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MathGroup Archive 2006

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Re: Defining two functions at once?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70765] Re: Defining two functions at once?
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Thu, 26 Oct 2006 02:39:22 -0400 (EDT)
  • References: <ehmttr$sd0$1@smc.vnet.net>

Hi,

and it does not help to define a single function
that return a vector ??

FAndG[x_,y_]:=Module[{myEqns,myValues,mySolns},
          myEqns = {f1 - g1 == x1 + 3y1, f1 + g1 == 2x1 - 7y1};
          myValues = {x1 -> x, y1 -> y};
          mySolns = FindRoot[ myEqns /. myValues, {f1, 0} , {g1, 0}];
          {f1} /. mySolns]

f[x_,y_]:=FandG[x,y][[1]]
g[x_,y_]:=FandG[x,y][[2]]

Regards
   Jens

AES wrote:
> I want to define two functions at once, where both of their values 
> depend on the same two variables, and both of their values come out of a 
> single two-dimensional FindRoot which I'd rather not evaluate twice.  
> 
> The following approach seems to work fine for a *single* function
> 
>    f[x_, y_] := Module[{},    
>          myEqns = {f1 - g1 == x1 + 3y1, f1 + g1 == 2x1 - 7y1};    
>          myValues = {x1 -> x, y1 -> y};    
>          mySolns = FindRoot[ myEqns /. myValues, {f1, 0} , {g1, 0}];    
>          f1 /. mySolns]
> 
> But if I try to define two functions at once by replacing the first and 
> last lines with
> 
>    {f[x_, y_], g[x_,y_]} := Module[{},    
>          myEqns = {f1 - g1 == x1 + 3y1, f1 + g1 == 2x1 - 7y1};    
>          myValues = {x1 -> x, y1 -> y};    
>          mySolns = FindRoot[ myEqns /. myValues, {f1, 0} , {g1, 0}];    
>          {f1, g1} /. mySolns]
> 
> I get a message about "shapes not being the same".  In fact, if I just 
> make the first and last lines even a single element list, e.g.
> 
>    {f[x_, y_]} := Module[{},    
>          myEqns = {f1 - g1 == x1 + 3y1, f1 + g1 == 2x1 - 7y1};    
>          myValues = {x1 -> x, y1 -> y};    
>          mySolns = FindRoot[ myEqns /. myValues, {f1, 0} , {g1, 0}];    
>          {f1} /. mySolns]
> 
> this doesn't work either.
> 
> If a module supposedly returns the result of its (compound) expression, 
> and the final term in that expression is a list, shouldn't the final 
> example work?  More important (to me anyway): Is there a simple way to 
> define two functions that use a shared FindRoot evaluation in a way that 
> (a) evaluates the FindRoot only once, and (b) involves only a single 
> compound expression of some sort?
> 


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