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MathGroup Archive 2006

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Re: Problem with using /.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69273] Re: Problem with using /.
  • From: Peter Pein <petsie at dordos.net>
  • Date: Tue, 5 Sep 2006 05:30:35 -0400 (EDT)
  • References: <edadil$pd0$1@smc.vnet.net><eddqil$3uh$1@smc.vnet.net> <edg849$gel$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Nag schrieb:
> Peter Pein wrote:
>> Hi Nag,
>>
>> you try to assign a value to Sum[...]; this becomes Set[Sum[...],0] --
>> but Sum is a Protected function. Furthermore, Set is not Rule.
>>
>> I guess you wanted:
>>
>> M[j_, t_, k_] := FullSimplify[D[MGF[t, k], {t, j}] /. u -> 0,
>>      Sum[Subscript[p, i], {i, 0, k}] == 1];
>>
>> You can use Simplfy instead of FullSimplify too.
>>
>> Peter
> 
> Hi Peter:
> 
> The = was a typo. It should have been a  ->
> With  -> there is no error but the substitution doesn't happen.
> The substitution works when I use == as you suggested.
> 
> Why does -> not work but == work?
> 
> Thanks
> Nag
> 

Hi Nag,

if Sum[..]->1 is in the list of rules, the pattern matcher recognizes it 
only in exactly this form. If Sum[..]==1 is given as second parameter of 
(Full)Simplify, Mathematica tries to use the _information_, not the 
_pattern_ inside the call to Simplify.

expr = (x^2 - y^2)/(1 + x);

The pattern "x+y" doesn't occur, so the outcome is expr:

expr /. x + y -> 1
--> (x^2 - y^2)/(1 + x)

FullSimplify uses x+y==1, to get rid of the squares:

FullSimplify[expr, x + y == 1]
--> (1 - 2*y)/(1 + x)

You can isolate y in x+y==1 and use that as rule
(same as expr /. Solve[x + y == 1, y][[1]] // Together):

Together[expr /. y -> 1 - x]
--> (-1 + 2*x)/(1 + x)

or isolate x:

Together[expr /. x -> 1 - y]
--> (-1 + 2*y)/(-2 + y)

hth,
Peter


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