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Checking regular variance around 0, hypergeometric fucntion

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A function g is [itex]\alpha[/itex]-regularly varying around zero if for all [itex]\lambda > 0[/itex], [itex]\lim_{x\to 0} \frac{g(\lambda x)}{g(x)}=\lambda^{\alpha}[/itex]
    For real s and [itex]\alpha \in (0,1)[/itex], define f:
    [itex]f(s)=1-\alpha \int_{0}^{\infty} e^{\alpha t} \frac{\frac{1}{1+s^2}}{e^t(1-\frac{1}{1+s^2})+\frac{1}{1+s^2}} dt = 1 + \alpha \sum_{n=1}^{\infty} \frac{(-s^{-2})^{n}}{n+\alpha}[/itex].
    Show f is [itex]2\alpha[/itex]-regularly varying around zero.


    2. Relevant equations
    Note [itex]\frac{1}{1+s^2}[/itex] is a characteristic function of Laplace distribution.


    3. The attempt at a solution
    I am not familiar with hypergeometric series. For example, I do not know how to show that the series converges to [itex]-\frac{1}{\alpha}[/itex] when s goes to 0. Are there any properties of hypergeometric series that might be useful for proving this? Attempt to directly compute the limit failed.
     
    Last edited: Jun 20, 2011
  2. jcsd
  3. Jun 20, 2011 #2

    lanedance

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    Homework Helper

    so, i may be reading it wrong but can you make the following simplifications
    [tex] f(s)
    = 1-\alpha \int_0^{\infty}
    e^{\alpha t}
    \frac{\frac{1}{1+s^2}}
    {e^t(1- \frac{1}{1+s^2}) + \frac{1}{1+s^2}}dt[/tex]

    [tex]
    = 1-\alpha \int_0^{\infty}
    e^{(\alpha -1)t}
    \frac{1}
    {(1+s^2- 1) + 1}dt[/tex]

    [tex]
    = 1-\alpha \int_0^{\infty}
    \frac{e^{(\alpha -1)t}}
    {1+s^2}dt
    [/tex]
     
    Last edited: Jun 21, 2011
  4. Jun 20, 2011 #3

    lanedance

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    Homework Helper

    then i would be thinking consider expanding in the function in either integral or summation for around s=0 in terms of s

    substitute [itex] s = s\lambda [/itex] and look at taking the limit or something along those lines...
     
  5. Jun 20, 2011 #4
    Thank you for the reply, but the [itex]e^t[/itex] on the denominator is multiplied to the first term only, so the simplification doesn't quite work.
     
  6. Jun 21, 2011 #5

    lanedance

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    Homework Helper

    ok so that points you in the direction of using the sum, which is looking easier anyway
    [tex] f(s)
    = 1 + \alpha \Sum_{n=1}^{infty} \frac{(-s^{-2})^n}{n+\alpha}
    = 1 + \alpha \Sum_{n=1}^{infty} \frac{(-1)^n s^{-2n}}{n+\alpha}
    [/tex]

    now when s is small you can approximate f(s) by
    [tex] f(s) \approx f(0) + \frac{df(0)}{ds}s[/tex]

    try and come up with a similar approximation for [itex]f(s\lambda) [/itex] then consider the limit of the ratio
     
    Last edited: Jun 21, 2011
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