Bug introduced in 9.0 or earlier and fixed in 10.3.0

This is a documentation mistake in MathieuCharacteristicA, MathieuCharacteristicB, and MathieuCharacteristicExponent.

According to the documentation, the Mathieu characteristic function generates parameter a:

The characteristic value Subscript[a, r] gives the value of the parameter a in y′′+(a-2q cos(2z))y=0 for which the solution has the form e^(i r z) f(z), where f(z) is an even function of z with period 2π.

However, I get the function f that are periodic of π instead of 2π. Here is the construction of the periodic function f (followed from Input 76 on page 1105 of The Mathematica Guidebook for Symbolics):

f[k_, q_, z_] := (MathieuC[MathieuCharacteristicA[k, q], q, z] + I Sign[k] MathieuS[MathieuCharacteristicB[k, q], q, z])/Exp[I k z]
Plot[{Abs@f[3, -1, z], Abs@f[1/3, -1, z]}, {z, -2 π, 2 π}, Axes -> False, Frame -> True, GridLines -> {π/2 Range[-3, 3, 2], {}}]

enter image description here

So why does the periodic function f have period of π instead of 2π ?

  • $\begingroup$ According to Wikipedia, MathWorld, and DLMF, the period is supposed to be Pi. I guess the documentation is wrong. -- Now, should we close this as a "simple mistake in the documentation"? ;P (Unless, I'm wrong, of course.) $\endgroup$
    – Michael E2
    Sep 3, 2014 at 3:47
  • $\begingroup$ @MichaelE2 Thanks for the information, it's really helpful! But I guess I would not agree it's a simple mistake. It has been consistently wrong in the documentation page of MathieuCharacteristicA , MathieuCharacteristicB and MathieuCharacteristicExponent. I was too faithful about the documentation, that I wasted a whole weekend doing completely wrong things :( $\endgroup$ Sep 3, 2014 at 20:10
  • 1
    $\begingroup$ The "simple mistake" remark was a joke. I appreciate that the frustration and waste of time it has caused you is no joke. $\endgroup$
    – Michael E2
    Sep 3, 2014 at 20:12
  • $\begingroup$ @MichaelE2 Ah, I see. By the way, have you used MathieuA function in Maple? It's the counterpart of MathieuCharacteristicA in Mathematica but I'm wondering why it only accept integer number as argument. $\endgroup$ Sep 3, 2014 at 20:22
  • $\begingroup$ @MichaelE2 Here is my problem in detail. Could you show me the documentation page you are referring to? $\endgroup$ Sep 3, 2014 at 20:31

1 Answer 1


Actually, the characteristic value $a_r$ gives the value of the parameter $a$ in $y′′+(a-2q \cos(2z))\,y=0$ for which the solution has the form $e^{i r z} f(z)$, where $f(z)$ is an even function of $z$ with period $\pi$, not $2\pi$ as stated in the documentation.

See, for instance, Wikipedia, MathWorld, or the Digitial Library of Mathematical Functions.

The documentation is wrong.


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