Skip to main content

numerics - Higher order periodic interpolation (curve fitting)


I have a list of points in 3D, and I want to get a smooth interpolation or curve fit (it is more for illustration) of these points such that the first and second derivatives at the start and end points agree. With


ListInterpolation[pts1, {0, 1}, InterpolationOrder -> 4, PeriodicInterpolation -> True]

even the first derivatives do not agree.



Answer



It does seem that the options PeriodicInterpolation -> True and Method -> "Spline" are incompatible, so I'll give a method for implementing a genuine cubic periodic spline for curves. First, let's talk about parametrizing the curve.


Eugene Lee, in this paper, introduced what is known as centripetal parametrization that can be used when one wants to interpolate across an arbitrary curve in Rn. Here's a Mathematica implementation of his method:



parametrizeCurve[pts_List, a : (_?NumericQ) : 1/2] /; MatrixQ[pts, NumericQ] := 
           FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]]

The default setting of the second parameter for parametrizeCurve[] gives the centripetal parametrization. Other popular settings include a == 0 (uniform parametrization) and a == 1 (chord length parametrization).


Now we turn to generating the derivatives needed for periodic cubic spline interpolation. This is done through the solution of an appropriate cyclic tridiagonal system, for which the functions LinearSolve[], SparseArray[], and Band[] come in handy:


periodicSplineSlopes[pts_?MatrixQ] := 
 Module[{n = Length[pts], dy, ha, xa, ya}, {xa, ya} = Transpose[pts];
        ha = {##, #1} & @@ Differences[xa];
        dy = ({##, #1} & @@ Differences[ya])/ha;
        dy = LinearSolve[SparseArray[{Band[{2, 1}] -> Drop[ha, 2], 

                                      Band[{1, 1}] -> ListConvolve[{2, 2}, ha], 
                                      Band[{1, 2}] -> Drop[ha, -2],
                                      {1, n - 1} -> ha[[2]], {n - 1, 1} -> ha[[-2]]}], 
                         3 MapThread[Dot[#1, Reverse[#2]] &,
                                     Partition[#, 2, 1] & /@ {ha, dy}]];
        Prepend[dy, Last[dy]]]

Using Sjoerd's example:


sc = Table[{Sin[t], Cos[t], Cos[t] Sin[t]}, {t, 0, 2 π, π/5}] // N;
tvals = parametrizeCurve[sc]

   {0, 0.102805, 0.196242, 0.303758, 0.397195, 0.5, 0.602805, 0.696242,
    0.803758, 0.897195, 1.}

cmps = Transpose[sc];
slopes = periodicSplineSlopes[Transpose[{tvals, #}]] & /@ cmps;
{f1, f2, f3} = MapThread[Interpolation[Transpose[{List /@ tvals, #1, #2}], 
                                       InterpolationOrder -> 3, Method -> "Hermite",
                                       PeriodicInterpolation -> True] &,
                         {cmps, slopes}];


Plot the space curve:


Show[ParametricPlot3D[{f1[u], f2[u], f3[u]}, {u, 0, 1}],
     Graphics3D[{AbsolutePointSize[6], Point[sc]}]]

interpolated space curve


Individually plotting the components and their respective derivatives verify that the interpolating functions are C2, as expected of a cubic spline:


plots of components and derivatives


Comments

Post a Comment

Popular posts from this blog

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...
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

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more