Given a power law equation: $$ y= a*b^x $$ Is it possible to approximate this equation with series of exponentials similar to the following? $$ y = c*\sum_i e^{k_i * x} $$ Thank you
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$\begingroup$ For optimal approximation of power law by exponentials see the following paper: arxiv.org/pdf/physics/0605149.pdf $\endgroup$– MoustacheCommented Feb 5, 2019 at 23:47
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Yes, even a single exponential, because $a b^x$ is the same as $c e^{kx}$ with $a=c$ and $b=e^k$. But "power law" usually means a multiple of $x^{-r}$, not $b^x$. That can't be written as a sum of exponential decays, but can be written as a weighted average (= integral) of exponentials $e^{-xt}$ using the Gamma function: $$ x^{-r} = \frac1{\Gamma(r)} \int_0^\infty t^{r-1} e^{-xt} dt. $$
answered Oct 8, 2016 at 4:50
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$\begingroup$ Dear Noam, could you please take a look at mathoverflow.net/questions/252071/… ? I think my answer is correct, but I also think you would know for sure. $\endgroup$ Commented Oct 15, 2016 at 15:56
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$\begingroup$ Um, surely this request doesn't belong on this completely unrelated query... (and you have my e-mail address.) To your question, that seems right (though I didn't check the numbers), though probably unnecessary because splitting into several spinor genera requires discriminants of high valuation, and the Leech/Niemeier genus has dicsriminant whose valuations are as smaller as possible! But I've not studied this aspect of the theory carefully enough to be certain of this argument. $\endgroup$ Commented Oct 15, 2016 at 21:01
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1$\begingroup$ @NoamD.Elkies could you add a derivation? $\endgroup$ Commented Feb 26, 2021 at 19:11
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1$\begingroup$ This integral is well known and should be in most integral tables. The change of variable $xt=u$ converts the integral to $x^{-r} \int_0^\infty u^{r-1} e^{-u} \, du$, and $\int_0^\infty u^{r-1} e^{-u} \, du$ is the definition of $\Gamma(r)$. $\endgroup$ Commented Feb 26, 2021 at 20:22
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1$\begingroup$ @RylanSchaeffer If you restrict $t$ to be $\in [0,1]$, your equation gives $\operatorname{Beta}(r,1)$ distribution $\endgroup$ Commented Jun 19, 2023 at 2:40