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Recursive formula for zeta function of positive even integers

  1. Oct 21, 2005 #1

    StatusX

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    I was working with fourier series and I found the following recursive formula for the zeta function:

    [tex]\frac{p \\ \pi^{2p}}{2p+1} + \sum_{k=1}^{p} \frac{(2p)! (-1)^k \pi^{2(p-k)}}{(2(p-k)+1)!} \zeta(2k) = 0[/tex]

    where [itex]\zeta(k)[/tex] is the Riemann zeta function and p is a positive integer. I know this has already been found, but I was wondering who found it and whether it had a name, and whether it is interesting at all. You can use it to calculate the values of the zeta function pretty easily. For example, for p=1, we have:

    [tex]\frac{\pi^2}{3} -\frac{2!}{1!}\zeta(2)=0[/tex]

    [tex]\zeta(2)=\frac{\pi^2}{6} [/tex]

    for p=2:

    [tex]\frac{2\pi^4}{5} -\frac{4! \\ \pi^2}{3!}\zeta(2)+\frac{4!}{1!}\zeta(4)=0[/tex]

    [tex]\zeta(4)=\frac{1}{24}(4\pi^2 (\frac{\pi^2}{6})-\frac{2\pi^4}{5}) =\frac{\pi^4}{90}[/tex]

    etc.
     
    Last edited: Oct 21, 2005
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  3. Oct 23, 2005 #2

    shmoe

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    This looks to be a form of Euler's [tex]\zeta(2n)=\frac{(-1)^{k+1}(2\pi)^{2n}B_{2n}}{2(2n)!}[/tex], and I suspect it wouldn't be difficult to get from one to the other using the usual properties of the Bernoulli numbers, [tex]B_n[/tex] (see their generating series).

    If I have the history correct Euler first proved the special cases for n=1 to 13, then a few years later proved the general formula.

    It's definitely interesting in that it proves zeta is a rational multiple of a power of pi for even positive integers (and gives a corresponding result for zeta at negative integers via the functional equation). Compare with what's known about zeta at other values and this looks all the more impressive.
     
  4. Oct 25, 2005 #3
    ramsey2879

    This is one reason why I consider [tex]\pi[/tex] the most interesting constant in math. For instance, Math World at http://mathworld.wolfram.com/RelativelyPrime.html gives the probability that two numbers are prime to each other as
    [tex]{\zeta(2)}^{-1} = \frac{6}{\pi^2}[/tex]
     
    Last edited: Oct 25, 2005
  5. Oct 25, 2005 #4

    shmoe

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    If you haven't already, it's worth the time to consider why the probability that two randomly chosen numbers are relatively prime is given by [tex]\zeta(2)^{-1}[/tex] as mathworld says.

    For that matter, it's worth considering what this actually means in precise terms (i.e. how are we handling the idea of randomly selecting numbers from an infinite set).
     
    Last edited: Oct 25, 2005
  6. Oct 25, 2005 #5
    Good question, I imagine that in all probability, most numbers compared with each other would be less than [tex]2^{32}[/tex] or some lower power of 2, thus would not be an infinite set. For sake of argument though, lets assume that any number in an infinitely large set is equally likely to be chosen. If we can assume that the question then can be broken down by considering each prime separately. The probability that a prime number is divisible by p is 1/p and the probability that both numbers are divisible by p then becomes [tex]1[/tex]/[tex]p^2[/tex]. Thus the probability that this doesn't happen for p is then
    [tex]\frac{p^2-1}{p^2}[/tex]
    What I am not sure about is whether one can take the next step of multiplying the above probabilities for each prime together to find the total probability as a whole, or how this involves the zeta function.
     
    Last edited: Oct 25, 2005
  7. Oct 25, 2005 #6

    shmoe

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    Under any reasonable definition of "most", most numbers are larger than 2^32, or any finite bound you'd like. We can not worry about this for now and go on you next assumption (which is a fair place to start):

    Very good. it might help to rewrite this as [tex]1-\frac{1}{p^2}[/tex]

    Yea, multiplying the probabilities together is fine. Note that you'd be making the assumption that the probability a random number is divisible by the prime p and the probability it's divisible by the prime q are independant (a fair assumption).

    Have you seen the Euler product form of Zeta? If not, don't look it up before you try to convince yourself that the infinite product we get over primes is in fact equal to 1/zeta(2).
     
  8. Sep 24, 2008 #7

    Using a somewhat different method (still using trigonometric series though), I ended up with a recursion as well here.
    I do remember a Fourier series method for both the Riemann zeta at positive even integers and the Dirichlet L at positive odd integers in a paper called 'Recursive Formulas for [itex]\zeta(2k)[/tex] and [itex]L(2k-1)[/tex] by Xuming Chen though. As I recall the relation in that paper was somewhat different from yours though (I'd like to see how you derived it).

    Hope this helped!
     
  9. Sep 25, 2008 #8

    CRGreathouse

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    I have to admit, the standard informal proof of the probability that two numbers are relatively prime always bothered me. Each 'number' has a 1/p chance of being divisible by p, hence each 'number' has an infinite number of factors (since Euler proved that the sum of the reciprocals of the primes diverges).

    Of course, seeing it as the limit of p-smooth numbers as p -> infty makes it sensible again.
     
  10. Sep 25, 2008 #9
    I'm a bit confused why the coprime probability is being discussed here (I wanted to see more recursions like the first post mentions)- but its also true that [itex]n[/tex] randomly chosen integers have probability of being coprime to each other of [itex]\frac{1}{\zeta(n)}[/tex].
    I once tried to show this after seeing the base case for two integers in Havil's Gamma: Exploring Euler's Constant by generalising the proof there, but I forgot all about it.

    Incidentally, I wonder if an inductive method will work- it might even come back full circle to recursions....
     
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