Difference between revisions of "Sech"
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− | \mathrm{sech}(z)=\frac{ |
+ | \mathrm{sech}(z)=\frac{1}{\cosh(z)}=\frac{2}{\exp(z)+\exp(-z)} =\frac{2}{e^z+e^{-z}} \) |
[[Sech]] is a [[bell function]]. It is used to describe the smooth distribution of some continuous quantities. |
[[Sech]] is a [[bell function]]. It is used to describe the smooth distribution of some continuous quantities. |
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where $G$ is Catalan constant <ref>https://en.wikipedia.org/wiki/Catalan%27s_constant |
where $G$ is Catalan constant <ref>https://en.wikipedia.org/wiki/Catalan%27s_constant |
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\(G = 0.915965594177219015054603514932384110774\)</ref>, |
\(G = 0.915965594177219015054603514932384110774\)</ref>, |
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− | \(\displaystyle \int_0^\infty \frac{x^2 \mathrm d x}{\cosh(x)}=\frac{\pi^3}{8} \) |
+ | \(\displaystyle \int_0^\infty \frac{x^2 \mathrm d x}{\cosh(x)}=\frac{\pi^3}{8} ~\) ; |
+ | \(~ \displaystyle \int_{-\infty}^\infty \frac{x^2 \mathrm d x}{\cosh(x)}=\frac{\pi^3}{4} ~\) |
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\(\displaystyle \int_0^\infty \frac{x^3 \mathrm d x}{\cosh(x)}=\frac{\pi^4}{8} \) |
\(\displaystyle \int_0^\infty \frac{x^3 \mathrm d x}{\cosh(x)}=\frac{\pi^4}{8} \) |
Latest revision as of 10:11, 4 April 2023
Sech is short name for elementary function,
\(\displaystyle \mathrm{sech}(z)=\frac{1}{\cosh(z)}=\frac{2}{\exp(z)+\exp(-z)} =\frac{2}{e^z+e^{-z}} \)
Sech is a bell function. It is used to describe the smooth distribution of some continuous quantities.
Sech appears in models of generation of light pulses in lasers, it describes the exponential growth of the seed and then, since the inversion of population os depleted, the exponential decay of the output signal.
Some formulas for sech are collected below.
Integration
\(\displaystyle {\rm sech}(z)=\frac{1}{\cosh(z)}=\frac{2}{e^x+e^{-x}}\)
In math texts, it is common eseption for constant \(e=\exp(1)\) : often, it appears with Italics font, while common rule for constants is to write them in Roman font.
\(\displaystyle \int_{-\infty}^{\infty} \frac{\mathrm d x}{e^x+e^{-x}} = \int_{-\infty}^{\infty} \frac{e^x\ \mathrm d x}{e^{2x}+1} = \int_{0}^{\infty} \frac{\mathrm d z}{z^2+1} = \arctan(z) \Big|_{z=0}^{z=\infty} = \frac{\pi}{2} \) ; \(\displaystyle \int_0^\infty \frac{ \mathrm d x}{\cosh(x)}=\pi \)
\(\displaystyle \int_0^\infty \frac{x\ \mathrm d x}{e^x+e^{-x}}=2G=\pi\ln(2)-4L(\pi/4) \approx 1.831931188 \)
where $G$ is Catalan constant [1],
\(G \approx 0.915965594177219015054603514932384110774\):
\( G = \beta(2) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2} = \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \frac{1}{9^2} - \cdots \)
\(\displaystyle \int_0^\infty \frac{x^2 \mathrm d x}{\cosh(x)}=\frac{\pi^3}{8} ~\) ;
\(~ \displaystyle \int_{-\infty}^\infty \frac{x^2 \mathrm d x}{\cosh(x)}=\frac{\pi^3}{4} ~\)
\(\displaystyle \int_0^\infty \frac{x^3 \mathrm d x}{\cosh(x)}=\frac{\pi^4}{8} \)
\(\displaystyle \int_0^\infty \frac{x^4 \mathrm d x}{\cosh(x)}=\frac{5}{32} \pi^5 \)
\(\displaystyle \int_0^\infty \frac{x^5 \mathrm d x}{\cosh(x)}=\frac{\pi^6}{4} \)
\(\displaystyle \int_0^\infty \frac{x^6 \mathrm d x}{\cosh(x)}=\frac{61}{128} \pi^7 \)
\(\displaystyle \int_0^\infty \frac{x^7 \mathrm d x}{\cosh(x)}=\frac{17}{16} \pi^8 \)
Warning
Some misprints may appear in TORI. Please test any content before to use it.
References
https://en.wikipedia.org/wiki/Hyperbolic_secant_distribution
http://fisica.ciens.ucv.ve/~svincenz/TISPISGIMR.pdf Gradstyein, Ryzhik, Hyperbolic Functions, page 422, 3.521 and page 425, 3.523
Keywords
«Elementary function», «Cosh», «[[]]»,
- ↑ https://en.wikipedia.org/wiki/Catalan%27s_constant \(G = 0.915965594177219015054603514932384110774\)