Simplify made more intelligent
- To: mathgroup at yoda.ncsa.uiuc.edu
- Subject: Simplify made more intelligent
- From: jacobson at cello.hpl.hp.com
- Date: Wed, 21 Nov 90 09:17:12 PST
Ken Yip asked for suggestions regarding Integrate[Cosh[x],x].
Basically, Mathematica always turns hyperbolics into exponentials. So
in theory it already knows how to do this. But note that, even with
the help of Simplify, it cannot determine that the result is equal to
Sinh[x].
In[5]:= Integrate[Cosh[x],x]==Sinh[x]
x -x x
-1 E -E + E
Out[5]= ---- + -- == ---------
x 2 2
2 E
In[6]:= Simplify[%]
2 x -x x
-1 + E -E + E
Out[6]= --------- == ---------
x 2
2 E
On the other hand, if I subtract that two from each other, Simplify can
determine that the result is zero:
In[13]:= Integrate[Cosh[x],x]-Sinh[x]
x -x x
-1 E -E + E
Out[13]= ---- + -- - ---------
x 2 2
2 E
In[14]:= Simplify[%]
Out[14]= 0
Conclusion: If you want to check that two symbolic things are equal,
it is better to test the difference for zero than the two things for
equality. I managed to teach Simplify to do this:
In[16]:= Unprotect[Simplify]
Out[16]= {Simplify}
In[17]:= Simplify[x_ == y_] := True /; Simplify[x-y]==0
In[18]:= Protect[Simplify]
Out[18]= {Simplify}
In[19]:= Simplify[Integrate[Cosh[x],x] == Sinh[x]]
Out[19]= True
-- David Jacobson