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Add example with more math formulas (#46)
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\begin[papersize=a5]{document} | ||
\use[module=packages.math] | ||
\nofolios | ||
\neverindent | ||
\define[command=section]{\medskip\noindent\font[size=12pt]{\strong{\process}}\medskip} | ||
\font[family=Libertinus Serif] | ||
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\section[numbered=false]{Laplace’s method} | ||
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Suppose \math{f(x)} is a twice continuously differentiable function on \math{[a,b]}, and there exists a unique point \math{x_0 \in (a,b)} such that: | ||
\math[mode=display]{f(x_0) = \max_{x \in [a,b]} f(x) \quad \text{and} \quad f''(x_0) < 0.} | ||
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Then: | ||
\math[mode=display, numbered=true]{\lim_{n\to\infty} \frac{\int_a^b e^{nf(x)} \, dx}{e^{nf(x_0)} \sqrt{\frac{2\pi}{n(-f''(x_0))}}}= 1.} | ||
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\section[numbered=false]{Euler Product Formula} | ||
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Let’s take \math{s \in \mathbb{C}}. | ||
The Euler Product Formula, when \math{\Re(s) > 1}, is given by: | ||
\math[mode=display, numbered=true]{\prod_{p \in \mathbb{P}} (\frac{1}{1 - p^{-s}}) | ||
= \prod_{p \in \mathbb{P}} (\sum_{k=0}^{\infty}\frac{1}{p^{ks}}) | ||
= \sum_{n=1}^{\infty} \frac{1}{n^s} | ||
= \zeta (s) | ||
= \frac{1}{\Gamma(s)} \int_0^\infty \frac{x ^ {s-1}}{e ^ x - 1} \, \mathrm{d}x} | ||
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Where: | ||
\math[mode=display]{\Gamma (s) = \int_0^\infty x^{s-1}\,e^{-x} \, \mathrm{d}x} | ||
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\section[numbered=false]{Stirling’s formula} | ||
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It is also called Stirling’s approximation for factorials: | ||
\math[mode=display, numbered=true]{\lim_{n\to +\infty} \frac{n\,!}{\sqrt{2 \pi n} \; (n/e)^{n}} = 1} | ||
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Also frequently written as: | ||
\math[mode=display]{n\,!\sim \sqrt{2\pi n}\, (\frac{n}{e})^n} | ||
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One can easily derive the following limit from Stirling’s formula: | ||
\math[mode=display]{\lim_{n\to\infty} \frac{(n!)^{1/n}}{n} = \frac{1}{e}} | ||
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\end{document} |
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