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Re: Re: Hold and Equal
*To*: mathgroup at smc.vnet.net
*Subject*: [mg73918] Re: [mg73799] [mg73799] Re: Hold and Equal
*From*: Murray Eisenberg <murray at math.umass.edu>
*Date*: Sat, 3 Mar 2007 01:00:58 -0500 (EST)
*Organization*: Mathematics & Statistics, Univ. of Mass./Amherst
*References*: <erufqm$s7j$1@smc.vnet.net> <200702271048.FAA24024@smc.vnet.net> <es3ib9$nus$1@smc.vnet.net> <200703021058.FAA02239@smc.vnet.net>
*Reply-to*: murray at math.umass.edu
By itself, yes, that works. But not in a sufficiently robust way if I
want to call formEquation within, say, a Table expression where there's
a parameter n that is part of the first argument to formEquation, as in:
Table[formEquation[Integrate[x^n,x],Identity],{n,1,2}]
Table[formEquation[(a+b)^n,Expand],{n,1,2}
However, the methods suggested by albert in mg 73863 -- using a With
construct within Table, or using a replacement for n -- handle this
situation nicely.
My question seems to have been provocative, and I thank all who responded.
Jens-Peer Kuska wrote:
> Hi,
>
> and
>
> SetAttributes[formEquation, HoldFirst]
>
> formEquation[expr_, op_] := HoldForm[expr = z] /. z -> op[expr]
>
> will work as expected.
>
> Regards
> Jens
>
> Murray Eisenberg wrote:
>> Aha! I believe this approach _almost_ allows me to accomplish what I
>> was REALLY trying to accomplish. It certainly works in the example I
>> gave. If I encapsulate this in a function...
>>
>> formEquation[expr_, op_]:= HoldForm[expr=z]/.z\[Rule]op[expr]
>>
>> ... then
>>
>> formEquation[(a+b)^2,Identity]
>>
>> will produce exactly what I want.
>>
>> However, if I try something like the example I was really after (which I
>> didn't mention in my original post, since I gave something simpler), it
>> works in the direct version...
>>
>> HoldForm[Integrate[x^2,x] = z] /. z\[Rule]Integrate[x^2,x]
>>
>> but not with the encapsulating function:
>>
>> formEquation[Integrate[x^2, x], Identity]
>>
>> The latter produces the equation
>>
>> x^3/3 = x^3/2
>>
>> whereas I want the left-hand side to be the unevaluated integral expression.
>>
>> You can tell I'm struggling with Hold! (One of the "last frontiers" in
>> my Mathematica education.)
>>
>> bghiggins at ucdavis.edu wrote:
>>> Murray,
>>>
>>> Try this
>>>
>>>
>>> HoldForm[(a + b)^2 = z] /. z -> Expand[(a + b)^2]
>>>
>>>
>>> (a + b)^2 = a^2 + 2*a*b + b^2
>>>
>>> Cheers,
>>>
>>> Brian
>>>
>>>
>>>
>>> On Feb 26, 3:20 am, Murray Eisenberg <mur... at math.umass.edu> wrote:
>>>> How can I produce in an Output cell (under program control) an
>>>> expression like the following,
>>>>
>>>> (a+b)^2 = a^2+ 2 a b + b^2
>>>>
>>>> where instead of the usual Equal (==) I get a Set (=), as in traditional
>>>> math notation? I want to input the unexpanded (a+b)^2 and have the
>>>> expansion done automatically.
>>>>
>>>> Of course, I can try something like the following:
>>>>
>>>> (a+b)^2 == Expand[(a+b)^2])
>>>>
>>>> So how do I convert the == to =? Of course
>>>>
>>>> ((a + b)^2 == Expand[(a + b)^2]) /. Equal -> Set
>>>>
>>>> gives a Set::write error. And
>>>>
>>>> (Hold[(a + b)^2 == Expand[(a + b)^2]]) /. Equal -> Set
>>>>
>>>> doesn't actually evaluate the Expand part and leaves the "Hold" wrapper.
>>>>
>>>> --
>>>> Murray Eisenberg mur... at math.umass.edu
>>>> Mathematics & Statistics Dept.
>>>> Lederle Graduate Research Tower phone 413 549-1020 (H)
>>>> University of Massachusetts 413 545-2859 (W)
>>>> 710 North Pleasant Street fax 413 545-1801
>>>> Amherst, MA 01003-9305
>>>
>
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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