Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

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

Search the Archive

Re: another bug...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47667] Re: another bug...
  • From: ab_def at prontomail.com (Maxim)
  • Date: Wed, 21 Apr 2004 05:23:05 -0400 (EDT)
  • References: <c62jsh$rgh$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Enrique Zeleny <ezeleny at fismat1.fcfm.buap.mx> wrote in message news:<c62jsh$rgh$1 at smc.vnet.net>...
> Hi,
> 
> 
> In[1]:= expr = FunctionExpand[Cos[Pi/22]];
> 
> The next input doesn't work in 5.0, the expression that I want to replace
> by a appears many times in expr
> 
> In[2]:= expr /. (1/2)*(-1 + Sqrt[5]) + I*Sqrt[2 - (1/2)*(-1 - Sqrt[5])] -> a
> 
> only works doing this
> 
> In[3]:= expr //. (1/2)*(-1 + Sqrt[5]) + I*Sqrt[2 - (1/2)*(m_)] -> a
> 
> If you look at FullForm
> 
> In[4]:= FullForm[(1/2)*(-1 + Sqrt[5]) + I*Sqrt[2 - (1/2)*(-1 - Sqrt[5])]]
> 
> Out[4]:= Plus[Times[Rational[1, 2], Plus[-1, Power[5, Rational[1, 2]]]],
>   Times[Complex[0, 1],
>     Power[Plus[2, Times[Rational[1, 2], Plus[1, Power[5, Rational[1, 2]]]]],
>       Rational[1, 2]]]]
> 
> but if you copy and paste the result of expr and you use //.
> (1/2)*(Sqrt[5] - 1) + I*Sqrt[(1/2)*(Sqrt[5] + 1) + 2] -> a to replace works well!!!
> 
> In Version 4.1 only this works well (but not in 5.0), but the expression
> that appears many times is written with +'s
> 
> In[2]:= expr //. (1/2)*(-1 + Sqrt[5]) + I*Sqrt[2 + (1/2)*(1 + Sqrt[5])] ->
> a
> 
> If you copy and paste the result of expr and you use only /. instead of
> //. to replace (1/2)*(-1 + Sqrt[5]) + I*Sqrt[2 + (1/2)*(1 + Sqrt[5])] ->a
> works fine!!!
> 
> 
> Clearly the problem is recognizing
> 
> - (1/2)*(-1 - Sqrt[5])
> 
> 
> 
> Enrique Zeleny
> Universidad Autonoma de Puebla
> Mexico

I'd say there are three questions here. First is that we have to know
the internal representation of the expression:

In[1]:=
MatchQ[1/2, HoldPattern[1/2]]

Out[1]=
False

because 1/2 is Rational[1,2] but with HoldPattern preventing
evaluation 1/2 is stored as Times[1, Power[2,-1]]. Second is
understanding that Mathematica can perform some transformations
automatically:

In[2]:=
MatchQ[-1/2(-a-b), k_(-a-b)]

Out[2]=
False

because -1/2(-a-b) is rewritten as 1/2(a+b). This is more annoying;
for example,

In[3]:=
MatchQ[-2(-a-b), k_(-a-b)]

Out[3]=
True

works differently -- no transformation is done here. And the third
thing is that in the case of FunctionExpand[Cos[Pi/22]] there is
another complication: -1/2(-1-Sqrt[5]) in the output is not rewritten
as 1/2(1+Sqrt[5]). I think this violates the principle of infinite
evaluation (I like the term fixed point evaluation), because when you
copy and paste the output and reevaluate it, you get a different
expression, now with -1/2(-1-Sqrt[5]) converted to 1/2(1+Sqrt[5]).
ReplaceAll outcome will also be different.

Maxim Rytin
m.r at prontomail.com


  • Prev by Date: Undo and Plot3D
  • Next by Date: Re: Elliptic Curves and Cryptography Questions
  • Previous by thread: another bug...
  • Next by thread: Adding your own Help