# Expanding the complex conjugate of a Mathieu function as a series

In my calculations, I found this equation:

Re[-Conjugate[MathieuCPrime[-4, -2 \[Epsilon], 0.7853981633974483]] Conjugate[MathieuS[-4, -2 \[Epsilon], 0.7853981633974483]] + MathieuC[-4, -2 \[Epsilon], 0.7853981633974483] MathieuSPrime[-4, -2 \[Epsilon],
0.7853981633974483]


As $$\epsilon$$ is small, I expanded it in a series and tried to obtain the coefficient of the $$\epsilon^2$$ term, but Mathematica gave me this:

In[1083]:= Coefficient[Series[ Re[-Conjugate[
MathieuCPrime[-4, -2 \[Epsilon],
0.7853981633974483]] Conjugate[
MathieuS[-4, -2 \[Epsilon], 0.7853981633974483]] +
MathieuC[-4, -2 \[Epsilon],
0.7853981633974483] MathieuSPrime[-4, -2 \[Epsilon],
0.7853981633974483]], {\[Epsilon], 0, 2}], \[Epsilon], 1]

Out[1083]= 0


As getting an output of $$0$$ doesn't make sense to me, I tried expanding the complex conjugate of a Mathieu function as a series. However, this is where I am having trouble:

In[1078]:= \$Assumptions = \[Epsilon] > 0

Out[1078]= \[Epsilon] > 0

In[1079]:= Normal[Series[Refine[Conjugate[MathieuC[1, \[Epsilon], 3.]]], {\[Epsilon], 0, 2}]]

Out[1079]= Conjugate[MathieuC[1, \[Epsilon], 3.]]


However, if I instead do the series expansion first, then apply the complex conjugate, I find:

In[1080]:= Refine[Conjugate[Normal[Series[MathieuC[1, \[Epsilon], 3.], {\[Epsilon], 0, 2}]]]]

Out[1080]= -0.989992 - (1.11614 + 1.31405 I) \[Epsilon] + (0.416372 -
0.0728342 I) \[Epsilon]^2


How do I get Out[1079] to appear the same as Out[1080]? Or more generally, how do I get Mathematica to expand the complex conjugate of a Mathieu function (and their derivatives) as a series?

If it is important, I am using Mathematica 11.2.0.0

• I believe the problem is with Re and Conjugate, which are not differentiable. Simpler example of failure: Series[Re@Conjugate[E^x], {x, 0, 2}] – Michael E2 May 20 at 18:43
• I applied this solution so that Mathematica knows not to differentiate Conjugate (mathematica.stackexchange.com/a/137443/8459). Can I do the same thing for Re? – user85503 May 20 at 18:58