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MathGroup Archive 1998

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Re: Assuming n is even


  • To: mathgroup@smc.vnet.net
  • Subject: [mg10327] Re: Assuming n is even
  • From: hello@there.com (Rod Pinna)
  • Date: Thu, 8 Jan 1998 23:40:50 -0500
  • Organization: UWA
  • References: <68l3e7$mf6@smc.vnet.net> <68q770$202@smc.vnet.net>

 Paul Abbott <paul@physics.uwa.edu.au> wrote:

>Good to see a posting from the University of Western Australia!  It is a
>FAQ but the answer is, briefly, no.  A recent and related question was:

Just trying to keep busy over the break....

Thanks for the responses

>In my opinion, the best way to is using pattern-matching and replacement
>rules (see The Mathematica Journal 2(4): 31).  E.g., for n integral, we
>have
>
>        {Cos[(n_)*Pi] -> (-1)^n, Sin[(n_)*Pi] -> 0}; 
>
>Please post your integral so that perhaps readers can make other
>suggestions.

The above integral is pretty close to what I'm looking at actually. 

Say I have

w1=A1*Sin[n*t]*(Cos[(1/2)*m*\[Pi]*x/L]-1)
v1=A3*Cos[n*t]*Sin[m*\[Pi]*x/L]

Then with

Et1=(1/a)*(D[v1,t]+w1)

The integral is

\!\(V12 = 
    \[Integral]\_0\%L
        \(\[Integral]\_0\%\(2*\[Pi]\)Et1\^2\ \[DifferentialD]t 
          \[DifferentialD]x\)\)

(Apologies for the rather horrid formatting above)

i.e.  Integrate[Integrate[Et^2,{x,0,L}],{t,0,2*pi}]

And a few integrals of that type. Some of the results given then have 
cosine terms which are equivalent.

I've used ReplaceAll to replace one with the other. If there isn't a 
better way, that should be ok.

Thanks for the help.

 

Rod Pinna
(rpinnaX@XcivilX.uwa.edu.au  Remove the X for email)




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