Math Help Forum - View Single Post

chisigma · #3 July 9th, 2009, 08:01 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

$\chi$ $\sigma$