plotting - Table and ListPlot3D

I would like to do a ListPlot3D of an empirical function where variables c and d are my x and y axes and tkappa is my z.

I either get an empty plot or axis ranges that do not correspond to values of c, d or tkappa.

If I use the code below; I get an empty plot but the axis ranges are correct.

tkappa[a_,b_,c_,d_]:= 0.67 - 0.07*a - 0.17*b + 0.40*c + 0.32*d + 0.16*a*b + 0.13*a*c-8.093*10^-4*a*d + 0.13*b*c + 0.084*b*d + 0.27*c*d - 0.10*a*b*c + 0.016*a*b*d+0.051*a*c*d + 0.16*b*c*d -0.059*a*b*c*d;
timingData = Table[{c,d,tkappa[0,0,c,d]},{c,0,1,0.1},{d,0,1,0.1}]
ListPlot3D[timingData]

If I use a Table function as an argument for ListPlot3D, I get a surface but the axis ranges don't seem to correspond to c, d or values for tkappa.

ListPlot3D[Table[tkappa[0,0,c,d],{c,0,1,0.1},{d,0,1,0.1}]]

What is the best approach for ListPlot3D for x,y,z plots?

Answer

The data feed to ListPlot3D should be of the form {{x1,y1,z1},{x2,y2,z2},....}

 timingData = Partition[
              Flatten[Table[{c, d, tkappa[0, 0, c, d]}, {c, 0, 1, 0.1}, {d, 0, 1,0.1}]],
              3];



 ListPlot3D[timingData]

Comments

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...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...

Blog

Search This Blog