Skip to main content

mathematical optimization - Why does Maximize return a wrong answer?


I want to find the maximum of the following function, but get a wrong result. What am I doing wrong?


Clear[x, y, xx, yy]
peaks[x_, y_] := 
  3*(1 - x)^2.*Exp[-(x^2) - (y + 1)^2] - 

   10*(x/5 - x^3 - y^5) Exp[-x^2 - y^2] - 1/3 Exp[-(x + 1)^2 - y^2];
N[Maximize[peaks[xx, yy], {xx, yy}]]
NMaximize[{peaks[x, y], x >= -2, x <= 2, y >= -2, y <= 3}, {x, y}]
peaks[0, 2]
peaks[1.28568, -0.00484756]

When you look at these graphs, you realize that the results are strange:


peaks[x_, y_] := 3*(1 - x)^2*Exp[-(x^2) - (y + 1)^2] - 
                 10*(x/5 - x^3 - y^5) Exp[-x^2 - y^2] - 1/3 Exp[-(x + 1)^2 - y^2];
Plot3D[peaks[x, y], {x, -2, 2}, {y, -2, 3}, 

    ColorFunction -> "DarkRainbow", 
    AxesLabel -> Automatic]
ContourPlot[peaks[x, y], {x, -2, 2}, {y, -2, 3},  
    ColorFunction -> "DarkRainbow", 
    ContourLines -> False,  
    Contours -> 25]

Mathematica graphics



Answer



You should use FindMaximum:



FindMaximum[peaks[xx, yy], {xx, yy}]


{8.106, {xx -> -0.009, yy -> 1.581}}



plot = Plot3D[peaks[x, y], {x, -2, 2}, {y, -2, 2}, PlotRange -> All];

point = Graphics3D[{PointSize@0.05, Point[{-0.009, 1.581, 8.106}]}];

Show[plot, point]


enter image description here


On the other hand, Maximize returns in this case the last local maximum it can find:


enter image description here


Comments

Post a Comment

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more