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Math Help - Dirichlet function

  1. #1
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    Dirichlet function

    could some one explain about the dirichlet function ? I'm confused.
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  2. #2
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    Quote Originally Posted by srinathmurthy View Post
    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?
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  3. #3
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    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.
    [IMG]file:///C:/DOCUME%7E1/Admin/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG][IMG]file:///C:/DOCUME%7E1/Admin/LOCALS%7E1/Temp/moz-screenshot-2.jpg[/IMG]
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  4. #4
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    Please post a link to that article. I cannot see the images you tried to post.
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  5. #5
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    Nowhere continuous function - Wikipedia, the free encyclopedia

    here's the link , please explain. Who's the admin here? could anybody tell me?
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  6. #6
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    You mean \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 \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 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.
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  7. #7
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    Thank you.
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