Skip to main content

How to speed up integration of interpolation function?


I have a list of data, and I interpolate it to a function. Then I need to do an integration with the interpolating function. But I found that the speed is unacceptably slow.


My data is here (a little long, but don't go away :) ):


data = {-0.00799443, -0.00581522, -0.00557107, -0.00543862, -0.00528042, \

-0.00508618, -0.00486091, -0.00461279, -0.00435009, -0.00408028, \
-0.0038098, -0.00354416, -0.00328805, -0.00304535, -0.00281906, \
-0.0026108, -0.00242045, -0.00224588, -0.00208332, -0.00192845, \
-0.00177801, -0.00163133, -0.00149092, -0.00136145, -0.0012474, \
-0.00115024, -0.00106716, -0.000992246, -0.000919878, -0.000848073, \
-0.000779184, -0.000717175, -0.000663667, -0.000616407, -0.00057173, \
-0.000528424, -0.000488614, -0.000454547, -0.000425288, -0.000397686, \
-0.000370268, -0.00034446, -0.000321488, -0.00030009, -0.000278782, \
-0.000258483, -0.000240725, -0.000224931, -0.000209972, -0.000196452, \
-0.000184918, -0.000174195, -0.000163592, -0.000153752, -0.000144418, \

-0.000134884, -0.000125771, -0.000117444, -0.000109436, -0.000102175, \
-0.0000959463, -0.0000902133, -0.000085125, -0.0000806452, \
-0.0000762082, -0.0000719591, -0.0000677566, -0.000063368, \
-0.0000591507, -0.0000549953, -0.0000510613, -0.0000475951, \
-0.0000444333, -0.0000417897, -0.0000394736, -0.0000373836, \
-0.0000354749, -0.0000334705, -0.0000314543, -0.0000292503, \
-0.0000269879, -0.0000247026, -0.0000224853, -0.0000204942, \
-0.0000187118, -0.0000172668, -0.0000160166, -0.0000149913, \
-0.0000139883, -0.0000129844, -0.000011827, -0.0000105289, \
-9.06132*10^-6, -7.50783*10^-6, -5.94092*10^-6, -4.46213*10^-6, \

-3.15097*10^-6, -2.0399*10^-6, -1.13236*10^-6, -3.57489*10^-7, 
 3.57489*10^-7, 1.13236*10^-6, 2.0399*10^-6, 3.15097*10^-6, 
 4.46213*10^-6, 5.94092*10^-6, 7.50783*10^-6, 
 9.06132*10^-6, 0.0000105289, 0.000011827, 0.0000129844, \
0.0000139883, 0.0000149913, 0.0000160166, 0.0000172668, 0.0000187118, \
0.0000204942, 0.0000224853, 0.0000247026, 0.0000269879, 0.0000292503, \
0.0000314543, 0.0000334705, 0.0000354749, 0.0000373836, 0.0000394736, \
0.0000417897, 0.0000444333, 0.0000475951, 0.0000510613, 0.0000549953, \
0.0000591507, 0.000063368, 0.0000677566, 0.0000719591, 0.0000762082, \
0.0000806452, 0.000085125, 0.0000902133, 0.0000959463, 0.000102175, \

0.000109436, 0.000117444, 0.000125771, 0.000134884, 0.000144418, \
0.000153752, 0.000163592, 0.000174195, 0.000184918, 0.000196452, \
0.000209972, 0.000224931, 0.000240725, 0.000258483, 0.000278782, \
0.00030009, 0.000321488, 0.00034446, 0.000370268, 0.000397686, \
0.000425288, 0.000454547, 0.000488614, 0.000528424, 0.00057173, \
0.000616407, 0.000663667, 0.000717175, 0.000779184, 0.000848073, \
0.000919878, 0.000992246, 0.00106716, 0.00115024, 0.0012474, \
0.00136145, 0.00149092, 0.00163133, 0.00177801, 0.00192845, \
0.00208332, 0.00224588, 0.00242045, 0.0026108, 0.00281906, \
0.00304535, 0.00328805, 0.00354416, 0.0038098, 0.00408028, \

0.00435009, 0.00461279, 0.00486091, 0.00508618, 0.00528042, \
0.00543862, 0.00557107, 0.00581522, 0.00799443}

Interpolate it:


f = Interpolation[data];
Plot[f[x], {x, 1, 200}, PlotRange -> All]

It looks like:


interpolating function plot


Now I define a function:



Clear[b];
b[x_, y_] := NIntegrate[
   Cross[{0, f[s], 0}, {x - s, 0, y}][[{1, 3}]]/((x - s)^2 + y^2),
   {s, 1, 200}
];

The integration is so slow!!


b[1., 2.] // AbsoluteTiming

(*{1.17772, {-0.00827965, -0.0104805}}*)


What I want to do is a vector plot:


VectorPlot[b[x, y], {x, -10, 210}, {y, -3, 3}]

But with this kind of slow integration, this is painful. Are there better ways to speed up the integration?



Answer



The way to deal with this is to use the special setting Method -> "InterpolationPointsSubdivision" of NIntegrate[], which will automagically split the integrand so that an integration rule (by default, "GlobalAdaptive") is only applied within each piecewise polynomial interval of the InterpolatingFunction[] involved. This is akin to the functionality of the old package NumericalMath`NIntegrateInterpolatingFunct`​.


As a demonstration:


With[{x = 5, y = 5}, 
     NIntegrate[Cross[{0, f[s], 0}, {x - s, 0, y}][[{1, 3}]]/((x - s)^2 + y^2),

                {s, 1, 200}]] // AbsoluteTiming
   {0.619479, {-0.00929476, -0.00291246}}

With[{x = 5, y = 5}, 
     NIntegrate[Cross[{0, f[s], 0}, {x - s, 0, y}][[{1, 3}]]/((x - s)^2 + y^2),
                {s, 1, 200}, Method -> "InterpolationPointsSubdivision"]] // AbsoluteTiming
   {0.0798281, {-0.00929476, -0.00291246}}

Options[NIntegrate`InterpolationPointsSubdivision] displays the suboptions that can be fed to this method.


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