Re: Simplification question

• To: mathgroup at smc.vnet.net
• Subject: [mg59108] Re: Simplification question
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Fri, 29 Jul 2005 00:41:47 -0400 (EDT)
• Organization: The Open University, Milton Keynes, U.K.
• References: <dc76pp\$jvg\$1@smc.vnet.net> <dc9vdl\$cm6\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Jean-Marc Gulliet wrote:
> snoofly wrote:
>
>>Can someone explain this to me please.
>>
>>Clear[m]
>>
>>Simplify[Sin[3 m Pi], Assumptions -> m \[Element] Integers]
>>0
>>
>>Simplify[Cos[3 m Pi], Assumptions -> m \[Element] Integers]
>>
>>Cos[3 m \[Pi]]
>>
>>I'm not sure why Mathematica cannot deduce that the second simplification
>>should be (1)^m.
>>
>
> Because it is a wrong answer!
>
> In[1]:=
> Simplify[Cos[m*Pi], Assumptions -> m \[Element] Integers]
>
> Out[1]=
> (-1)^m
>
> In[2]:=
> Simplify[Cos[m*Pi], Assumptions -> m \[Element] Integers] /. m -> 3*m
>
> Out[2]=
> (-1)^(3*m)
>
> If m is even then
>
> In[3]:=
> Simplify[Cos[2*k*Pi], Assumptions -> k \[Element] Integers]
>
> Out[3]=
> 1
>
> If m is odd then
>
> In[4]:=
> Simplify[Cos[(2*k + 1)*Pi], Assumptions -> k \[Element] Integers]
>
> Out[4]=
> -1
>
> Regards,
> /J.M.
>
Sorry, the last two input lines should have been the following ones:

In[1]:=
Simplify[Cos[3*(2*k + 1)*Pi], Assumptions ->
k \[Element] Integers]

Out[1]=
-1

In[2]:=
Simplify[Cos[3*(2*k)*Pi], Assumptions ->
k \[Element] Integers]

Out[2]=
1

If you want to keep your original expression you an use a replacement
rule for m, depending on whether m is odd or even:

In[3]:=
Simplify[Cos[3*m*Pi] /. m -> 2*k + 1,
Assumptions -> k \[Element] Integers]

Out[3]=
-1

In[4]:=
Simplify[Cos[3*m*Pi] /. m -> 2*k,
Assumptions -> k \[Element] Integers]

Out[4]=
1

Best regards,
/J.M.

```

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