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Equating terms in summations, simple question!

  1. Oct 18, 2011 #1
    Today my physics professor briefly skipped over this during a derivation:

    We started with
    [itex]2 \sum F_{n}(x) = \sum G_{n}(x)[/itex] , summed from n=0 to [itex]\infty[/itex]

    which she then concluded
    [itex]2F_{n}(x) = G_{n}(x)[/itex]

    where F and G are functions of x, and different functions for different values of n. (she was using a generating function)

    What proves this is true?

    I think this is just equating terms of the sums for values of n, but how do we know this is valid? I provide the counter example:
    [itex]\sum x = \sum x^{2} [/itex] for x=0 to [itex]\infty[/itex], which is true
    however, equating individual terms is false: x[itex]\neq x^{2}[/itex]

    Is there some criteria for equating terms?

  2. jcsd
  3. Oct 18, 2011 #2


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    I have absolutely no idea what you mean by
    [tex]\sum x= \sum x^2[/tex]
    If x is an integer index, from 0 to infinity, neither of those sums exists so it makes no sense to say they are equal.

    But you are right that, in general, two sums being equal does not mean individual terms are equal- an easy counter-example: 1+ 2+ 1= 3+ 1+ 0.

    However, if you series is a series of "independent functions", then it is true.

    For example, it is true that if two power series are equal
    [tex]\sum_{n=0}^\infty a_nx^n= \sum_{n=0}b_n x^n[/tex]
    then "corresponding terms" are equal: [itex]a_n= b_n[/itex].
  4. Oct 18, 2011 #3
    Ok I was trying to be lazy but I'll just write out what I've got in my notes.

    Sums are from L=0 to [itex]\infty[/itex] and Integral limits are x = -1 to 1.

    [itex]2 \sum \frac{h^{2L}}{2L+1} = \sum(h^{2L}\int P_{L}^{2}(x)dx)[/itex]


    [itex]\frac{2}{2L+1} = \int P_{L}^{2}(x)dx[/itex]

    Why is equating terms valid here? I don't see a reason to just assume that it is ok, and she didn't provide any reason for it. So how do I know if this is true?

    If needed, here's more info. [itex] P_{L}(x) [/itex] are the legendre polynomials. She used a generating function [itex]\Phi(x,h)=\frac{1}{\sqrt{1-2xh+h^{2}}}[/itex]. We began with the relation [itex]\int\Phi^{2}(x,h)dx = \int (\sum h^{L}P_{L}(x))^{2}dx[/itex] ( where sum is from L=0 to [itex]\infty[/itex] and integral limits are x=-1 to 1)
    Integrating [itex]\Phi^{2}(x,h)[/itex] and doing a series expansion of natural log gives the LHS of the equation at the beginning of my post.
  5. Oct 18, 2011 #4


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    And the Legendre Polynomials are "independent functions" as I said.
  6. Oct 19, 2011 #5
    So operating on a set of lin. indep. functions gives another set of lin. indep. functions? I know the the legendre polynomials are indep., but I don't know that integrating and squaring them means they will still be indep.
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