Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ramanujan Summation and Divergent series in relation to the Riemann Zeta function.

  1. Dec 26, 2006 #1

    Gib Z

    User Avatar
    Homework Helper

    I was wondering if anyone could tell me more about the Riemann Zeta function, esp at negative values. Especially when [tex] \sum_{n=1}^{\infty}n= \frac{-1}{12} R[/tex] where R is the Ramanujan Summation Operator. Could anyone post a proof?
     
  2. jcsd
  3. Dec 27, 2006 #2
    Actually... I have thought of a proof for that.. but without that "R". I hope this helps...

    You first consider the "-" summation (from i = 1 to infinity) of i(-r)^i. Let this sum be equal to S. Then,

    S = r - 2r^2 + 3r^3 - 4r^4 + ... - ...
    Sr = 0 + r^2 - 2r^3 + 3r^4 - ... + ...

    S + Sr = r - r^2 + r^3 - r^4 + ... - ...
    S(1+ r) = r / [1+r]
    S = r/(1+r)^2

    When you take the limit of both sides of the equation as r -> 1, we have:

    Lim S (r-> 1) = Lim r/(1+r)^2 (r-> 1)

    1 - 2(1)^2 + 3(1)^3 - 4(1)^4 + ... - ... = 1/4
    (remember how I defined my S awhile ago? )

    or simply, 1 - 2 + 3 - 4 + .. - ... = 1/4 (remember this)

    Now, If I let T = 1 + 2 + 3 + ... then,

    2T = 2(1 + 2 + 3 + ... ).. then,
    4T = 4(1 + 2 + 3 + ... ) = 2(2 + 4 + 6 + ... ) (remember this too)

    Therefore, 1/4 + 4T = ?.
    Look...

    1 - 2 + 3 - 4 + .. - ...
    +
    2(2) + 2(4) + 2(6) + 2(8) +...

    As you can see, the odd numbers are left as is, but the even numbers are combined,
    say 2(2) - 2 and 2(4) -4.

    Thus, 1/4 + 4T = T.
    Solving for T, you got -1/12.

    So,

    [tex] \sum_{n=1}^{\infty}n= \frac{-1}{12} [/tex]
     
    Last edited: Dec 27, 2006
  4. Dec 27, 2006 #3

    Gib Z

    User Avatar
    Homework Helper

    That is excellent, thanks. The R merely shows that you have to derive the expression specially like you did, rather than the conventional thought that it diverges. The Riemann zeta function: [tex] Zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}[/tex] at even negative s, ie -2,-4,-6... are equal to zero, can anyone prove that?
     
  5. Dec 27, 2006 #4
  6. Dec 29, 2006 #5

    ssd

    User Avatar

     
    Last edited: Dec 29, 2006
  7. Dec 30, 2006 #6

    Gib Z

    User Avatar
    Homework Helper

    True, Maybe this proof wasn't the most rigorous, but none the less achieves the desired result. We know in forehand that that answer is correct, even though umm...illogical lol. The result that the series is equal to -1/12 is usual in string theory, if you check the link above.
     
  8. Dec 30, 2006 #7

    ssd

    User Avatar

    LOL, I am not a man of Physics, unable to understand the physical interpretation of String theory facts.
     
  9. Dec 30, 2006 #8

    Gib Z

    User Avatar
    Homework Helper

    lol i Don't know Strings so much either, just the theory and the most basic math.
    I know enough though to understand the link. Theres 2 proofs for the question, look there if you like.
     
  10. Dec 31, 2006 #9

    ssd

    User Avatar

    'Gib Z' I dont know what to say.....
    My konwledge goes in this way,
    r - r^2 + r^3 - r^4....ad inf
    = r/(1+r) if and only if "-1< r <1" holds. The inequalities being strict.
    When one uses the limit r->1, that means he calculates limit r->1- and limit r ->1+ and found them to be equal. But by taking limit r ->1+ one violets the necessary and sufficient condiditon |r|<1 for this summation.
     
  11. Dec 31, 2006 #10

    Gib Z

    User Avatar
    Homework Helper

    Well yes, as I said its not a very rigorous proof, but its sufficient for showing a high school class and thats what I originally intended this for. The Link has 2 rigorous proofs, look there. I know your disbelief, its not logical.
     
  12. Sep 30, 2007 #11
    A more formal derivation!

    Forgive the lack of LaTeX but the notation is self-explanatory! The following combined with the analytic extension theorems of complex analysis provide suitable Euler-Abel sums for negative values of the Riemann Zeta function.

    Note that <B*_k> stands for the old Bernoulli numbers(the k th one here) and <B_k> the new ones(I explain in the end)

    1 + e^iy + e^2iy + e^3iy + ... = (1 - e^iy)^-1 = 1/2 + (1/2)i.cot y/2
    Equating real and imaginary parts we have,

    cos y + cos 2y + cos 3y + ... = -1/2 --------(1)
    and,
    sin y + sin 2y + sin 3y + ... = (1/2)cot y/2 ------------(2)

    Replace y by (y + pi) and we have,
    cos y - cos 2y + cos 3y - ... = 1/2

    differentiate 2k times and we have
    Sum [n=1 to infinity] (-1)^(n-1) n^2k cos ny = 0
    differentiate 2k-1 times and we have,
    Sum [n=1 to infinity] (-1)^(n-1) n^(2k-1) sin ny = 0

    for (2) we get,
    sin y - sin 2y + sin 3y - .... = (1/2)tan y/2
    since the Taylor expansion of (1/2)tan y/2 is,
    Sum [k=1 to infinity] {(2^2k - 1) <B*_k> y^(2k-1)} / (2k)!

    Differentiate 2k-1 times, put y=1 and equate to get,

    Sum [n=1 to infinity] (-1)^(n-1) n^(2k-1) = [(-1)^(k-1) (2^2k - 1) <B*_k>] / 2k


    Now recall Z(1- 2k) = 1^(2k-1) + 2^(2k-1) + 3^(2k-1) + ....
    where Z is Riemann's Zeta function, we see that the function we handled,
    N(1-2k) = Sum [n=1 to infinity] (-1)^(n-1) n^(2k-1) = 1^(2k-1) - 2^(2k-1) + 3^(2k-1) - ...

    satisfies,
    (1 - 2^2k) Z(1-2k) = N(1-2k)
    Then,
    Z(1-2k) = (-1)^k <B*_k> / 2k

    But with the modern notation for the Bernoulli numbers,
    <B_2n> = (-1)^(n+1) <B*_n>

    therefore,
    Z(1-2k) = (-1)^(2k+1) <B_2k> / 2k = - <B_2k> / 2k

    Quite generally,
    Z(1-n) = (-1)^(n-1) <B_n> / n

    for all positive integer k(even zero in a sense, more below), defining the Zeta function for odd negative integers (the case we considered first obviously shows that even negative integers make the function zero, these are its trivial zeros, the other zeros constitute the Riemann Hypothesis, the greatest intellectual challenge faced by man)
    Now the fun bit, put k=1 and we have the magical,
    Z(-1 ) = 1 + 2 + 3+ 4 + .... = -1 / 12 , since B<2> = 1/6

    This value actually holds a meaning(though we are hard put to explain in any connotation of the word brief with regard to use in physics, but it has to do with what may be termed 'measure' of infinities, thankfully in math definition is all that matters!) and is extensively used with the case k=2 in theoretical physics(namely the Casimir Effect and Perturbation theory on quantum fields), but applications are always of secondary importance!

    Rigorous enough eh? Yes, but only if we consider it in a special summation definition (the extended A definition due to Euler, though the Ramanujan sum coincides too)and note the analyic extention of the zeta function to the whole plane.

    Actually this is most awe-inspiring since it fits with the (seemingly independant) analytic extention perfectly. As in how k=0 gives the simple pole!

    The 'trickery' per se is that le^iyl = 1 for all y, but since we avoid the actual pole there is no damage done!!
     
  13. Oct 3, 2007 #12
    So I came across a connection to Taylor's series, imagine my surprise when I learnt that one can think of $f(t + n)$ as the result of the operator ${\exp}{n{\delta}}$ applied on $f(t)$
    Then we have,
    $\sum_{n = 0}^{\infty} f(n) = \sum_{n = 0}^{\infty} {\exp}{n{\delta}} f(0)$

    When t is limited to 0.
    We can sum the exponential series as a geometric progression(this works even for the operator)and,
    $\sum_{n = 0}^{\infty} f(n) = (1 - {\exp}{{\delta}})^{ - 1} f(0)$

    since,
    $\frac {{\delta}}{{\exp}{{\delta}} - 1} = \sum_{n = 0}^{\infty} \frac {B_{n}}{n!} {\delta}^{n}$
    we substitute and we have,

    $\sum_{n = 0}^{\infty} f(n) = - \sum_{n = 0}^{\infty} \frac {B_{n}}{n!} {\delta}^{n - 1}f(0)$

    (the operations on the operator are valid)

    put f(n) = n and we have 1+2+3+4+...= -1/12
    etc.
     
  14. Oct 7, 2007 #13

    Gib Z

    User Avatar
    Homework Helper

    That is some interesting working yasiru89, i will need to look at it in more detail. By the way, on these forums instead of $ for tex, at the start of the code write [ tex ] without the space between the brackets, and [ /tex ] without spaces, when the code is finished.
     
  15. Oct 7, 2007 #14

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    And [ itex ] for stuff inlined with ordinary text.
     
  16. Oct 7, 2007 #15
    Yeah, thanks for pointing that out, I'm mostly a pen and paper guy and my tex is restricted to posts on mathforums!(my first post isn't tex)

    The beauty of mathematics is that logic can at times have very little bearing and all that matters is definition!(if you can follow it up that is)
     
  17. Oct 7, 2007 #16

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Of course, that's where you need logic. :tongue:
     
  18. Oct 20, 2007 #17
    I meant that our logic is bound by geometries and the freedom of definition absolves us. Yet we look the other way!
     
  19. Oct 21, 2007 #18
    wait a second, does this mean that the stuff I just reviewed about infinite series is in fact WRONG?! If there do exist rigorous proofs of the fact that 1 + 2 + 3 + .... = -1/12 then calculus lives inside of an inconsistent theory!?!?

    please can someone clarify whether or not this is a FACT?!
     
  20. Oct 21, 2007 #19

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The common meaning of the symbols you wrote is equivalent to "the infinite summation operation you learned in calculus". And in that meaning, there does not exist a proof of that statement.

    Only when you consider different kinds of 'summation' operations can statements like that be true.
     
  21. Oct 21, 2007 #20

    Gib Z

    User Avatar
    Homework Helper

    I didn't put an R in that equation for no reason you know =]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Ramanujan Summation and Divergent series in relation to the Riemann Zeta function.
  1. Riemann Zeta Function (Replies: 8)

Loading...