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Determining A Finite Value of Infinity

  1. Nov 19, 2014 #1
    okay.....if you accept that the sequence
    1+2+3+4.....=-1/12,
    I think I have determined a finite value of infinity.
    To find the value of the sums of all natural numbers up to a number, you can use the equation
    ((x^2)+x)/2.
    An example would be 4.
    4+3+2+1=10.
    ((4^2)+4)/2 also equals 10.
    following this logic,
    ((x^2)+x)/2=-1/12
    is true for the above sequence. this can be rearranged to
    (x^2)+x+1/6
    This is the resulting quadratic equation. Using the quadratic formula, one obtains the x intercepts as
    (-3+-(sqrt3))/6.
    and since x is infinity in this situation (since the highest value is infinity), the x intercepts are the values of infinity in the equation. Therefore, by assigning a value of -1/12 to riemann zeta(-1), you also assign finite values to infinity, approximately
    -0.211324....
    and
    -0.788675....
    If there is any faulty reasoning, please remember I'm 15 and I most likely have no clue what I'm talking about.
     
  2. jcsd
  3. Nov 19, 2014 #2

    phinds

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    Since this is a contraction in terms, I stopped reading right there.
     
  4. Nov 19, 2014 #3

    jbriggs444

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    http://en.wikipedia.org/wiki/Divergent_series

    There is no sum of the integers from one to infinity. The series is divergent. The assignment of the value -1/12 as the "sum" of this series does not correspond to anything which you should regard as being an actual sum.
     
  5. Nov 19, 2014 #4
    Most people would stop reading when someone says that 1+2+3.....=-1/12, that's counterintuitive but generally accepted.....so thanks for the ignorance
     
  6. Nov 19, 2014 #5
    Makes sense
     
  7. Nov 19, 2014 #6

    pwsnafu

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    No. The series 1+2+3+... is divergent. So it is infinite.
    It's Zeta regularization has value -1/12.
    That is not the same thing as claiming that the series is "equal" to -1/12

    I fail to see how. Just because a statement is true for finite values of x does not mean it is valid for infinite values of x. You need to prove that "((x^2)+x)/2=-1/12" is a valid statement for some x. You have not.

    This is not possible because (1) infinity is not a real number, (2) infinite values cannot be manipulated in the same way as finite values.

    Unfortunately, everything.
     
  8. Nov 19, 2014 #7
    Good to know. Thanks for the reply
     
  9. Nov 19, 2014 #8

    phinds

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    No, as has been pointed out, it is NOT generally accepted.
     
  10. Nov 19, 2014 #9
    It is generally accepted as a representation of Riemann zeta (-1)
     
  11. Nov 20, 2014 #10

    pwsnafu

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    No you have this backwards.
    ##\zeta(-1)## is a possible regularization of ##\sum_{n=1}^{\infty} n##.
    ##\sum_{i=1}^{\infty} n## is not a representation of ##\zeta(-1)##.
    Regularization and representation are separate concepts.
     
  12. Nov 20, 2014 #11
    Y-value of the vertex in the resulting quadratic equation is -1/12. Any significance do you think?
    good to know thanks
     
  13. Nov 21, 2014 #12
    I don't.
     
  14. Nov 21, 2014 #13

    Mark44

    Staff: Mentor

    This thread has run its course, so I'm closing it.
     
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