Re: DiscreteDelta Evaluation Question

• To: mathgroup at smc.vnet.net
• Subject: [mg49808] Re: DiscreteDelta Evaluation Question
• From: Paul Abbott <paul at physics.uwa.edu.au>
• Date: Sun, 1 Aug 2004 18:48:40 -0400 (EDT)
• Organization: The University of Western Australia
• References: <ced83t\$q7d\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```In article <ced83t\$q7d\$1 at smc.vnet.net>, Ben <serpent11 at hotmail.com>
wrote:

> In evaluating Poisson brackets, I've gotten Mathematica to simplify
> the brackets down to an infinite sum of deltafunctions, but though the
> sum should be very easy to evaluate, I can't convince it to simplify
> it.  Here's a typical case that it won't simplify:
>
> Sum[(-Sqrt[-2] indx DiscreteDelta[2 - indx] -
>           0.5 Sqrt[-1] DiscreteDelta[indx] DiscreteDelta[
>               2 - indx])((1 - Sqrt[2]) DiscreteDelta[indx]
> DiscreteDelta[
>               2 - indx] +
>           Sqrt[2] DiscreteDelta[2 - indx]) - (Sqrt[-2] indx
> DiscreteDelta[
>               2 - indx] +
>           0.5 Sqrt[-1] DiscreteDelta[indx] DiscreteDelta[
>               2 - indx])((1 - Sqrt[2]) DiscreteDelta[indx]
> DiscreteDelta[
>               2 - indx] + Sqrt[2] DiscreteDelta[2 - indx]), {indx, 1,
>     Infinity}]
>
> You can cut and paste it into a notebook.
>
> I've tried
>
> Unprotect[DiscreteDelta, Times];
>
> DiscreteDelta[i_] * DiscreteDelta[i_] := DiscreteDelta[i];
>
> Protect[DiscreteDelta, Times];
>
> but it still won't evaluate this seemingly simple expression.

You need to expand the summand and tell Mathematica that
DiscreteDelta[i_]^2 -> DiscreteDelta[i]. Also, for exact computations,
it is a good idea to replace 0.5 by 1/2.

Sum[Expand[((-(1/2)) Sqrt[-1] DiscreteDelta[indx]*
DiscreteDelta[2 - indx] + indx*DiscreteDelta[2 - indx]*(-Sqrt[-2]))*
(DiscreteDelta[2 - indx]*DiscreteDelta[indx]*(1 - Sqrt[2]) +
DiscreteDelta[2 - indx]*Sqrt[2]) - (indx*DiscreteDelta[2 - indx]*
Sqrt[-2] + (1/2)*DiscreteDelta[2 - indx]*DiscreteDelta[indx]*
Sqrt[-1])*(DiscreteDelta[2 - indx]* DiscreteDelta[indx]*(1 - Sqrt[2]) +
DiscreteDelta[2 - indx]*Sqrt[2])], {indx, 1, Infinity}] /.
DiscreteDelta[i_]^2 -> DiscreteDelta[i]

Cheers,
Paul

--
Paul Abbott                                   Phone: +61 8 9380 2734
School of Physics, M013                         Fax: +61 8 9380 1014
The University of Western Australia      (CRICOS Provider No 00126G)
35 Stirling Highway
Crawley WA 6009                      mailto:paul at physics.uwa.edu.au
AUSTRALIA                            http://physics.uwa.edu.au/~paul

```

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