Re: Sum of terms --> list
- To: mathgroup at smc.vnet.net
- Subject: [mg127213] Re: Sum of terms --> list
- From: Dana DeLouis <dana01 at me.com>
- Date: Sun, 8 Jul 2012 19:06:28 -0400 (EDT)
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- Delivered-to: mathgroup-newout@smc.vnet.net
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> LB[z_] := List @@ B[z]
> LB[3] (* NOT OK*)
> LB[3] gives 11/3 - this is a sum of term again - why ?
Hi. I believe B[3] is evaluated first to
1/(2-3) + 3/(4-3) + 5/(6-3)
Which is immediately reduced to 11/3.
Then the function -List- is applied to this reduced form, which returns just 11/3
One needs to hold the evaluation, or perhaps just:
B[z_]:=1/(2-z)+3/(4-z)+5/(6-z)
LB[z_]:=(List@@B[#1]) /. #1 -> z
LB[x]
{1/(2-x), 3/(4-x), 5/(6-x)}
LB[3]
{-1, 3, 5/3}
= = = = = = = = = =
HTH :>)
Dana DeLouis
Mac & Math 8
= = = = = = = = = =
On Jul 7, 5:31 am, b... at ANTYSPAM.ap.krakow.pl wrote:
> I have found this thread, an I heve additional question about List@@
>
> see on example:
>
> ----------
> ClearAll["`*"];
> B[z_] := 1/(2 - z) + 3/(4 - z) + 5/(6 - z);
> B[z] (*OK*)
> LB[z_] := List @@ B[z]
> LB[z] (*OK*)
> LB[z] /. z -> 3 (*OK*)
>
> LB[3] (* NOT OK*)
> ------------------------
>
> LB[3] gives 11/3 - this is a sum of term again - why ?
> Is a simply way to convert B[z] ---> LB[z]
> where LB[z] is a list of simple functions of z ?
>
> greetings, Olaf