# possible orders of elliptic functions

The order of a non-trivial elliptic function^{} cannot be zero. This is a
simple consequence of Liouville’s theorem. Were the order of an elliptic
function zero, then the function^{} would have no poles. By definition,
an elliptic function has no essential singularities and is doubly
periodic. Hence, if the degree were zero, the function would be
continuous^{} everywhere and hence, being doubly periodic, would be
bounded^{} (since continuous functions on a compact domain (like the
closure of the fundamental parallelogram) are bounded). By Liouville’s
theorem, this would imply that the function is constant.

The order of an elliptic function cannot be 1. This follows from the
fact that the residues^{} at the poles of an elliptic function within a
fundamental parallelogram must sum to zero — if the function were of
degree 1, it would have exactly one first-order pole in the
fundamental parallelgram but any first-order pole must have a non-zero
residue.

Any number greater than one is possible as the order of an elliptic function. As an example of an elliptic function of order two, we may take the Weierstass $\mathrm{\wp}$-function, which has a single pole of order 2 in the fundamental domain. The $n$-th derivative of this function will have a single pole of order $n+2$ in the fundamental domain, hence be of order $n+2$, so is an example showing that, for every integer greater than 2, there exists an elliptic function having that integer as its order.

Title | possible orders of elliptic functions |
---|---|

Canonical name | PossibleOrdersOfEllipticFunctions |

Date of creation | 2013-03-22 15:48:30 |

Last modified on | 2013-03-22 15:48:30 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 7 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 33E05 |

Related topic | PeriodicFunctions |