# Dirichlet function

• Nov 12th 2008, 10:11 AM
srinathmurthy
Dirichlet function
could some one explain about the dirichlet function ? I'm confused.
• Nov 12th 2008, 11:01 AM
HallsofIvy
Quote:

Originally Posted by srinathmurthy
could some one explain about the dirichlet function ? I'm confused.

What do you want explained about it? There are several "Dirichlet" functions but the standard one is:
f(x)= 0 if x is rational, 1 if x is irrational.

That has the property that it is discontinuous for all x because in any interval around x, there exist both rational and irrational numbers, so values of 0 and 1 arbitrarily close to x. What is it that is confusing you?
• Nov 13th 2008, 07:13 AM
srinathmurthy
there in an article in WIKI , [IMG]file:///C:/DOCUME%7E1/Admin/LOCALS%7E1/Temp/moz-screenshot-3.jpg[/IMG]
lim ( lim (cos k!pi) ^ 2j )
k-->inf j-->inf

I'd like to know how this reduces to the dirichlet function. form.
• Nov 13th 2008, 07:44 AM
HallsofIvy
Please post a link to that article. I cannot see the images you tried to post.
• Nov 16th 2008, 05:14 AM
srinathmurthy
Nowhere continuous function - Wikipedia, the free encyclopedia

here's the link , please explain. Who's the admin here? could anybody tell me?
• Nov 16th 2008, 06:05 AM
HallsofIvy
You mean $\displaystyle \lim_{k\rightarrow \infty}\left(\lim_{j\rightarrow \infty}cos(k!\pi x)^{2j}\right)$?
Suppose x is a rational number. Then x= m/n for integers m and n and, for k> n, k!/n is an integer. cosine of an integer times $\displaystyle \pi$ is either -1 or 1 and raised to an even power, 2j, is 1. Thus, for sufficiently large, k, that function is just 1 to the 2j power= 1. If x is irrational, then $\displaystyle k!\pi x$ is never an integer so [tex]-1< cos(k!\pi x)< 1[tex]. Raising that to the 2j power, as j goes to infintiy gives 0.
• Nov 19th 2008, 08:05 AM
srinathmurthy
Thank you.