Skip to main content

equation solving - Apart for complex roots?


Let p[x] be a polynomial in x and consider the partial fraction decomposition of 1/p[x].


The function Apart[] fails in simple cases like this



Apart[1/(1 + x^2)]

(* Out[37]= 1/(1 + x^2) *)

We could define a function


cApart[invp_] := Module[{z, c},
   z = x /. Solve[0 == 1/invp, x];
   c[k_] := 
    Product[If[i != k, 1/(z[[k]] - z[[i]]), 1], {i, 1, Length[z]}];
   Sum[c[i]/(x - z[[i]]), {i, 1, Length[z]}]];


which does the job


cApart[1/(1 + x^2)]

(* Out[36] = -(I/(2 (-I + x))) + I/(2 (I + x)) *)

But my question: is there an option for Apart[] or another standard facility in Mathematica which gives the decomposition in general, i.e. in the complex domain?


EDIT #1.1 Standard solution using Extension


The hints given so far can be codensed in this example


With[{d = 1 + x + x^2}, 

 Apart[1/Factor[d, Extension -> (x /. Solve[d == 0, x])]]]

(* -(1/((-1 + 2 (-1)^(1/3)) (-1 + (-1)^(1/3) - x))) - 1/((-1 + 
    2 (-1)^(1/3)) ((-1)^(1/3) + x)) *)

But it turns out that this procedure is not useful in practical applications as it takes extremely long calculation times (e.g. 1+x+x^4 took to Long to wait for it).


The following form (or something similar) would be nice to have


Apart[1/p[x], Extension -> Complexes] (* proposal, not available *)

EDIT #1.2 Other applications of cApart



It is interesting to apply cApart to a polynomial of higher degree


cApart[1/(1 + x + x^6)]

(* 
Out[64]= 
 1/((x - Root[1 + #1 + #1^6 &, 1]) (Root[1 + #1 + #1^6 &, 1] -
       Root[1 + #1 + #1^6 &, 2]) (Root[1 + #1 + #1^6 &, 1] - 
      Root[1 + #1 + #1^6 &, 3]) (Root[1 + #1 + #1^6 &, 1] - 
      Root[1 + #1 + #1^6 &, 4]) (Root[1 + #1 + #1^6 &, 1] - 
      Root[1 + #1 + #1^6 &, 5]) (Root[1 + #1 + #1^6 &, 1] - 

      Root[1 + #1 + #1^6 &, 6])) + 
 1/((x - Root[1 + #1 + #1^6 &, 2]) (-Root[1 + #1 + #1^6 &, 1] + 
      Root[1 + #1 + #1^6 &, 2]) (Root[1 + #1 + #1^6 &, 2] - 
      Root[1 + #1 + #1^6 &, 3]) (Root[1 + #1 + #1^6 &, 2] - 
      Root[1 + #1 + #1^6 &, 4]) (Root[1 + #1 + #1^6 &, 2] - 
      Root[1 + #1 + #1^6 &, 5]) (Root[1 + #1 + #1^6 &, 2] - 
      Root[1 + #1 + #1^6 &, 6])) + 
 1/((x - Root[1 + #1 + #1^6 &, 3]) (-Root[1 + #1 + #1^6 &, 1] + 
      Root[1 + #1 + #1^6 &, 3]) (-Root[1 + #1 + #1^6 &, 2] + 
      Root[1 + #1 + #1^6 &, 3]) (Root[1 + #1 + #1^6 &, 3] - 

      Root[1 + #1 + #1^6 &, 4]) (Root[1 + #1 + #1^6 &, 3] - 
      Root[1 + #1 + #1^6 &, 5]) (Root[1 + #1 + #1^6 &, 3] - 
      Root[1 + #1 + #1^6 &, 6])) + 
 1/((x - Root[1 + #1 + #1^6 &, 4]) (-Root[1 + #1 + #1^6 &, 1] + 
      Root[1 + #1 + #1^6 &, 4]) (-Root[1 + #1 + #1^6 &, 2] + 
      Root[1 + #1 + #1^6 &, 4]) (-Root[1 + #1 + #1^6 &, 3] + 
      Root[1 + #1 + #1^6 &, 4]) (Root[1 + #1 + #1^6 &, 4] - 
      Root[1 + #1 + #1^6 &, 5]) (Root[1 + #1 + #1^6 &, 4] - 
      Root[1 + #1 + #1^6 &, 6])) + 
 1/((x - Root[1 + #1 + #1^6 &, 5]) (-Root[1 + #1 + #1^6 &, 1] + 

      Root[1 + #1 + #1^6 &, 5]) (-Root[1 + #1 + #1^6 &, 2] + 
      Root[1 + #1 + #1^6 &, 5]) (-Root[1 + #1 + #1^6 &, 3] + 
      Root[1 + #1 + #1^6 &, 5]) (-Root[1 + #1 + #1^6 &, 4] + 
      Root[1 + #1 + #1^6 &, 5]) (Root[1 + #1 + #1^6 &, 5] - 
      Root[1 + #1 + #1^6 &, 6])) + 
 1/((x - Root[1 + #1 + #1^6 &, 6]) (-Root[1 + #1 + #1^6 &, 1] + 
      Root[1 + #1 + #1^6 &, 6]) (-Root[1 + #1 + #1^6 &, 2] + 
      Root[1 + #1 + #1^6 &, 6]) (-Root[1 + #1 + #1^6 &, 3] + 
      Root[1 + #1 + #1^6 &, 6]) (-Root[1 + #1 + #1^6 &, 4] + 
      Root[1 + #1 + #1^6 &, 6]) (-Root[1 + #1 + #1^6 &, 5] + 

      Root[1 + #1 + #1^6 &, 6]))
*)

Which gives the solution in a very regular pattern involving the function Root[]. The numeric evaluation gives


% // N

(*
Out[65]= -((
  0.0965468 + 0.0295033 I)/((-0.945402 - 0.611837 I) + x)) - (
 0.0965468 - 0.0295033 I)/((-0.945402 + 0.611837 I) + x) - (

 0.084438 + 0.114801 I)/((0.154735 - 1.03838 I) + x) - (
 0.084438 - 0.114801 I)/((0.154735 + 1.03838 I) + x) + (
 0.180985 - 0.279696 I)/((0.790667 - 0.300507 I) + x) + (
 0.180985 + 0.279696 I)/((0.790667 + 0.300507 I) + x)
*)

We can even continue to use these symbolic Root[] expressions in more complicated environments such as


g[a_] = Integrate[
  Exp[-a x]/(x - Root[1 + #1 + #1^6 &, 1]), {x, 0, \[Infinity]}, 
  Assumptions -> a > 0]


(* Out[69]= E^(-a Root[1 + #1 + #1^6 &, 1]) (-I \[Pi] - 
   CoshIntegral[a Root[1 + #1 + #1^6 &, 1]] - 
   SinhIntegral[a Root[1 + #1 + #1^6 &, 1]]) *)

It is gratifying that the integral is evaluated symbolically. We can now easily calculate numerical values, e.g.


g[1.]


(* Out[70]= 0.653737 - 0.158332 I *)


Regards,
Wolfgang





Comments

Post a Comment

Popular posts from this blog

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}]
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

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.
Post a Comment
Read more