Skip to main content

special functions - Why does N[Re@f] give complex result?


Consider this code:


BlochΚ[κ_, V0_, z_] := 
 MathieuC[MathieuCharacteristicA[κ, 2 V0], 2 V0,  z/2] + 
  Sign[κ] I MathieuS[MathieuCharacteristicB[κ, 2 V0], 

    2 V0, z/2]

Block[{$MaxExtraPrecision = 500, ϵ = 10^-10}, N[Re@BlochΚ[-2 + ϵ, -1, -10]]]
Block[{$MaxExtraPrecision = 1000, ϵ = 10^-20}, N[Re@BlochΚ[-2 + ϵ, -1, -10]]]
Block[{$MaxExtraPrecision = 500, ϵ = 10^-10}, N[Re@BlochΚ[-2 + ϵ, -1, -10]]]

On a fresh kernel I get


(*
-0.484175
-0.993753

-1.38778*10^-16+6.17104*10^-9 I
*)

Why I get complex number at the third time even I used Re explicitly? And why the results are different for the first and third time? Did I made a stupid mistake or what?



Answer



Update: I think this is a numeric precision problem rather than a matter of the behavior of Re.
I don't know if I should leave my original answer below for reference or remove it.


Consider:


expr = MathieuC[MathieuCharacteristicA[-(19999999999/10000000000), -2], -2, 5];


N[expr]
N[expr, 15]
SetPrecision[expr, 15]


-9.85323*10^-16 + 3.39211*10^-8 I

-0.484175231115992

-0.4841752311160


Only the machine precision calculation returns a complex value. I believe that puts this problem in the same class as:



Sorry for the earlier misdirection.


Note: I believe $MaxExtraPrecision has no effect upon a machine precision calculations.



Old, misleading answer


Intending to further illuminate Junho Lee's answer we may consider how Re handles symbolic expressions:


Re[a + b I]



-Im[b] + Re[a]

It performs this replacement whether or not a and b have a numeric equivalent. Therefore:


Re[
  MathieuC[MathieuCharacteristicA[-(19999999999/10000000000), -2], -2, 5] + 
   I MathieuS[MathieuCharacteristicB[-(19999999999/10000000000), -2], -2, 5]
]

Becomes:



Re[MathieuC[MathieuCharacteristicA[-(19999999999/10000000000), -2], -2, 5]] - 
 Im[MathieuS[MathieuCharacteristicB[-(19999999999/10000000000), -2], -2, 5]]

And:


Re[MathieuC[MathieuCharacteristicA[-(19999999999/10000000000), -2], -2, 5]]


MathieuC[MathieuCharacteristicA[-(19999999999/10000000000), -2], -2, 5]

Im[MathieuS[MathieuCharacteristicB[-(19999999999/10000000000), -2], -2, 5]]



0

In some manner Re did its job, nevertheless this symbolic expression has a complex numeric value.


If you want a function that operates only on explicit numbers you might use:


re[x_?NumberQ] := Re[x]

Now re will remain unevaluated until its argument is expressly a number:


re[Pi + 4 I]



re[4 I + π]

However N goes inside as re does not have NHoldFirst etc. therefore:


re[Pi + 4 I] // N


3.14159


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