Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2010

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

Search the Archive

Re: assuming certain properties about variables

  • To: mathgroup at smc.vnet.net
  • Subject: [mg111513] Re: assuming certain properties about variables
  • From: Murray Eisenberg <murray at math.umass.edu>
  • Date: Tue, 3 Aug 2010 06:43:26 -0400 (EDT)

I don't see what $Assumptions has to do with the situation at hand, 
since using

   $Assumptions = t>0

does not change the result of leaving f[t] unevaluated; nor does either of

   Assuming[t>0, f[t]]
   Simplify[f[t], t>0]

The only really workable thing I've seen posted in response to the 
O.P.'s query -- sorry, I cannot find any previous posting by you to 
MathGroup on this subject -- is Leonid Shifrin's [mg111421], using UpValues:

   t /: Positive[t] = True

And I guess that, except for employing user-defined heads to create new 
types of objects, that's the closest one can come to declaring types of 
variables.  Or am I missing something?

On 8/2/2010 4:26 AM, Christoph Lhotka wrote:
> hello,
>
> there is the symbol $Assumptions exactly for this purpose (please see my
> previous response for a working definition of f, such that it will
> return t, whenever it is assumed to be positive)...
>
> best,
>
> christoph
>
> On 01/08/2010 10:57, Murray Eisenberg wrote:
>> But once you put the restriction on the argument of f, you've told
>> Mathematica not to carry out the evaluation of f unless the input
>> supplied is actually positive.
>>
>> If now t is just a symbol, then it is not positive (and not negative,
>> either) -- it's just a symbol. So why would you expect to be able to
>> tell Mathematica that it's positive? In general, Mathematica variables
>> don't really have types.
>>
>> If it's just the particular symbol t that you want to supply to f, then
>> you could do this (I'm changing your function definition for clarity) --
>> not that you probably want SetDelayed (:=) instead of Set (=):
>>
>> f[x_?Positive] := x^2
>> f[t] = t^2;
>>
>> f[3]
>> 9
>> f[t]
>> t^2
>> t = -5;
>> f[t]
>> f[-5]
>>
>> f
>>
>>
>> On 7/31/2010 2:40 AM, Benjamin Hell wrote:
>>> let's say I have defined the following function:
>>> f[x_?Positive] = x
>>> Now I want to evaluate f with a variable t:
>>> f[t]
>>> As mathematica knows nothing about t, the output is f[t] instead of t.
>>>
>>> How can I tell mathematica, that t should be a positive number so that
>>> Positive[t] evaluates true and then f[t] evaluates to t?
>>> Of course this is just an example, which should present what I would
>>> like to know.
>

-- 
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


  • Prev by Date: Re: ODE solution transfer
  • Next by Date: Re: A distribution problem using Mathematica
  • Previous by thread: Re: Re: assuming certain properties about variables
  • Next by thread: Re: assuming certain properties about variables