Skip to main content

numerical integration - What is the correct way to use NIntegrate inside the FindMinimum function?


I'm having minor issues with the FindMinimum function when using NIntegrate inside. The functions work perfectly well but I get warning messages and I was wondering if maybe I could be enlightened on the correct usage of these two functions together.


For the sake of illustration I provide the full set of functions I used on a simple example.


phi[t_, k_, h_] := (1/h)^3*
Piecewise[{{(h (1 - k) + t)^2 (h (1 + 2 k) - 2 t), (k - 1) h <= t <=

k*h}, {(h (1 + k) - t)^2 (h (1 - 2 k) + 2 t),
k*h <= t <= (k + 1) h}}];
psi[t_, k_, h_] := (1/h)^3*
Piecewise[{{(t - k*h) (h + t - k*h)^2, (k - 1) h <= t <=
k*h}, {(t - k*h) (h - t + k*h)^2, k *h <= t <= (k + 1)*h}}] ;
phipp[t_, k_, h_] := (1/h)^3*
Piecewise[{{2 (h (1 + 2 k) - 2 t) - 8 (h (1 - k) + t), (k - 1) h <=
t <= k*h}, {-8 (h (1 + k) - t) + 2 (h (1 - 2 k) + 2 t),
k*h <= t <= (k + 1) h}}];
psipp[t_, k_, h_] := (1/h)^3*

Piecewise[{{2 (-h k + t) + 4 (h - h k + t), (k - 1) h <= t <=
k*h}, {-4 (h + h k - t) + 2 (-h k + t),
k*h <= t <= (k + 1) h}}];
alpha[t_, k_, h_] := phi[t, k, h] + phipp[t, k, h];
beta[t_, k_, h_] := psi[t, k, h] + psipp[t, k, h];
T = Pi;
n = 2;
h = T/n;
FindMinimum[
NIntegrate[(h*beta[t, 0, h] - h*beta[t, n, h] +

a.Table[alpha[t, i, h], {i, 1, n - 1}] +
b.Table[h*beta[t, i, h], {i, 1, n - 1}])^2, {t, 0,
T}], {{a, {0.76}}, {b, {0.4}}}, Method -> "ConjugateGradient"]

I get the following warning message :


NIntegrate::inumr: "The integrand (a.{(8 Piecewise[{<<2>>},0])/\[Pi]^3+(8 Piecewise[{<<2>>},0])/\[Pi]^3}+b.{1/2\\[Pi]\(8\Power[<<2>>]\Piecewise[<<2>>]+8\Power[<<2>>]\Piecewise[<<2>>])}-1/2\\[Pi]\((8 Piecewise[{{<<2>>},{<<2>>}},0])/\[Pi]^3+(8 Piecewise[{{<<2>>},{<<2>>}},0])/\[Pi]^3)+1/2\\[Pi]\((8 Piecewise[{{<<2>>},{<<2>>}},0])/\[Pi]^3+(8 Piecewise[{{<<2>>},{<<2>>}},0])/\[Pi]^3))^2 has evaluated to non-numerical values for all sampling points in the region with boundaries {{0,3.14159}}"

I ask about this because FindMinimum takes unexpectedly long time to converge to the solution (which is a right one). But I thought that maybe using the functions correctly will accelerate the process.



Answer



The problem is indeed that Mathematica tries to do numerical operations before the symbols have value. You can fix this as follows.



Preliminary definitions, unchanged from yours


phi[t_, k_, h_] := (1/h)^3*
Piecewise[{{(h (1 - k) + t)^2 (h (1 + 2 k) - 2 t), (k - 1) h <=
t <= k*h}, {(h (1 + k) - t)^2 (h (1 - 2 k) + 2 t),
k*h <= t <= (k + 1) h}}];
psi[t_, k_, h_] := (1/h)^3*
Piecewise[{{(t - k*h) (h + t - k*h)^2, (k - 1) h <= t <=
k*h}, {(t - k*h) (h - t + k*h)^2, k*h <= t <= (k + 1)*h}}];
phipp[t_, k_, h_] := (1/h)^3*
Piecewise[{{2 (h (1 + 2 k) - 2 t) - 8 (h (1 - k) + t), (k - 1) h <=

t <= k*h}, {-8 (h (1 + k) - t) + 2 (h (1 - 2 k) + 2 t),
k*h <= t <= (k + 1) h}}];
psipp[t_, k_, h_] := (1/h)^3*
Piecewise[{{2 (-h k + t) + 4 (h - h k + t), (k - 1) h <= t <=
k*h}, {-4 (h + h k - t) + 2 (-h k + t),
k*h <= t <= (k + 1) h}}];
alpha[t_, k_, h_] := phi[t, k, h] + phipp[t, k, h];
beta[t_, k_, h_] := psi[t, k, h] + psipp[t, k, h];
T = Pi;
n = 2;

h = T/n;

Pull the integral out of the FindMinimum and make it only evaluate for numerical lists


Here's the integral, defined so as to only evaluate for numeric lists. This is cumbersome because of the way a is defined (to be a list).


ClearAll[fn];
fn[a_?(VectorQ[#, NumericQ] &), b_?(VectorQ[#, NumericQ] &)] :=
NIntegrate[(h*beta[t, 0, h] - h*beta[t, n, h] +
a.Table[alpha[t, i, h], {i, 1, n - 1}] +
b.Table[h*beta[t, i, h], {i, 1, n - 1}])^2, {t, 0, T}]


And now FindMinimum has no problem:


FindMinimum[fn[a, b], {{a, {0.76}}, {b, {0.4}}}, 
Method -> "ConjugateGradient"]

Mathematica graphics


Comments

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...

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...

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}]