RE: Simple question on whether function is defined or not

*To*: mathgroup at smc.vnet.net*Subject*: [mg35136] RE: [mg35093] Simple question on whether function is defined or not*From*: "DrBob" <majort at cox-internet.com>*Date*: Tue, 25 Jun 2002 19:55:16 -0400 (EDT)*Reply-to*: <drbob at bigfoot.com>*Sender*: owner-wri-mathgroup at wolfram.com

>>BracketDefine[e[3],e[5]] for instance returns nothing That's the whole point. If BracketDefine were undefined with those arguments, it would return BracketDefine[e[3],e[5]] Instead, it returns Null, because none of the conditions in the Which statement are true. A legally formed Which statement ALWAYS returns a value, and when you see "nothing", that means there IS something (a Null). If you want BracketDefine to be undefined except for those explicit cases, you could write it this way: ClearAll[BracketDefine] BracketDefine[a_.*e[i_], b_.*e[j_]] /; i + j > n := 0 BracketDefine[a_.*e[i_], b_.*e[i_]] := 0 BracketDefine[a_.*e[1], b_.*e[j_]] := a*b*e[j + 1] BracketDefine[a_.*e[2], b_.*e[3]] := -a*b*e[5] BracketDefine[a_.*e[3], b_.*e[4]] := 2*a*b*e[7] ValueQ[BracketDefine[e[3], e[5]]] False Or: ClearAll[BracketDefine] BracketDefine[a_.*e[i_], b_.*e[j_]] /; i + j > n := 0 BracketDefine[a_.*e[i_], b_.*e[i_]] := 0 BracketDefine[a_.*e[1], b_.*e[j_]] := a*b*e[j + 1] BracketDefine[a_.*e[2], b_.*e[3]] := -a*b*e[5] BracketDefine[a_.*e[3], b_.*e[4]] := 2*a*b*e[7] BracketDefine[a_.*e[i_], b_.*e[j_]] := Undefined BracketDefine[e[3], e[5]] === Undefined True (Undefined isn't anything special here--it's just a symbol we can check for to see if none of the other cases fit. I could have called it anything, or used a string like "undef" or something, so long as it isn't a value that BracketDefine could evaluate to in the other cases.) You probably could do better by defining Bracket all at once, but with these hints, I'll leave you to it. Bobby Treat -----Original Message----- From: Geoff Tims [mailto:Geoff at swt.edu] To: mathgroup at smc.vnet.net Subject: [mg35136] [mg35093] Simple question on whether function is defined or not I am trying to make a function in mathematica. I have two questions on this and an answer to either would be wonderful. For me, this looks complicated, but for you it probably does not. However, even if it does, my question, I believe, is still simple and I only give what I have to clarify if that is needed. What I have so far should define the function completely as far as I can tell except for one small problem. n := 15 BracketDefine[a_.*e[i_], b_.*e[j_]] := Which[ i + j > n, 0, i == j, 0, i == 1, a*b*e[j + 1], i == 2 && j == 3, -a*b*e[5], i == 3 && j == 4, 2*a*b*e[7] ]; Bracket[mul_, expr_Plus] := Map[Bracket[mul, #] &, expr]; Bracket[expr_Plus, mul_] := Map[Bracket[#, mul] &, expr]; Bracket[e[i_], e[j_]] := Which[ ValueQ[BracketDefine[e[i], e[j]]], BracketDefine[e[i], e[j]], j - i == 2, -Bracket[e[1], Bracket[e[i], e[i + 1]]], True, -Bracket[e[i + 1], e[j - 1]] + Bracket[e[1], Bracket[e[i], e[j - 1]]] ] I have BracketDefine[] and then Bracket[] which, in theory, takes values from BracketDefine. Well, it does take values from BracketDefine[], but BracketDefine only has certain values of the entire function Bracket[]. That's why in the first condition and statement of Which, I see if it's already defined in BracketDefine, and if so, it takes on that value. That's my problem. Mathematica, at least with ValueQ, believes that BracketDefine is defined for all values of i and j. However, it's only defined for a few values. BracketDefine[e[3],e[5]] for instance returns nothing. This brings us to my questions Is there another test which would return false if the function returns nothing for certain values, but true if it returns something for other values? Or, to avoid the problem altogether, is there a way to just define the entire Bracket function all at once? I tried defining Bracket the way BracketDefine is defined and then again defining Bracket the way Bracket is now defined, but it seemed to lose all knowledge of the first definition. Thanks Geoff Tims