list manipulation - Failure in integrating from an interpolating function!

Consider a list like below:

ttable={{0, 2.6596 - 66.137 I}, {1/9, 2.45339 - 65.3148 I}, {2/9, 
  1.82053 - 62.8922 I}, {1/3, 0.720006 - 58.9982 I}, {4/
  9, -0.911205 - 53.8382 I}, {5/9, -3.15056 - 47.6797 I}, {2/
  3, -6.08057 - 40.8346 I}, {7/9, -9.77826 - 33.6382 I}, {8/
  9, -14.3047 - 26.4282 I}, {1, -19.6947 - 19.5216 I}}`

by defining "f" as interpolating function we'll have:

f = Interpolation[ttable]

now we define ff as the following integration:

ff[\[Xi]_]:=NIntegrate[(f[rr]) Cos[rr \[Xi]], {rr, 0, 1}]

the first problem is that this function can not be calculated and I have no idea why the followoing error happens:

In[87]:= ff[.1]
During evaluation of In[87]:= NIntegrate::inumr: The integrand Cos[rr \[Xi]] InterpolatingFunction[{{0.,1.}},{4,15,0,{10},{4},0,0,0,0,Automatic,{},{},False},<<1>>,{
Developer`PackedArrayForm,{0,<<10>>},{2.6596 -66.137 I,<<8>>,-19.6947-19.5216 I}},{Automatic}][rr] has evaluated to non-numerical values for all sampling points in the region with boundaries {{0,0.111111}}. >>
Out[87]= NIntegrate[f[rr] Cos[rr \[Xi]],{rr,0,1}]

The other thing should be mentioned is that I have to use ff in integrand of another integration like below:

NIntegrate[ff[\[Xi]] * \[Xi],{\[Xi],0,3}]

I've manipulated the former equations in many ways but none of them made an accurate output for the last integration, so I would be greatly thankful if somebody out there could help me!

Answer

The most important information is that you used your function ff inside another NIntegrate, because this is the source of confusion. What you have to know is that NIntegrate doesn't start right away with the numerical calculation when you call

NIntegrate[ff[ξ]*ξ, {ξ, 0, 3}]

It will try to do some analysis of your integrand and most likely, it will try to evaluate ff[ξ] without putting in numbers. And what happens then? Right, you call the NIntegrate of ff without proper numerical value of ξ:

The solution is pretty simple: Change your definition of ff so that it only calls its NIntegrate body when the argument is indeed numeric:

ClearAll[ff];
ff[ξ_?NumericQ] := NIntegrate[(f[rr]) Cos[rr ξ], {rr, 0, 1}]
NIntegrate[ff[ξ]*ξ, {ξ, 0, 3}]

(* 4.28747 - 124.522 I *)

Comments

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}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....

Blog

Search This Blog