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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
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