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Re: Sum of terms --> list

  • To: mathgroup at smc.vnet.net
  • Subject: [mg127280] Re: Sum of terms --> list
  • From: bar at ANTYSPAM.ap.krakow.pl
  • Date: Sat, 14 Jul 2012 01:26:48 -0400 (EDT)
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Dana DeLouis <dana01 at me.com> wrote:
> > LB[z_] := List @@ B[z]
> > LB[3]  (* NOT OK*)

[...]
> 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
> = = = = = = = = = =

After help from Thomas G. i have found a simple solution
I have used 

B[z_] := 1/(2 - z) + 3/(4 - z) + 5/(6 - z);
LB[z_] = List @@ B[z] (* = instead := in function definition*)
LB[3]

It gives good solution, 

I'm not sure if this way can have bad consequences ..

Olaf



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