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)