Explanation to Hold?

*To*: mathgroup at smc.vnet.net*Subject*: [mg113931] Explanation to Hold?*From*: fd <fdimer at gmail.com>*Date*: Thu, 18 Nov 2010 07:06:35 -0500 (EST)

Dear Group, While doing some work in Mathematica I stumbled into something that has let me quite intrigued. Indeed, I'm having some difficulties to understand how hold works. I'm an unexperienced user, so any clarification on this would be greatly appreciated. I'm using Mathematica to generate a report where I display points of minimum for two different function and a set of 4 parameters. I start by defining the functions like this fnA = Function[{x}, x^2 + x] fnB = Function[{x}, x^4 - x^2] results = {fnA, fnB}; pointsOfInterest = {1, 2, 3, 4}; And then I generate the points of minimum using the following command Thread[NArgMin[#, x] & /@ ((# - (pointsOfInterest))^2) & /@ Through[results[x]]] // Grid All works fine, but then, when I use compiled functions fnAC = Compile[{x}, fnA[x]]; fnBC = Compile[{x}, fnB[x]]; resultsC = {fnAC, fnBC} and use the same command as before Thread[NArgMin[#, x] & /@ ((#[x] - (pointsOfInterest))^2) & /@ resultsC] I get the following message, along with the answer I had before. CompiledFunction::cfsa: Argument x at position 1 should be a machine- size real number. Well, I thought that functions like NArgMin would first assign a real value to x and then evaluate the function, as Plot would do. When I hold the compiled function NArgMin[Hold[(fnAC[x] - 3)^2], x] I do get the result without any warning message. If I plot the compiled functions I don't get any message either. Funnier still, when I use the mapping to plot everything at once Thread[Plot[#, {x, -10, 10}] & /@ ((#[x] - (pointsOfInterest))^2) & /@ resultsC] I do get the same message CompiledFunction::cfsa: Argument x at position 1 should be a machine- size real number But in this case I can remedy by using Hold and ReleaseHold Thread[Plot[ ReleaseHold[#], {x, -10, 10}] & /@ ((Hold[#[x]] - (pointsOfInterest))^2) & /@ resultsC] But this same trick will not work with NArgMin, which is puzzling to me. Well, any comments? Best Regards

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