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Sums: Substituting two sums

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    Lets assume that we know the following:

    [tex]
    \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t) = a_0 + \sum\limits_{n = 1}^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t),
    [/tex]

    where a0 is the contribution for n=0. Now I have an expression for a function f given by the following:

    [tex]
    f = \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}.
    [/tex]

    Am I allowed to write f as this?:

    [tex]
    f = a_0 \frac{1}{{Z(\omega _0 )}} + \sum\limits_{n = 1}^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}},
    [/tex]

    i.e. substitute the sum? Personally, I think yes, but I am a little unsure, which is why I thought it would be best to check here. Thanks in advance.

    Best regards,
    Niles.
     
  2. jcsd
  3. Jan 11, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This appears to be saying that
    [tex]\sum\limites_{n=-\infty}^{-1}\varepsilon_n(t)}\exp(-i\omega_n t)= 0[/itex]

    But that does NOT mean that

    [tex]\sum\limits_{n = -\infty}^{-1} {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}= 0[/tex]
    since the [itex]Z(\omega_n)[/itex] term may change the contribution of each indvidual term in the sum.
     
  4. Jan 11, 2009 #3
    Aw man, I made an error in my first post. Very stupid of me, because now I made you look at something which was not my intention; I am very sorry for that.

    What I meant to write is the following (I'm still really sorry, I should have taken a closer look before posting):

    Lets assume that we know the following:
    [tex]
    \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t) = a_0 + \sum\limits_{n = 1}^\infty {n^2} \exp ( - i\omega _n t).
    [/tex]

    Now we look at an expression for a function f given by the following:
    [tex]
    f = \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}.
    [/tex]

    Am I allowed to write f as:
    [tex]
    f = a_0 \frac{1}{{Z(\omega _0 )}} + \sum\limits_{n = 1}^\infty {n^2} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}.
    [/tex]

    I have double-checked, and the formulas are correct now. Sorry, again.

    Best regards,
    Niles.
     
  5. Jan 11, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    And the answer is still "No". Because [itex]Z(\omega_n)[/itex] may be different for different n, it can completely change the sum.
     
  6. Jan 11, 2009 #5
    I had actually hoped that you would answer "Yes", because the "No" means that my calculations are wrong. But the quantity [itex]Z(\omega_n)[/itex] does depend on n, so it is different for different n.

    I will have to think about this. I guess I made a wrong assumption along the way.. I do that sometimes (as we can also see from this thread :smile:).

    Thanks for helping, I really do appreciate it.
     
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