x) {foo(x)+1e-4}
>  adapt(2, lo = c(-1,-1), up = c(1,1), functn = bar)
      value      relerr      minpts      lenwrk       ifail
      4e-04 9.23213e-05         165          73           0


	-thomas

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