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Cauchy product of series

  1. Aug 5, 2014 #1
    Hey guys,

    I was just doing some independent study on products of series and I'm trying to understand/derive the following form of the Cauchy product of series:
    [tex] \left(\sum_{n=0}^{N} a_{n}\right) \left(\sum_{m=0}^{N} b_{m}\right) = \sum_{n=0}^{N} \left(\sum_{k=0}^{n} a_{k}b_{n-k}\right) [/tex]

    Is it that one redefines the dummy variable [itex] m[/itex] in the following manner [itex]m\rightarrow n-k[/itex] with the following constraint on [itex] k[/itex], [itex] 0\leq k\leq n[/itex], or is there some other reasoning behind it that I'm missing?

    Thanks in advance.
  2. jcsd
  3. Aug 5, 2014 #2


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    Consider [tex]
    \sum_{n=0}^\infty \sum_{m=0}^\infty a_nb_m.
    [/tex] This is a sum over a two-dimensional grid. Assuming suitable convergence properties, you can rearrange the order of the terms so that instead of proceeding by row and then by column, you proceed by diagonals, starting with all terms with [itex]n + m = 0[/itex], then [itex]n + m = 1[/itex], etc. The points lying on the diagonal [itex]n + m = p[/itex] are [itex](0,p), (1,p-1),\dots, (q,p-q), \dots, (p,0)[/itex]. This yields [tex]
    \sum_{n=0}^\infty \sum_{m=0}^\infty a_nb_m = \sum_{p=0}^\infty \sum_{q=0}^p a_qb_{p-q}.
    [/tex] This idea must be adapted if [itex]\infty[/itex] is replaced by a finite upper limit [itex]N[/itex], because you're then not taking the whole of the diagonal when [itex]N < n+m \leq 2N[/itex]. Finding an expression for the sum in that case is left as an exercise (consider the reflection of [itex](n,m)[/itex] in the line [itex]n + m = N[/itex]).
    Last edited: Aug 5, 2014
  4. Aug 5, 2014 #3


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    Another way of looking it at is this. Say they are power series$$
    \sum a_nx^n = a_0+a_1x+a_2x^2+ ...$$and$$
    \sum b_nx^n= b_0+b_1x+b_2x^2+...$$Look at what happens if you take the expanded series and start writing out the product, grouping on powers of ##x##. You get$$
    a_0b_0 + (a_0b_1+a_1b_0)x + (a_0b_2+a_1b_1+a_2b_0)x^2 +...$$It's the same idea. Put ##x=1## for the same result.
  5. Aug 5, 2014 #4
    Ah ok, thanks for the help on the subject, both explanations have really helped my understanding.

    One further question on the matter though - is there any way that you can "derive" the relation by just redefining the dummy summation variables (analogously to how one can prove the binomial theorem by shifting the index), without expanding the sum, or giving a heuristic argument?
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