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MathGroup Archive 2004

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Re: Re: How to reprensent the result in vector form?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg51696] Re: [mg51657] Re: How to reprensent the result in vector form?
  • From: DrBob <drbob at bigfoot.com>
  • Date: Fri, 29 Oct 2004 03:39:48 -0400 (EDT)
  • References: <cl9t1d$7m0$1@smc.vnet.net> <200410230422.AAA26310@smc.vnet.net> <clndek$o4s$1@smc.vnet.net> <200410280343.XAA09777@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

>> However, I think the original question was to do with spotting a
>> particular type of pattern in the algebra, and then rewriting it
>> in a more compact and familiar notation.

If such rules work only for a very limited set of situations, it's arguably easier to type the new answer (because we've recognized what it should be) rather than write a rule that retrieves it in the specific case (which amounts to concocting rules on the fly).

Your new rule seems general enough to be genuinely useful, however.

Bobby

On Wed, 27 Oct 2004 23:43:55 -0400 (EDT), Steve Luttrell <steve_usenet at _removemefirst_luttrell.org.uk> wrote:

> I fully agree with you that my suggestion was not a general solution. I had
> in mind the sort of piecemeal replacement(s) you sometimes have to do at the
> end of a derivation to coax the result into the form you want. I usually
> concoct these on the fly and store them as rules.
>
> A quick (and very dirty!) fix to the problem of matching to ALL possible
> sign combinations is
>
> crossrule2=
>   Flatten[Table[{Sign[ai3 bi2]a3_ b2_ -Sign[ai2 bi3]a2_ b3_,
>           Sign[ai1 bi3]a1_ b3_ -Sign[ai3 bi1]a3_ b1_,
>           Sign[ai2 bi1]a2_ b1_ -Sign[ai1 bi2]a1_ b2_}->
>         myCross[{bi1 b1,bi2 b2,bi3 b3},{ai1 a1,ai2 a2,ai3 a3}],{ai1,-1,1,
>         2},{ai2,-1,1,2},{ai3,-1,1,2},{bi1,-1,1,2},{bi2,-1,1,2},{bi3,-1,1,2}],
>     5];
>
> Now your 3 counterexamples behave thus:
>
> Cross[{-u1,u2,u3},{v1,v2,v3}]/.crossrule2
>
> myCross[{u1,-u2,-u3},{-v1,-v2,-v3}]
>
> Cross[{u1,-u2,u3},{v1,v2,v3}]/.crossrule2
>
> myCross[{-u1,u2,-u3},{-v1,-v2,-v3}]
>
> Cross[{u1,u2,-u3},{v1,v2,v3}]/.crossrule2
>
> myCross[{-u1,-u2,u3},{-v1,-v2,-v3}]
>
> I agree with your final two remarks about the non-uniqueness of the
> representation. However, I think the original question was to do with
> spotting a particular type of pattern in the algebra, and then rewriting it
> in a more compact and familiar notation. Because this sort of
> pseudo-inversion is non-unique I tend to make up the replacement rules on an
> ad hoc basis as I said earlier.
>
> Steve Luttrell
>
> "DrBob" <drbob at bigfoot.com> wrote in message
> news:clndek$o4s$1 at smc.vnet.net...
>> This won't always work, however.
>>
>> crossrule = {a3_ b2_ - a2_ b3_, a1_ b3_ - a3_ b1_,
>>   a2_ b1_ - a1_ b2_} -> myCross[{b1, b2, b3}, {a1, a2, a3}];
>>
>> Cross[{u1, u2, u3}, {v1, v2, v3}] /. crossrule
>> myCross[{u1,u2,u3},{v1,v2,v3}]
>>
>> Fine so far, but these should return cross products too:
>>
>> Cross[{-u1,u2,u3},{v1,v2,v3}]/.crossrule
>> {-u3 v2+u2 v3,u3 v1+u1 v3,-u2 v1-u1 v2}
>>
>> Cross[{u1,-u2,u3},{v1,v2,v3}]/.crossrule
>> {-u3 v2-u2 v3,u3 v1-u1 v3,u2 v1+u1 v2}
>>
>> Cross[{u1,u2,-u3},{v1,v2,v3}]/.crossrule
>> {u3 v2+u2 v3,-u3 v1-u1 v3,-u2 v1+u1 v2}
>>
>> Besides this problem there are other issues, such as:
>>
>> (1) EVERY vector is the cross product of two vectors,
>>
>> and
>>
>> (2) The vectors are never unique.
>>
>> Good luck!!
>>
>> Bobby
>>
>> On Sat, 23 Oct 2004 00:22:49 -0400 (EDT), Steve Luttrell
>> <steve_usenet at _removemefirst_luttrell.org.uk> wrote:
>>
>>> You can use a replacement rule to convert your expression into the
>>> compact
>>> form that you want:
>>>
>>> Define the replacement rule.
>>>
>>> crossrule = {a3_ b2_ -a2_ b3_,a1_ b3_ -a3_ b1_,a2_ b1_ -a1_ b2_} ->
>>> myCross[{b1,b2,b3},{a1,a2,a3}];
>>>
>>> Apply the rule to your expression.
>>>
>>> result = {-u3 v2 + u2 v3, u3 v1 - u1 v3, -u2 v1 + u1 v2}/.crossrule
>>>
>>> which gives the output
>>>
>>> myCross[{u1,u2,u3},{v1,v2,v3}]
>>>
>>> You could make the notation more compact by doing this
>>>
>>> result /. {{u1,u2,u3} -> u,{v1,v2,v3} -> v}
>>>
>>> which gives the output
>>>
>>> myCross[u,v]
>>>
>>> Another (related but more sophisticated) method is to use the
>>> Utilities`Notation` package to implement the reformatting of your output
>>> automatically. This includes a way of defining infix operators so that
>>> you
>>> can get an output that looks like u x v rather than myCross[u,v]
>>>
>>> Steve Luttrell
>>>
>>> "melon" <sweetmelon at gmail.com> wrote in message
>>> news:cl9t1d$7m0$1 at smc.vnet.net...
>>>> I used mathmetica5 to solve a vector equation. Mathmetica expands the
>>>> result into sub-variable form. But I want to get the vector form. How
>>>> to do that?
>>>>
>>>> ex:
>>>> Solution:
>>>> {-u3 v2 + u2 v3, u3 v1 - u1 v3, -u2 v1 + u1 v2}
>>>> I want mathematica give me {u cross v}
>>>>
>>>> p,s,  Could you add me into the forum? Thank you very much. Regards.
>>>>
>>>> ShenLei
>>>>
>>>
>>>
>>>
>>>
>>>
>>
>>
>>
>> --
>> DrBob at bigfoot.com
>> www.eclecticdreams.net
>>
>
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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