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Re: Question: how to get Sin[n*Pi]=0 (n integer)
*To*: mathgroup at christensen.cybernetics.net
*Subject*: [mg1962] Re: Question: how to get Sin[n*Pi]=0 (n integer)
*From*: Scott.A.Hill at williams.edu (Lancelot)
*Date*: Sun, 27 Aug 1995 23:23:59 -0400
*Organization*: Williams College, Williamstown MA
In article <DDuM67.IuF at wri.com>, Ilija I Zovko <izovko at dominis.phy.hr> wrote:
>Hi,
>How can one tell Mathematica to simplify Sin[n Pi]=0 or
>Cos[n Pi]=(-1)^n and similar kind of stuff.
In such cases I tend just to substitute them by hand. E.g.:
In[4]:=RandomFunction[5]
Out[4]= 3 + 5 Cos [n Pi] + Sin[n Pi]
In[5]:=%4/.{Sin[n Pi]->0,Cos[n Pi]->(-1)^n}
n
Out[5]= 3 + 5 (-1)
Of course, you should make sure n is really an integer first.
>Also, how does one tell it "A" & "B" are matrices so it doesn't
>commute them (AB.not equal.BA).
Mathematica has a non-commutative multiplication command,
represented by **. Thus:
In[6]:= B*A
Out[6]= A B
In[7]:= B**A
Out[7]= B ** A
/
:@-) Scott
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