I'm trying to understand this PDF, which is in English but has Polish authors, so there may be translation issues confusing me. The subject is the frequencies of sentence lengths in novels.

Eq. (1), $S(f)=\left|\sum_{j=1}^{j_\max}l(j)e^{-2\pi ifj}\right|^2$, claims to present a power spectrum "of the series $l(j)$ representing lengths of the consecutive sentences $j$". It is then claimed empirical $S$-against-$f$ plots exhibit Eq. (7), $S(f)=1/f^\beta$.

I expect that might be meant to be a proportionality relation, but let's put that aside for the moment. To my mind, these equations are incompatible, because (1) implies $S(f+1)=S(f)$, with a finite peak in $S$ at integer $f$, including $f=0$, by the triangle inequality in $\Bbb C$. By contrast, a $1/f^\beta$ law is monotonic decreasing on $f\ge0$, with a divergence at $f=0$.

Figure 1 suggests values of $f$ from about $10^{-3}$ to $10^{-1}$ are considered, which on log-log axes allows one to make a power-law approximation, but I don't see the merit in that. But supposing this is a standard procedure, how would I estimate $\beta$ for $l$ from my own time series? Is there, for example, a computationally cheap library function to estimate $\beta$?

  • $\begingroup$ Would you be able to expand on why you think $S(f+1) = S(f)$? $\endgroup$ – Stephen Jackson Sep 23 '19 at 13:46
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    $\begingroup$ @StephenJackson Because $e^{-2\pi i(f+1)j}=e^{-2\pi ifj}e^{-2\pi ij}$, the last factor being $1$ because $j\in\Bbb Z$. $\endgroup$ – J.G. Sep 23 '19 at 13:51
  • $\begingroup$ @StephenJackson Wait: is the value of $j_\max$ dependent on $f$? $\endgroup$ – J.G. Sep 23 '19 at 13:57
  • $\begingroup$ No I don't think so, I think $j_{\text{max}}$ is just the number of sentences, and I also cant fault your logic here, usually I would expect that (1) would be written as $$S(f) = |\sum_{j=1}^{j_{\text{max}}}l(j)e^{-2\pi i f \frac{j}{j_{\text{max}}}}|^2$$ so usually $S(f+1) \neq S(f)$ but this doesn't seem to be the case here $\endgroup$ – Stephen Jackson Sep 23 '19 at 14:21
  • $\begingroup$ @StephenJackson In which case $S(f+j_\max)=S(f)$. $\endgroup$ – J.G. Sep 23 '19 at 14:54

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