# An interesting function

• Aug 26th 2010, 03:49 AM
TerenceCS
An interesting function
-Challenging question-
Let $\displaystyle c(x)$ define on all real numbers and
$\displaystyle \displaystyle c(x)=\lim_{m\to \infty} \left [\lim_{n\to \infty} \cos^n (\pi m! x)\right ].$
Show that $\displaystyle c(x)$ is nowhere continuous.

I suppose I was sent the solution so that I could approve this, but you are relying on me and/or the other senior staff being psychic rather than the more usual psychotic which as Sir Humphry would say is "Brave". In future try saying why you have sent such material.

Approved CB
• Aug 26th 2010, 01:18 PM
TheEmptySet
I think you have a typo in your question.

Note if $\displaystyle x=1$ you get

$\displaystyle \displaystyle \lim_{m \to \infty} \lim_{n \to \infty}\cos^n(\pi m!)= \lim_{m \to \infty} \lim_{n \to \infty}[(-1)^{m!}]^n$

This limit does not exists.

I think you mean the function

$\displaystyle \displaystyle \lim_{m \to \infty} \lim_{n \to \infty}\cos^{2n}}(\pi m!)=\boldsymbol{\chi}_{\mathbb{Q}}(x)$
• Aug 26th 2010, 07:34 PM
TerenceCS
No typo indeed. And your deduction is part of the proof - the function is not continuous at x=1. Below is a valid proof, not very rigorous though.

Let $\displaystyle f(m, n, x)=\cos^n (\pi m! x)$. If $\displaystyle x$ is a rational number other than $\displaystyle 1$, that is, it can be rewritten in the form $\displaystyle p/q$, where $\displaystyle (p, q)=1$ and $\displaystyle q\ne 0$. Once $\displaystyle m$ increases to become greater than $\displaystyle q$, it forces $\displaystyle n(x)=1$.

On the other side, if $\displaystyle x$ is irrational, $\displaystyle n(x)$ will approach $\displaystyle 0$. Hence given any real number $\displaystyle x\ne 1$, the limits do not agree when approach from irrationals and rationals respectively so the function is almost nowhere continuous and only proving the case $\displaystyle x=1$ is left.

In the case of $\displaystyle x=1$, the limit does not exists, so the function is nowhere continuous.
• Mar 26th 2011, 11:13 PM
bkarpuz
Quote:

Originally Posted by TerenceCS
-Challenging question-
Let $\displaystyle c(x)$ define on all real numbers and
$\displaystyle \displaystyle c(x)=\lim_{m\to \infty} \left [\lim_{n\to \infty} \cos^n (\pi m! x)\right ].$
Show that $\displaystyle c(x)$ is nowhere continuous.

See [Example 7.4, 1] for the solution.
Probably, [1] is the book which most of the MHF members read. (Happy)

References
[1] W. Rudin, Principles of Mathematical Analysis, 3rd Ed., McGraw-Hill Book Co., New York, 1976.