Re: Use of Mathematica with Rule-based Equation Derivations
- To: mathgroup at smc.vnet.net
- Subject: [mg61935] Re: [mg61914] Use of Mathematica with Rule-based Equation Derivations
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sat, 5 Nov 2005 01:52:18 -0500 (EST)
- References: <200511041011.FAA14913@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 4 Nov 2005, at 19:11, mmorriss at gcn.ou.edu wrote:
> Hi All - I'm a many-year user of mathematica, but have always had one
> particular problem with it that I may have just simply missed reading
> about.
>
> Mathematica Version: 5.2
>
> Problem: I would like to develop a set of re-rewite rules to apply
> to the
> Expected Value operator. For example:
>
> E[a x] = a E[x] a -> constant, x -> variable
> E[b + f[x]] = b + E[f[x]] , etc.
>
> The issue is how does one using Mathematica distinguish a 'constant
> variable (i.e. a and b)' from a variable 'variable' (i.e. 'x')? The
> head
> of a, b and x is 'Symbol' and neither a, b nor x contain a number so I
> can't use a_?NumberQ to identify it as a constant.
>
> This actually goes to the wider question of how does one use
> Mathematica
> for symbolic derviations where numbers are not actually substituted
> in the
> derviation?
>
> E.g. E[a + E[b x]] /. Rule2
> out= a + b E[x] etc.
>
> Thanks all - Mark Morrissey
> University of Oklahoma
>
>
First of all, it is not a good idea to use a protected symbol (E) in
this way. But concerning your question: there is no reason why you
shouldn't use NumberQ or NumericQ in the way you indicated.
First define the rule exactly in the way you said you could not:
Expectation[b_?NumericQ* x_]:=b*Expectation[x]
Now set:
NumericQ[a]=True;
Now you get
Expectation[a x]
a Expectation[x]
and of course as a bonus you automatically get:
Expectation[3 x]
3 Expectation[x]
This gives me a chance to offer a puzzle (for which I offer no prize)
for those who like this sort of things. Note that
NumericQ[a]
True
but where is this information stored by Mathematica? It can't be a
DownValue for NumericQ, since NumericQ has the Attribute Protected
and I did not unprotect it. And indeed:
DownValues[NumericQ]
{}
Or, we could try to give a an UpValue:
Expectation[b_?NumericQ* x_]:=b*Expectation[x]
a/:NumericQ[a]=True;
NumericQ[a]
True
Expectation[a x]
a Expectation(x)
Everything works fine yet:
UpValues[a]
{}
So now again the puzzle: where is this information stored?
Note also that NumberQ behaves quite differently. In fact you can't
set NumberQ[a]=True without first unprotecting NumberQ. On the other
hand if you use an UpValue to make a into a "number"
a/:NumberQ[a]=True;
then as expected:
UpValues[a]
{HoldPattern[NumberQ[a]] :> True}
Well, any guesses?
Andrzej Kozlowski
P.S. All the above with Mathematica 5.1 but I expect it;s the same in
5.2
- Follow-Ups:
- Re: Re: Use of Mathematica with Rule-based Equation Derivations
- From: János <janos.lobb@yale.edu>
- Re: Re: Use of Mathematica with Rule-based Equation Derivations
- References:
- Use of Mathematica with Rule-based Equation Derivations
- From: mmorriss@gcn.ou.edu
- Use of Mathematica with Rule-based Equation Derivations