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-   -   Summation from -infinity to infinity (https://www.askmehelpdesk.com/showthread.php?t=390335)

  • Aug 25, 2009, 09:44 AM
    radiation
    summation from -infinity to infinity
    how do u find the value of 'sigma' from n=-infinity to infinity for (1/2)^(2 mod n)..
    sorry wasn't able to put it in the right format..
  • Aug 25, 2009, 10:00 AM
    Unknown008

    Is that it?



    You can try the LaTeX tutorial...
  • Aug 26, 2009, 04:12 AM
    moonfish1985
    The most logical interpretation I can come up with for (2 mod n) is 2 modulo n, which makes no sense whatsoever. Unless the sum is supposed to diverge, then it's prefectly logical. Because for |n|>to the expression (2 mod n) is simply 2, meaning the sum is the same as (and excuse me for being too lazy to use LaTeX):
    [the sum from n=-inf to -3 for (1/4)] + [the sum from n=-2 to 2 for (1/2)^(2 mod n)] + [the sum from n=3 to inf for (1/4)]

    The first and past part of that expression diverge to infinity.
  • Aug 26, 2009, 03:59 PM
    galactus
    If |n| were an absolute value, then we could write:



    What we have is a geometric series.

    Just a thought. I am uncertain of the 'mod' as well.
  • Aug 28, 2009, 06:40 AM
    radiation

    The notation given by jerry is exactly right.. The answer given is 5/3... I don't know how to arrive at it..
  • Aug 28, 2009, 07:05 AM
    jcaron2
    Quote:

    Originally Posted by radiation View Post
    The notation given by jerry is exactly right..The answer given is 5/3...i dont know how to arrive at it..

    Galactus almost had it! I'm guessing he sees his mistake now that you say the right answer.

    In doubling the sum from 0 to infinity (once to account for the negative values of n and once for the positive), he inadvertently counted n=0 twice. Hence, the real answer is actually
  • Aug 28, 2009, 07:58 AM
    galactus
    Good catch, JCaron. Yes, it was an oversight.

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