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Math Help - About a 'function of function'...

  1. #1
    MHF Contributor chisigma's Avatar
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    About a 'function of function'...

    Does exist a real number \nu >0 for which is...

    \cos (\omega * \ln t) = t^{-\nu} \sum_{n=0}^{\infty} a_{n}\cdot t^{n}

    Any help will be appreciated very much!...

    Kind regards

    \chi \sigma
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  2. #2
    MHF Contributor chisigma's Avatar
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    We can start remembering the well known identity...

    t^{\omega} = e^{\omega \cdot \ln t} = \sum_{n=0}^{\infty} \frac{(\omega \cdot \ln t)^{n}}{n!} (1)

    ... where both t and \omega are real or even complex variables. Also well known is the Taylor expansion of the fuction  \cos (*) that permits us to write...

    \cos (\omega \cdot \ln t) = \sum_{n=0}^{\infty} (-1)^{n} \frac{(\omega \cdot \ln t)^{2n}}{(2n)!} (2)

    Combining (1) and (2) we arrive to the 'simple' identity...

    \cos (\omega\cdot \ln t) = \frac{t^{i\cdot \omega} + t^{-i\cdot \omega}}{2} (3)

    An easy question to You before proceeding: is it all right?...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chisigma's Avatar
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    The identity...

    \cos (\omega\cdot \ln t)= \frac{t^{i \omega} + t^{-i\omega}}{2} (1)

    ... we have established il last post permits us to expand the (1) in Taylor series around t=1. To perform this we have to know the derivatives of (1) at t=1 and to do that first we introduce a sequence of complex polynomials in the variable \omega defined as follows...

    p_{n} (\omega) = p_{n-1} (\omega) (i \omega-n+1) ,  p_{0} (\omega) = 1 (2)

    It is not diffcult to demonstrate that is...

    \frac {d^{n}}{dt^{n}}_{t=1} \cos (\omega\cdot \ln t)= Re \{p_{n}(\omega)\} = a_{n} (\omega) (3)

    Some a_{n} are …

    a_{0}=1 ,

    a_{1}=0 ,

    a_{2}= -\omega^{2} ,

    a_{3}= 3 \cdot \omega^{2} ,

    a_{4}= \omega^{4} - 11\cdot \omega^{2} ,

     \dots (4)

    Finally we can write...

     \cos(\omega\cdot \ln t)= \sum_{n=0}^{\infty} a_{n} (\omega) \frac {(t-1)^{n}}{n!} (5)

    ... that confirms the hypothesis made at the beginning that is satisfied by  \nu=0 ...

    Kind regards

    \chi \sigma
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