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Show the equality of two expressions

  1. Apr 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that

    [tex]\sum_{k=0}^{N-1}e^{\gamma \tau k}\int_{0}^{\tau}F'(k\tau+s)ds[/tex]

    can be written as

    [tex]\int_{0}^{t}e^{\gamma t'}F'(t')dt'[/tex]

    2. Relevant equations

    1. [itex]t=N\tau[/itex]

    2. [itex]\int_{0}^{\tau}F'(k\tau+s)ds[/itex] has the same statistical properties for each interval of length [itex]\tau[/itex], and is statistically independent with respect to [itex]k[/itex].

    3. The attempt at a solution
    I barely know where to start. As a first step, I'm thinking that perhaps "same statistical properties" means that the integral is the same regardless of [itex]k[/itex], so that

    [tex]\int_{0}^{\tau}F'(k\tau+s)ds=\int_{0}^{\tau}F'(s)ds,[/tex]

    i.e., [itex]k=0[/itex]. Is this correct?
     
  2. jcsd
  3. Apr 12, 2015 #2

    mfb

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    Staff: Mentor

    Without the exponential, the first expression would just be a piecewise definition of the second integral, and a substitution would transform them into each other (can you see how? Hint: which argument values are used in F in the first, second, ... integral?). I guess you need the second relevant equation to get the same result including the exponential, but it could be still interesting to make that substitution.
     
  4. Apr 12, 2015 #3
    Thanks. That gives me

    [tex]\sum_{k=0}^{N-1}e^{\gamma \tau k}\int_{k\tau}^{(k+1)\tau}F'(t')dt'[/tex].

    What's left now is to move the exponential into the integrand, but I'm not sure how that can be justified.
     
  5. Apr 13, 2015 #4

    mfb

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    Staff: Mentor

    That I don't know. It does not work for general functions F', but it works if you take the limit N -> infinity (with finite t), and it might work for some special F' even with finite N.
     
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