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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.