About a 'function of function'...

chisigma · #1 $Old$ July 2nd, 2009, 05:13 AM

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

$\chi$ $\sigma$

chisigma · #2 $Old$ July 3rd, 2009, 11:47 PM

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

$\chi$ $\sigma$

chisigma · #3 $Old$ July 9th, 2009, 07:01 AM

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

$\chi$ $\sigma$