1. ## 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

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

2. 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

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

3. 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

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