Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2000
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2000

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Evaluate and HoldAll

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24222] Re: Evaluate and HoldAll
  • From: Andre Heinemann <andre at aflo4.ifw-dresden.de>
  • Date: Sat, 1 Jul 2000 03:21:53 -0400 (EDT)
  • Organization: IFW
  • References: <395C69F9.44E390C9@aflo4.ifw-dresden.de>
  • Sender: owner-wri-mathgroup at wolfram.com

Andre Heinemann wrote:

> Hi mathgroup
>
> Attributes[NIntegrate] gives: {HoldAll, Protected}
>
> In THE BOOK (or online Help) one can find:
>
> *****************************************************************************
>
> *You can use Evaluate to evaluate the arguments of a HoldAll function in
> a controlled way.  *
> *****************************************************************************
>
> So I built:
>
> rseq = Sequence[a1, a2, a3]
> lseq = Sequence[a1_?NumericQ, a2_?NumericQ, a3_?NumericQ]
>
> My Function f[r, rseq]
>
> gives the output: (e.g.) a1 Exp[a2 - a3 r]
> (I know that I can use Integrate here, instead of Nintegrate but it is
> for testing)
>
> Clear[test]
> test[q_, lseq] := NIntegrate[Evaluate[f[r, rseq]], {r, 0, 10^5}]
>
> test[0.1, 1, 0.2, 0.5 ]  presented:
>
> NIntegrate::"inum" :  ... Integrand ... is not numerical at r = .... and
> so on
>
> NIntegrate[f[r, rseq],{r, 0, 10^5}]
>
> Ok, I can see that the integrand is not numerical  and so I tried:
>
> Clear[test]
> test[q_, lseq] := NIntegrate[a1 Exp[a2 - a3 r] , {r, 0, 10^5}] and
>
> test[0.1, 1, 0.2, 0.5 ] works very well and finds: 2.442805516320 as the
> answer !
>
> What is the point here ? I think my understanding of Evaluate[] is false
> but
> how can I solve the problem ?
>
> I would be glad to have an explanation or a hint.

Hi

Sorry for this update, but I think I found a better formulation of my problem,
and I found out it has nothing to do with Evaluate and HoldAll.
Here is what I mean:

lseq = Sequence[a1_, a2_, a3_]
rseq = Sequence[a1, a2, a3]

Clear[gr]
gr[x_, lseq] := a1*Exp[a2 - a3 x]

In[1415]:=
gr[r, rseq]

Out[1415]=
\!\(a1\ \[ExponentialE]\^\(a2 - a3\ r\)\)

Clear[t]
t[q_, lseq] :=
  NIntegrate[(4\[Pi]/q)*r*Sin[q r]*gr[r, Sequence[a1, a2, a3]], {r, 0, 10^5},
    MinRecursion -> 5,
    MaxRecursion -> 30,
    PrecisionGoal -> 10,
    WorkingPrecision -> 64]

In[1406]:=
t[0.1, 1, 2, 3]

Out[1406]=
6.862786625992

Works very well ! But I need for further developments:

Clear[t]
t[q_, lseq] :=
  NIntegrate[(4\[Pi]/q)*r*Sin[q r]*gr[r, rseq], {r, 0, 10^5},
    MinRecursion -> 5,
    MaxRecursion -> 30,
    PrecisionGoal -> 10,
    WorkingPrecision -> 64]

... some errors...

\!\(NIntegrate[\(\((4\ \[Pi])\)\ r\ Sin[0.1`\ r]\ gr[r, rseq]\)\/0.1`, {r, 0,
      10\^5}, MinRecursion -> 5, MaxRecursion -> 30, PrecisionGoal -> 10,
    WorkingPrecision -> 64]\)

Where you can see the problem. The right side in the definition := has not been
evaluated,
so mathematica doesn't know somthing about a1, a2, a3.


Clear[t]
t[q_, lseq] := Module[{},
    core = r*Sin[q r]*gr[r, rseq]
    ]

t[0.1, 1, 2, 3]

\!\(125.66370614359172`\ a1\ \[ExponentialE]\^\(a2 - a3\ r\)\ r\ Sin[
      0.1`\ r]\)

-------------------------
AND the rigth answer should be:
-------------------------

Clear[t]
t[q_, lseq] := Module[{},
    core = (4\[Pi]/q)*r*Sin[q r]*gr[r, a1,a2,a3]
    ]

t[0.1, 1., 2., 3.]

\!\(1.\[ExponentialE]\^\(\(\(2.)\(\[InvisibleSpace]\)\) - 3. r\)\ r\ \
Sin[0.1 r]\)



 I have not the faintest idea how to find a solution for this.

Andre


  • Prev by Date: Re: making a plane
  • Next by Date: LaTeX (eps) file into notebook
  • Previous by thread: Evaluate and HoldAll
  • Next by thread: Re: Re: Evaluate and HoldAll