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Assuming n is even

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Rod Pinna

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Jan 3, 1998, 3:00:00 AM1/3/98
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Hopefully this isn't a FAQ.

Is it possible to get Mathematica (3.0) to assume that n is an even
number for an indefinite integral?


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

Paul Abbott

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Jan 5, 1998, 3:00:00 AM1/5/98
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Rod Pinna wrote:

> Hopefully this isn't a FAQ.
>
> Is it possible to get Mathematica (3.0) to assume that n is an even
> number for an indefinite integral?

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:

>I want to make an assignment T = k/omega and somehow cause Mathematica
>to know that k is an integer. How do I do this?

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.

Cheers,
Paul

____________________________________________________________________
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia Nedlands WA 6907
mailto:pa...@physics.uwa.edu.au AUSTRALIA
http://www.pd.uwa.edu.au/~paul

God IS a weakly left-handed dice player
____________________________________________________________________


sean...@worldnet.att.net

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Jan 6, 1998, 3:00:00 AM1/6/98
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Paul Abbott wrote:
>
> Rod Pinna wrote:
>
> > Hopefully this isn't a FAQ.
> >
> > Is it possible to get Mathematica (3.0) to assume that n is an even
> > number for an indefinite integral?
>
> 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:
>
> >I want to make an assignment T = k/omega and somehow cause Mathematica
> >to know that k is an integer. How do I do this?
>
> 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.
>
> Cheers,
> Paul


Another possibility is to use the Assumptions option in the Integrate
command. I have not experimented with it, but there may be a way. --
Remove the _nospam_ in the return address to respond.


Rod Pinna

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Jan 8, 1998, 3:00:00 AM1/8/98
to

Paul Abbott <pa...@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.

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