Re: Mistake in applying a rule

*To*: mathgroup at smc.vnet.net*Subject*: [mg80701] Re: Mistake in applying a rule*From*: dimitris <dimmechan at yahoo.com>*Date*: Wed, 29 Aug 2007 04:18:58 -0400 (EDT)*References*: <fb0f17$f05$1@smc.vnet.net>

On 28 , 09:23, Steven Siew <sie... at bp.com> wrote: > Hi, > > I'm trying to show that Integrate[Cos[x-a]] is Sin[x-a] + C > > I manage to get Simplify @ Integrate[Cos[x-a]] to give -Sin[a-x] > > This is fine because Sin[-x] is equal to -Sin[x] > > But when I apply the rule Times[-1,Sin[Plus[g_,Times[-1,h_]]]] -> > Sin[h-g] it fails. > > Here is the complete transcript. > > Mathematica 5.2 for Students: Microsoft Windows Version > Copyright 1988-2005 Wolfram Research, Inc. > > In[1]:= > Out[1]= {stdout} > > In[2]:= (* Write your mathematica code below *) > > In[3]:= Simplify @ Integrate[Cos[x-a],x] > > Out[3]= -Sin[a - x] > > In[4]:= FullForm[%] > > Out[4]//FullForm= Times[-1, Sin[Plus[a, Times[-1, x]]]] > > In[5]:= % /. Times[-1,Sin[Plus[g_,Times[-1,h_]]]] -> Sin[h-g] > > Out[5]= -Sin[a - x] > > In[6]:= (* End of mathematica code *) > > In[7]:= Quit[]; (The following is for Mathematica 5.2. But I think there belong to fundemental aspects of Mathematica to have changed to the new version) For Mathematica evaluator In[5]:= Sin[x - a] Out[5]= -Sin[a - x] In[6]:= -Sin[x - a] Out[6]= Sin[a - x] always. So there is not a trivial way to write -Sin[a - x] as Sin[x-a]. I think it has to do with the ordering of the symbols. (I think also you want to avoid any extraordinary rules/modifications). In[10]:= Head /@ {x, a, y} Out[10]= {Symbol, Symbol, Symbol} For Mathematica evaluator the ordering of the symbol x follows a. Just compare In[12]:= {-Sin[x - a], -Sin[a - x] /. a -> y} Out[12]= {Sin[a - x], Sin[x - y]} Dimitris