1
$\begingroup$

I have been given the following functions (where $\omega_0$ is some positive constant)

$$\begin{align} \Gamma(t) &= \frac{1}{t^2} \left(\; \frac{1}{2} \omega_0^2 t^2 - \cos(\omega_0 t) - \omega_0 t \sin(\omega_0 t) + 1 \;\right) \\[4pt] \gamma(t) = \frac{d \Gamma(t)}{dt} &= \frac{1}{t^3} \left(\; 2 \omega_0 t \sin(\omega_0 t) - (\omega_0^2 t^2 - 2) \cos(\omega_0 t) - 2 \;\right) \end{align}$$

and need to find the intervals $(a_n, b_n)$ such that $\gamma(t)<0$ for $ t \in (a_n,b_n)$.

I then need to calculate

$$N(\Phi)= \int_{\gamma(t)<0} - \gamma(t) e^{-\Gamma(t)}dt =\sum_n e^{-\Gamma(b_n)}-e^{-\Gamma(a_n)}$$

I have been working on this and can't get it to work. The question really only is if $N(\Phi)=\infty$ or if it is finite. Both are possible. I tried working on the roots but don't get anywhere for an explicit interval. I know that

$$\lim_{t \to \infty} \Gamma(t) = \frac12\omega_0^2$$

I am not sure though what this tells me about my sum. Clearly individually the sums are divergent. But if I take the difference I get that

$$\lim_{n \rightarrow \infty} e^{-\Gamma(b_n)} - e^{-\Gamma(a_n)}=0$$

and thus I have the possibility of convergence.

If it is not possible to calculate the roots explicitly but know that $N(\Phi)=\infty$ that is sufficient. Otherwise I guess I will have to use numerical methods.

I would really appreciate any kind of help or hint. Thank you.

Edit: What I gather so far from Maxims answer is this. $\gamma(t)$ behaves like $\frac{-\omega_0^2 \cos(\omega_0 t)}{t}$. This I assumes is the case as $\sin(t)/t^2$ and $2/t^3$ are "faster" decreasing then my other term. So finding the roots of $\frac{-\omega_0^2 \cos(\omega_0 t)}{t}$ gives me $$ r_k=\frac{\pi (2k +1)}{2 \omega_0} + O(k^{-3})$$ Shouldn't there be $O(k^{-3})$ as I have the term $2/t^3$? If not, why not?

The calulations for $\Gamma(r_k)$ are clear for the roots. But then calculating $$ exp(-\Gamma(r_{2k}) - exp(-\Gamma(r_{2k-1}) = e^{\frac{-\omega_0^2}{2}}(e^{\frac{\omega_0^2}{2 \pi k}}-e^{-\frac{\omega_0^2}{2 \pi k}}) + O(k^{-2}) $$ Thus leaving me with the following sum

$$e^{\frac{-\omega_0^2}{2}} \sum_k (e^{\frac{\omega_0^2}{2 \pi k}}-e^{-\frac{\omega_0^2}{2 \pi k}}) + O(k^{-2}) $$

But then $\sum_k O(k^{-2}) \rightarrow \frac{\pi^2}{6}$.and what happens to the other sum? Thank you for your time and effort.

$\endgroup$
  • 1
    $\begingroup$ What is $\gamma(t)$? Is it $\gamma_3(t)$? Likewise, are $\Gamma(t)$ and $\Gamma_3(t)$ the same? $\endgroup$ – Blue May 15 '18 at 19:57
  • $\begingroup$ Looks like the intervals are of approximately fixed length and regularly spaced, the derivative is of size $1/t$ and $\Gamma(t)$ is nearly constant at $+\infty$, so if you are talking about $(0,+\infty)$, the sum diverges like a harmonic series, does not it? $\endgroup$ – fedja May 15 '18 at 23:22
  • 1
    $\begingroup$ @ Blue, yes they are the same. I will edit it. $\endgroup$ – Maike Zinterl May 16 '18 at 4:56
  • $\begingroup$ You appear to have a typo in the definition of $\gamma(t)$. The coefficient of the $\sin(\omega_0 t)$ term should be $2\omega_0 t$, not $2\omega_0 t^2$. $\endgroup$ – Blue May 16 '18 at 5:18
  • $\begingroup$ @ Blue Yes, I corrected the typo. Thank you $\endgroup$ – Maike Zinterl May 16 '18 at 13:49
1
$\begingroup$

The general idea goes like this. $\gamma(t)$ behaves like $-\omega_0^2 \cos(\omega_0 t)/t$, and the position of the $k$th root of $\gamma(t)$ is $$r_k = \frac {\pi (2 k + 1)} {2 \omega_0} + O(k^{-1}).$$ (The next order term is $-2 / (\pi \omega_0 k)$, but we only need the $O$ estimate.) Then $$\Gamma(r_{2 k}) = \frac {\omega_0^2} 2 - \frac {\omega_0^2} {2 \pi k} + O(k^{-2}), \\ \Gamma(r_{2 k - 1}) = \frac {\omega_0^2} 2 + \frac {\omega_0^2} {2 \pi k} + O(k^{-2}), \\ \exp(-\Gamma(r_{2 k})) - \exp(-\Gamma(r_{2 k - 1})) = \frac {\omega_0^2 e^{-\omega_0^2 / 2}} {\pi k} + O(k^{-2})$$ (the remainder is in fact $O(k^{-3})$, but that isn't important either), and the sum over the first $n$ negative intervals grows as $\ln n$ times the coefficient at $k^{-1}$.

$\endgroup$
  • $\begingroup$ Thank you. I will look at it tomorrow. My day is over (in germany) $\endgroup$ – Maike Zinterl May 16 '18 at 14:26
  • $\begingroup$ Thank you. I am not very familiar with the big O notation. Would you mind explaining a bit more where the $O(k^{-1}$ is coming from in the $k$th root? Also, how does it work for $\Gamma(r_k)$? I guess this is simply a question of big O notation. Then the last step, I would have $= e^{-\omega_0^2/2}(e^{\frac{\omega_0^2}{2 \pi k}}-e^{-\frac{\omega_0^2}{2 \pi k}})$ I feel you are missing an exponential there. Thank you for your help $\endgroup$ – Maike Zinterl May 17 '18 at 9:38
  • 1
    $\begingroup$ Suppose we have $\cos t_0 + \phi(t_0)/t_0 = 0$, and, for large $t$ and for some constant $M$, we have $|\phi(t)| < M$. Then $|\cos t_0| < M/t_0$. From this we know how far $t_0$ can be from zeros of cosine. Next step is expanding $\exp(-\Gamma(t))$ around that point. The $O$ estimates are for $k \to \infty$; expand $\exp(\omega_0^2 / (2 \pi k))$ around $k = \infty$, and you'll get the same result. $\endgroup$ – Maxim May 17 '18 at 11:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.