Re: mathematica newbie trouble

*To*: mathgroup at smc.vnet.net*Subject*: [mg99317] Re: mathematica newbie trouble*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Sat, 2 May 2009 06:02:07 -0400 (EDT)

On 5/1/09 at 5:52 AM, yangshuai at gmail.com (Guapo) wrote: >i wrote the following mathematica code: >s := (a + b + c)/2; >k[a_, b_, c_] := Sqrt[s (s - a) (s - b) (s - c)]; >k[3, 4, 5] >k[5, 9, 12] >when run it, i can't get the write answer. but i change >setDelayed(:=) to set(=), everything works ok >s = (a + b + c)/2; >k[a_, b_, c_] = Sqrt[s (s - a) (s - b) (s - c)]; >k[3, 4, 5] >k[5, 9, 12] >i did a google search for the difference of set and setDelayed, >however, i still can't understand it for the upper described >problem, could anyone explain it? You would have gotten what you expected with SetDelayed had you defined s as: s[a_, b_, c_]:=(a + b + c)/2 When you write s:=(a+b+c)/2 you are telling Mathematica s is a function of *global* variables. When you write k[a_,b_,c_]:= you are telling Mathematica k a function of *local* variables. So, given no global assignment made to a,b and c, s gets evaluated to (a+b+c)/2 and is not evaluated further. But the other instances of a,b, and c in the definition of k are assigned the values supplied as arguments to k. Use your version with SetDelayed and do the following: In[16]:= a = b = c = 2; k[m, n, p] Out[17]= Sqrt[3] Sqrt[(3-m) (3-n) (3-p)] Since, I set a,b and c to 2, s evaluates to 3. So, 3 appears everywhere s is in the definition of k. Note: In[18]:= k[3, 4, 5] Out[18]= 0 Here s defined with global values of a,b and c gets evaluated to 3. And k uses the local value of a to be 3 per the definition. So the term s-a gets evaluated to 0 causing the entire product to be 0. Given the way s is used in k, I would have written the code as s = (a + b + c)/2; k[a_, b_, c_] := Sqrt[s (s - a) (s - b) (s - c)] or possibly k[a_, b_, c_] := Block[{s}, Sqrt[s (s - a) (s - b) (s - c)]/.s -> (a + b + c)/2] Note, I have not used a semicolon to terminate the definition of k using SetDelayed since there is no output to be suppressed. That is the right hand side is not evaluated by definition when k is defined using SetDelayed. Consequently, there is no output to be suppressed.