I have a question about the Dirichlet function, which is 1 at the rationals and 0 at the irrationals. I had thought this was an "arbitrary" definition, but then I saw the function given algebraically as f(x) = [tex] \lim_{k\rightarrow\infty}\left( \lim_{n\rightarrow\infty}\left( Cos( k! \pi x ) ^{2n} ) )[/tex] I understand how rational x is p/q and eventually the k! will multiply out the denominator leaving an even multiple of pi in the cosine for 1. An irrational x always produces a cosine less than 1, which the even exponent then drives to a strictly positive zero. Each point converges, but it's nowhere continuous. That seems clear enough, but still there's something about how all this works that I find disturbing.(adsbygoogle = window.adsbygoogle || []).push({});

If we were to modify this equation just a little bit, by substituting [tex]\Gamma(k)[/tex] instead of the factorial, we have a completely different situation. The limit no longer exists. There's a similar issue with the exponent n, which has to be integral because the cosine is negative half the time. Yet the second function contains the first. Especially as they approach infinity, why should this make any difference?

The more I think about it the more mixed up I get. Are there any rules about how to handle these mixing of cardinalities?

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Dirichlet function question

Loading...

Similar Threads - Dirichlet function question | Date |
---|---|

Limit of dirichlet function (from DSP) | Apr 12, 2014 |

Dirichlet's formula proof | Dec 21, 2010 |

Dirichlet integral | Jul 2, 2010 |

Heat Equation- Constant Dirichlet and Neumann BCs | Dec 23, 2008 |

Dirichlet function | Sep 22, 2005 |

**Physics Forums - The Fusion of Science and Community**