Results 1 to 8 of 8

Math Help - is this series of products zero?

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    3

    is this series of products zero?

    Hi,

    Given the following two certainties:

    is this series of products  zero?-absum.gif

    a and b are real numbers and may be positive or negative.
    Would it be possible to prove the following:

    is this series of products  zero?-lim.gif

    I tried product rule for limits and several other tricks, but I can't do it.
    Nonetheless, intuitively this limit seems to be zero. For every a or b in
    an infinite set there will be one with opposite sign (otherwise the sums
    of a and b are not zero, but they always are), in fact each of the a or b
    itself already occurs an infinite number of times. This means that also
    every product ab occurs an infinite number of times, with both signs,
    or the set would not be infinite. Am I right? Infinity is tricky business...
    Is there a basic proof?

    Thank you for any help...
    Last edited by joland; October 12th 2010 at 01:44 AM. Reason: GIF image did not show properly
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Welcome to MHF!

    1) For what N is the equality true (in the attached file)? Any such N, a big enough one, or a specific one?

    2) Have you seen the comparison test for infinite series?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by joland View Post
    Hi,

    Given the following two certainties:

    Click image for larger version. 

Name:	absum.gif 
Views:	24 
Size:	914 Bytes 
ID:	19290

    a and b are real numbers and may be positive or negative.
    Would it be possible to prove the following:

    Click image for larger version. 

Name:	LIM.gif 
Views:	47 
Size:	878 Bytes 
ID:	19289

    I tried product rule for limits and several other tricks, but I can't do it.
    Nonetheless, intuitively this limit seems to be zero. For every a or b in
    an infinite set there will be one with opposite sign (otherwise the sums
    of a and b are not zero, but they always are), in fact each of the a or b
    itself already occurs an infinite number of times. This means that also
    every product ab occurs an infinite number of times, with both signs,
    or the set would not be infinite. Am I right? Infinity is tricky business...
    Is there a basic proof?

    Thank you for any help...

    Take \{a_n\}=\{1,-1,1,-1,\ldots\}\,,\,\,\{b_n\}=\{-1,1,-1,1,\ldots\}. Then, for some values of N\in\mathbb{N} ,

    \sum\limits^N_{k_1}a_k=\sum\limits^N_{k=1}b_k=0 , but a_nb_n=-1\,,\,\forall n\in\mathbb{N} ...

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2010
    Posts
    3

    thank you

    Thank you very much for the response.

    My numbers a and b represent deviations with respect to an average value, so the global sum over all "a" (or all "b") is always zero, even for N = 1, N = 2, etc. In your counter example this is not the case, because if the first and last elements of the sets have the same sign, their overall sum is not zero. Only if the signs of the first and final elements are opposite (i.e. there are as many "+1" as "-1" in the sets) the sum of all "a" or "b" is zero. This does not really matter, because the sum of the products aj * bj will never be zero, in fact it will be +-N.

    So, your counter example shows that I can not assume that this limit is zero. This is annoying! I was really hoping that this term would cancel out in my (much larger) equation. Back to the drawing board, I guess...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by joland View Post
    Hi,

    Given the following two certainties:

    Click image for larger version. 

Name:	absum.gif 
Views:	24 
Size:	914 Bytes 
ID:	19290

    a and b are real numbers and may be positive or negative.
    Would it be possible to prove the following:

    Click image for larger version. 

Name:	LIM.gif 
Views:	47 
Size:	878 Bytes 
ID:	19289

    I tried product rule for limits and several other tricks, but I can't do it.
    Nonetheless, intuitively this limit seems to be zero. For every a or b in
    an infinite set there will be one with opposite sign (otherwise the sums
    of a and b are not zero, but they always are), in fact each of the a or b
    itself already occurs an infinite number of times. This means that also
    every product ab occurs an infinite number of times, with both signs,
    or the set would not be infinite. Am I right? Infinity is tricky business...
    Is there a basic proof?

    Thank you for any help...
    An example: let be...

    a_{n}=\left\{\begin{array}{ll}-1 ,\,\,n=1 \\{}\\\frac{1}{n\ (n-1)} = \frac{1}{n-1} - \frac{1}{n} ,\,\, n>1\end{array}\right.

    b_{n}=\left\{\begin{array}{ll}1 ,\,\,n=1 \\{}\\\- \frac{1}{n\ (n-1)} = \frac{1}{n} - \frac{1}{n-1} ,\,\, n>1\end{array}\right. (1)

    It is evident that...

    \displaystyle \sum_{n=1}^{\infty} a_{n} = -1 + 1 -\frac{1}{2} + \frac{1}{2} - \frac{1}{3} +... =0

    \displaystyle \sum_{n=1}^{\infty} b_{n} = 1 -1 +\frac{1}{2} - \frac{1}{2} + \frac{1}{3} -... =0 (2)

    ... but also that...

    \displaystyle \sum_{n=1}^{\infty} a_{n}\ b_{n} = -1 -\frac{1}{4} - \frac{1}{36} - \frac{1}{144} -... \ne 0 (3)

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by joland View Post
    Thank you very much for the response.

    My numbers a and b represent deviations with respect to an average value, so the global sum over all "a" (or all "b") is always zero, even for N = 1, N = 2, etc. In your counter example this is not the case, because if the first and last elements of the sets have the same sign, their overall sum is not zero. Only if the signs of the first and final elements are opposite (i.e. there are as many "+1" as "-1" in the sets) the sum of all "a" or "b" is zero. This does not really matter, because the sum of the products aj * bj will never be zero, in fact it will be +-N.

    So, your counter example shows that I can not assume that this limit is zero. This is annoying! I was really hoping that this term would cancel out in my (much larger) equation. Back to the drawing board, I guess...
    You will have to give a better definition of the sum, then -- the only way that \displaystyle \sum_{j=0}^{N} a_j = 0 \ \forall N \in \mathbb{N} is if a_n = 0 \ \forall n \in \mathbb{N}, but that is obviously not the case..
    Also, are the first sums always over a countable set of indices?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2010
    Posts
    3
    We have two countable sets of values, call them {x} and {y}.
    If X is the mean value of the set of x-values (= sum{x} / N) we can define {a} = {x - X} as set of deviations.
    Obviously the sum of all a-values is always zero, because any non-zero mean value (= sum{a} / N) has
    gone into the X value. Also, the distribution of a and b is Gaussian, not some highly exceptional set
    like {1, -1, 1, -1, ...}
    For two such sets of deviations a and b, the set of products aj * bj appears in another expression,
    and I'm trying to demonstrate that for large sets of values this term also has zero mean (= zero sum).
    Intuitively you would think so, because basically both sets a and b represent white noise, and the
    product of two white noise signals (...over a large enough set N) should not all at once contain some
    non-zero mean signal. I can't work out the maths though, because it's not really my specialist field.

    The more I think of it, the more it looks like a statistical problem that will have a statistical answer,
    so I may be in the wrong forum... Perhaps I'll check out some statistics sites, or even electronics.
    Those guys should have similar problems all the time.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    It's really hard (at least for me) to understand what's really going on in there.
    I think you'll have a much easier time solving this if you rigorously define the problem.

    Good luck
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sum and products of series
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 26th 2011, 09:42 AM
  2. dot products
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: January 9th 2011, 06:44 PM
  3. set products
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: August 5th 2010, 08:43 AM
  4. ...norm takes products to products.
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: April 23rd 2010, 08:55 PM
  5. Inner Products
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: April 17th 2010, 08:34 PM

Search Tags


/mathhelpforum @mathhelpforum