Skip to main content

numerical integration - a NIntegrate in the NIntegrate



Here is our problem:
enter image description here
Suppose g[y,z] and f[x,z] is complicated,we have to solve it numerically.



Writing this in Mathematica:


Clear["`*"]
g[y_, w_] := 1/(w - y + I*0.1)
f[x_, w_] := (x*x)/(w - x + I*0.1)
NIntegrate[(g[k2, w]/(1 - NIntegrate[f[k1, w], {k1, 0, 30}])), {k2, 0,
   30}, {w, -1, 2}]

the error message is


    NIntegrate::inumr: The integrand x^2/((2. +0.1 I)-x) has evaluated to non-numerical values 
for all sampling points in the region with boundaries {{0,30}}.


Answer



This might be a duplicate of one of the examples in User-defined functions, numerical approximation, and NumericQ, but there isn't one with nested NIntegrate.


ClearAll[f, g, h];
g[y_, w_] := 1/(w - y + I*0.1)             (* note use of patterns y_, w_ *)
f[x_?NumericQ, w_?NumericQ] := (x*x)/(w - x + I*0.1)
h[w_?NumericQ] := NIntegrate[f[k1, w], {k1, 0, 30}];
NIntegrate[(g[k2, w]/(1 - h[w])), {k2, 0, 30}, {w, -1, 2},
 PrecisionGoal -> 2, AccuracyGoal -> 8]    (* for speed over accuracy; adjust as desired *)
(*  -0.0249226 - 0.0125844 I  *)


The trouble is that NIntegrate evaluates the integrand once symbolically. When the outer one evaluates the inner one, the symbol w does not have a value, so the inner NIntegrate complains. Note this error does not prevent the integral from being evaluated; but it is annoying and the messages slow things down.


For the use of patterns in defining functions, see the tutorial Defining Functions and its related tutorials.


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