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Double series

  1. Mar 8, 2006 #1

    James R

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    Does anybody know how to prove the following?

    [tex]\sum\limits_{n=0}^{\infty} \sum\limits_{m=0}^{\infty} f(n,m) = \sum\limits_{p=0}^{\infty} \sum\limits_{q=0}^{p} f(p,p-q)[/tex]

    f(n,m) is any function of n and m.

    Note the change of the limiting values on the sums.
     
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  3. Mar 8, 2006 #2

    NateTG

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    I'd be very impressed if a suitable proof exists, considering that it's not true.

    For example, if
    [tex]f(x,y)=2^{y-2x}[/tex]
    then the LHS is clearly not convergent, but the RHS is.
     
  4. Mar 8, 2006 #3

    shmoe

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    That looks false to me, but it may be a little typo. Do you perhaps mean:

    [tex]\sum\limits_{n=0}^{\infty} \sum\limits_{m=0}^{\infty} f(n,m) = \sum\limits_{p=0}^{\infty} \sum\limits_{q=0}^{p} f(q,p-q)[/tex]

    This kind of rearrangement usually follows from absolute convergence pretty easily, are you sure there's no restrictions on f(n,m)?
     
  5. Mar 8, 2006 #4

    NateTG

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    That's still false.
    Consider, for example [itex]f(x,y)[/itex] is [itex]2^{-x!-y!}[/itex] if [itex]y>x[/itex] and [itex]0[/itex] otherwise. Both sums are absolutely convergent, but the LHS is non-zero, and the RHS is zero.

    If there is absolute convergence, then
    [tex]\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}f(m,n)=\sum_{p=0}^\infty\sum_{q=0}^p f(p+q,p-q)+\sum_{p=0}^\infty\sum_{q=0}^p f(p+q+1,p-q)[/tex]
    works. Although, I'd be inclined to write that
    [tex]\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}f(m,n)=\sum_{k=0}^{\infty}\sum_{m>0,n>0,m+n=k}f(m,n)[/tex]
     
    Last edited: Mar 8, 2006
  6. Mar 8, 2006 #5

    shmoe

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    The RHS is not zero. Take the p=1 term, it's f(0,1)+f(1,0), one of these is non-zero. (Maybe you missed where I changed f(p,p-q) to f(q,p-q))

    You can think of the double sum as a sum over the integer lattice (n,m) where n and m are >=0. The LHS sums along infinite vertical lines (e.g. (0,0), (0,1), (0,2),...) then sums the results. My version sums along the lines with slope -1 (the lines m+n=p), then sums the results. James version looks to sum everything under the line y=x say (though I'd expect it was a typo).

    Your version

    [tex]\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}f(m,n)=\sum_ {k=0}^{\infty}\sum_{m>0,n>0,m+n=k}f(m,n)[/tex]

    is the same as mine except you seem to be missing all points where one of n or m is zero. If you meant for n and m to be >= to zero instead of just >0, then it's the same as mine (take k=p, m=q)
     
    Last edited: Mar 8, 2006
  7. Mar 8, 2006 #6

    NateTG

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    Yeah, those should have been greater than or equal to.

    Yeah, I misread that as (q,q-p) which is no good.
     
  8. Mar 9, 2006 #7

    James R

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    Damn! Sorry about that. It was a typo, and should read:

    [tex]\sum\limits_{n=0}^{\infty} \sum\limits_{m=0}^{\infty} f(n,m) = \sum\limits_{p=0}^{\infty} \sum\limits_{q=0}^{p} f(q,p-q)[/tex]

    That is, it is f(q,p-q) rather than f(p,p-q).

    So, given that, does anybody have a proof?

    Here's one specific context, for example.

    We know that exp(x+y) = exp(x)exp(y).

    Writing exp(x+y) and exp(x) as Maclaurin series, I would like to prove that the left hand side equals the right hand side. If you try that, it leads to a double sum of the type given.
     
    Last edited: Mar 9, 2006
  9. Mar 9, 2006 #8

    shmoe

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    You will be able to find proofs of this sort of thing under various assumptions in most intro analysis texts, see Rudin's Principles of Mathematical Analysis for example.
     
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