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Thread: 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 $\displaystyle \nu >0$ for which is...

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

    Any help will be appreciated very much!...

    Kind regards

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

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

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

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

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

    $\displaystyle \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

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

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

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

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

    It is not diffcult to demonstrate that is...

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

    Some $\displaystyle a_{n} $ are …

    $\displaystyle a_{0}=1$ ,

    $\displaystyle a_{1}=0$ ,

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

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

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

    $\displaystyle \dots$ (4)

    Finally we can write...

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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