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? >