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Math Help - Riemann's Hypothesis

  1. #1
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    Riemann's Hypothesis

    Today I was speaking to a friend of mine, and he brought up something called Riemann's Hypothesis. He told me that it was a concept relating to the distribution of primes and has yet to be proven, and is evidently somehow related to modern financial security systems.

    However, upon researching the topic I've found very little in the area of practical application of the Hypothesis... and add to the fact I find the hypothesis confusing to understand.

    Would anyone care to explain the hypothesis to me in layman's terms, and what it means holistically speaking?
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  2. #2
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    A little help

    This involves the zeta function
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    No one in Particular VonNemo19's Avatar
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    The simply is no "lay" explaination for this particular hypothesis. The very nature of the conjecture requires that you know someting of several different areas of advanced mathematics. I'm not too bad at mathematics, and I've tried to interpret the meaning with no success.

    Good luck.
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  4. #4
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    Quote Originally Posted by wonderboy1953 View Post
    This involves the zeta function
    This I have read and am trying to do more research to understand it.

    Quote Originally Posted by VonNemo19 View Post
    The simply is no "lay" explaination for this particular hypothesis. The very anture of the conjecture requires that you know someting of several different areas of advanced mathematics. I'm not too bad at mathematics, and I've tried to interpret the meaning with no success.

    Good luck.
    You have confirmed my suspicions. I suppose I will lay this particular problem to rest.

    Thank you both for your replies.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    The Riemann Zeta function is defined by  \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}

    Note that  \zeta(s) is defined only for  s with  \Re(s) > 1 .

    One can extend  \zeta(s) by noticing  \zeta(s) = \frac{1}{2^{1-s}-1}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s} (this can be shown fairly easily). This is called an extension because now  \zeta(s) is defined for  \Re(s)>0 (except for  s=1 where there is a simple pole).

    The Riemann Hypothesis states that the only zeros of  \zeta(s) in the strip  0<\Re(s)<1 have real part equal to  \frac{1}{2} i.e.  s=\frac{1}{2}+it .
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chiph588@ View Post
    The Riemann Zeta function is defined by  \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}

    Note that  \zeta(s) is defined only for  s with  \Re(s) > 1 . <---What do you mean by this?

    One can extend  \zeta(s) by noticing  \zeta(s) = \frac{1}{2^{1-s}-1}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s} (this can be shown fairly easily). This is called an extension because now  \zeta(s) is defined for  \Re(s)>0 (except for  s=1 where there is a simple pole).

    The Riemann Hypothesis states that the only zeros of  \zeta(s) in the strip  0<\Re(s)<1 have real part equal to  \frac{1}{2} i.e.  s=\frac{1}{2}+it .
    .
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Let  s=\sigma+it ,  \Re(s) = \sigma .

     \Re(s) is the real part of  s .
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Let  s=\sigma+it ,  \Re(s) = \sigma .

     \Re(s) is the real part of  s .
    OK, so the sum in quetion parallels the p-series. So, can you provide an argument that

    \sum_{n=1}^{\infty}\frac{1}{n^s} converges if the real part of s>1?

    I'm sorry if I sound silly right now, it's just that I'd really lke to understand the hypothesis.
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    I'd be happy to explain.

    Let  s=\sigma+it .

    We want to know when  \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s} converges. For this we'll need a couple facts about complex numbers.

    1.)  \left|e^{x+iy}\right| =  \left|e^x\right|\cdot\left|e^{iy}\right| = e^x\cdot\left|\cos(y)+i\sin(y)\right| = e^x\cdot (\cos(y)^2+\sin(y)^2) = e^x

    2.)  n^s = e^{\log(n^s)} = e^{s\log(n)} = e^{(\sigma+it)\log(n)} = e^{\sigma\log(n)+it\log(n)} .
    Hence  \left|n^s\right| = n^{\sigma} .


    So  \sum_{n=1}^{\infty} \left|\frac{1}{n^s}\right| = \sum_{n=1}^{\infty} \frac{1}{n^\sigma} . Which is convergent for  \sigma >1 by the p-test.

    Thus  \zeta(s) is absolutely convergent for  \Re(s) > 1 .
    Last edited by chiph588@; December 22nd 2009 at 10:33 PM.
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by chiph588@ View Post
    I'd be happy to explain.

    Let  s=\sigma+it .

    We want to know when  \zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s} converges. For this we'll need a couple facts about complex numbers.

    1.)  \left|e^{x+iy}\right| = \left|e^x\right|\cdot\left|e^{iy}\right| = e^x\cdot\left|\cos(y)+i\sin(y)\right| = e^x\cdot (\cos(y)^2+\sin(y)^2) = e^x

    2.)  n^s = e^{\log(n^s)} = e^{s\log(n)} = e^{(\sigma+it)\log(n)} = e^{\sigma\log(n)+it\log(n)} .
    Hence  \left|n^s\right| = n^{\sigma} .


    So  \sum_{n=1}^{\infty} \left|\frac{1}{n^s}\right| = \sum_{n=1}^{\infty} \frac{1}{n^\sigma} . Which is convergent for  \sigma >1 by the p-test.

    Thus  \zeta(s) is absolutely convergent for  \Re(s) > 1 .
    Please don't laugh, but... How is it the case that

    |e^{iy}|=|\sin{y}+i\cos{y}| ?

    And moreover,

    |\sin{y}+i\cos{y}|=\sin^2{y}+\cos^2{y} ?

    I'm good with the rest of the argument.
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  11. #11
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    Hi. I think there is a layman's way of explaining it. Start with:

    When is x+3=0

    Alright, you're insulted right now. How then about: z+i=0. Same dif I know. Well, we can also consider integral functions like:

    f(x)=\int_0^x 6t+2 dt

    Then when is that one zero? Well, piece of cake. Integrate it, it's a quadratic in x, then find the roots via the quadratic formula. Well, then, that's what the Riemann Hypothesis is about except the integral function is more complicated and it deals with complex numbers:

    \zeta(s)=\Gamma(1-s)\frac{1}{2\pi i} \mathop\int\limits_{H^-}\frac{z^{s-1}}{e^{-z}-1}dz

    Now, for which values of s is that one zero? The Riemann Hypothesis claims that in the half-plane, \text{Re}(s)>0, all of them have the form s=1/2+it
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  12. #12
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    Quote Originally Posted by VonNemo19 View Post
    Please don't laugh, but... How is it the case that

    |e^{iy}|=|\sin{y}+i\cos{y}| ?

    And moreover,

    |\sin{y}+i\cos{y}|=\sin^2{y}+\cos^2{y} ?

    I'm good with the rest of the argument.

    It's the definition of modulus of complex number: if z=x+iy\in\mathbb{C}, then |z|=|x+iy|=\sqrt{x^2+y^2}

    Since, by definition, e^{iy}=\cos y+i\sin y\,,\,\,y\in\mathbb{R} , we get the above.

    Tonio


    Addition: of course, it should be |\cos y+i\sin y|=\sqrt{\cos^2y+\sin^2y}=\sqrt{1}=1
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  13. #13
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    Further comment

    "I've found very little in the area of practical application of the Hypothesis..."

    A simple check by Googling would confirm that (my attitude is that if it isn't on the internet, it doesn't exist). When you say "practical application", do you mean within math or outside of it?

    "...and is evidently somehow related to modern financial security systems."

    That would be because of the primes.

    It would help to know something of your math background to try to answer your questions.
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  14. #14
    No one in Particular VonNemo19's Avatar
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    Ok. I feel good about that, but

    (which can be shown fairly easily)

    Can you run me through this? Starting with

    \sum_{n=1}^{\infty}\frac{1}{n^s}
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  15. #15
    MHF Contributor chiph588@'s Avatar
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     \sum_{n=1}^{\infty} \frac{(-1)^n}{n^s} = \sum_{n=1}^{\infty} \frac{1}{(2n)^s}-\sum_{n=0}^{\infty} \frac{1}{(2n+1)^s}

     =\sum_{n=1}^{\infty} \frac{1}{(2n)^s}+\sum_{n=1}^{\infty} \frac{1}{(2n)^s}-\sum_{n=1}^{\infty} \frac{1}{(2n)^s}-\sum_{n=0}^{\infty} \frac{1}{(2n+1)^s}

     = 2\sum_{n=1}^{\infty} \frac{1}{(2n)^s} - \zeta(s)

     = \frac{2}{2^s}\sum_{n=1}^{\infty} \frac{1}{n^s}-\zeta(s)

     = (2^{1-s}-1)\zeta(s)

    Therefore  \zeta(s) = \frac{1}{2^{1-s}-1}\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s} .
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