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

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}]