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

Question

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 1

I think your problems are made by order of appling Re and N. Re@Bloch is not yet a state before the computation. So you have to apply the computation by Re@Norder.

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

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

Answer 2

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:

• Complex result for Real vectors in VectorAngle

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