- Questo topic ha 2 risposte, 3 partecipanti ed è stato aggiornato l'ultima volta 10 anni, 3 mesi fa da .
-
Topic
-
Di seguito un elenco di formule matematiche da inserire nella tesi di un mio amico.
Possibile ottimizzarle? (magari non tutte, ma quelle che lo richiedono?) O vanno bene così?`
\documentclass[11pt,a4paper,oneside]{book}\usepackage[T1]{fontenc} % lettere accentate
\usepackage[utf8]{inputenc} % codifica font
\usepackage[italian]{babel} % per scrivere in italiano\usepackage{amsmath} % simboli matematici
\usepackage{amssymb}
\usepackage{amsfonts}\usepackage{layaureo} % imposta i margini di pagina
\begin{document}
\chapter{Formule}
\[
f_z(z) = 2\phi(z) \Phi(\alpha z), \qquad z \in \mathbb{R}
\]\[
F_Z(z) = \int_{-\infty}^{z} 2\phi(z) \Phi(\alpha z) dx \qquad \ and \ \qquad 1 – F_Z(z) = \int_{z}^{\infty} 2\phi(z) \Phi(\alpha z) dx
\]\[
\frac{\phi(z)}{z} – \frac{\phi(z)}{z^3} < 1 - \Phi (z) < \frac{\phi(z)}{z} \] \[ 1 - F_Z(z) = \int_{z}^{\infty} 2\phi(z) \Phi(\alpha z) dx \leq 2 \Phi (\alpha z) \int_{z}^{\infty} \phi (x) dx = 2 \Phi (\alpha z) {1 - \Phi (z)} \] \[ \Phi (\alpha z) < \frac{\phi (\alpha z)}{\vert \alpha \vert z} \] \[ 1 - F_Z(z) < \sqrt{\frac{2}{\pi}} \frac{\phi (z \sqrt{1 + \alpha^2}) }{\vert \alpha \vert z^2} \] \[ 1 - F_Z(z) < \frac{\phi (z)}{z} \] \[ 1 - F_Z(z) = \int_{z}^{\infty} 2\phi(x) \Phi(\alpha x) dx \leq \int_{z}^{\infty} 2\phi(x) dx < 2 \frac{\phi (z)}{z} \] \newpage if $ \alpha < 0 $ \[ \sqrt{\frac{2}{\pi}} \frac{ \phi (z \sqrt{1 + \alpha^2}}{\vert \alpha \vert (1 + \alpha^2) z^2} - \sqrt{\frac{2}{\pi}} (2 + \frac{1 + \alpha^2}{\alpha^2}) \frac{ \phi (z \sqrt{1 + \alpha^2}}{\vert \alpha \vert (1 + \alpha^2)^2 z^4} < 1 - F_Z(z) < \sqrt{\frac{2}{\pi}} \frac{ \phi (z \sqrt{1 + \alpha^2}}{\vert \alpha \vert (1 + \alpha^2) z^2} \] if $ \alpha > 0 $
\[
2 \frac{\phi (z)}{z} – 2 \frac{\phi (z)}{z^3} – \sqrt{\frac{2}{\pi}} \frac{
\phi (z \sqrt{1 + \alpha^2}}{\vert \alpha \vert (1 + \alpha^2) z^2} < 1 - F_Z(z) < 2 \frac{\phi (z)}{z} - \sqrt{\frac{2}{\pi}} \frac{ \phi (z \sqrt{1 + \alpha^2}}{\vert \alpha \vert (1 + \alpha^2) z^2} + \sqrt{\frac{2}{\pi}} (2 + \frac{1 + \alpha^2}{\alpha^2}) \frac{ \phi (z \sqrt{1 + \alpha^2}}{\vert \alpha \vert (1 + \alpha^2)^2 z^4} \] \[ \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert x} dx - \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert^3 x^3} dx < 1 - F_Z(z) < \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert x} dx \] \[ \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert x} dx = \sqrt{\frac{2}{\pi}} \frac{1}{\vert \alpha \vert} \int_{z}^{\infty} \frac{1}{x^2} x \phi (x \sqrt{1 + \alpha^2)} dx \] \[ \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert x} dx = \sqrt{\frac{2}{\pi}} \frac{\phi (z \sqrt{1 + \alpha^2})}{\vert \alpha \vert (1 + \alpha^2)} \int_{z}^{\infty} 2 \frac{\phi (x \sqrt{1 + \alpha^2})}{x^3} dx \] \[ \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{x^3} dx = \sqrt{\frac{2}{\pi}} \frac{\phi (z \sqrt{1 + \alpha^2}))}{(1 + \alpha^2) z^4} - \sqrt{\frac{2}{\pi}} \frac{4}{(1 + \alpha^2)} \int_{z}^{\infty} \frac{\phi (x^5} dx \] \[ \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert x} dx > \sqrt{\frac{2}{\pi}} \frac{\phi (z \sqrt{1 + \alpha^2})}{\vert \alpha \vert (1 + \alpha^2) z^2} – 2 \sqrt{\frac{2}{\pi}} \frac{\phi (z \sqrt{1 + \alpha^2})}{\vert \alpha \vert (1 + \alpha^2)^2 z^4} = A
\]\[
\int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert x} dx < \sqrt{\frac{2}{\pi}} \frac{\phi (z \sqrt{1 + \alpha^2})}{\vert \alpha \vert (1 + \alpha^2) z^2} = B \] \[ \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert^3 x^3} dx < \sqrt{\frac{2}{\pi}} \frac{\phi (z \sqrt{1 + \alpha^2})}{\vert \alpha \vert^3 (1 + \alpha^2) z^4} = C \] \[ \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert x} dx - \int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert ^3 x^3} dx > A – C \qquad
\]\[
\int_{z}^{\infty} 2\phi (x) \frac{\phi (\alpha x)}{\vert \alpha \vert x} dx < B \] \[ 1 - F_Z(z) = 2 \int_{z}^{\infty} \phi (x) dx - \int_{z}^{\infty} 2 \phi (x) \Phi (- \alpha x) dx \] \[ 2 \frac{\phi (z)}{z} - 2 \frac{\phi (z)}{z^3} - \int_{z}^{\infty} 2\phi (x) \Phi (- \alpha x) dx< 1 - F_Z(z) < 2 \frac{\phi (z)}{z} - \int_{z}^{\infty} 2\phi (x) \Phi (- \alpha x) dx< 1 \] \[ \lim_{z \to +\infty} (1 - F_Z(z)) \sqrt{\frac{2}{\pi}} \frac{\vert \alpha \vert (1 + \alpha^2) z^2}{\phi (z \sqrt{1 + \alpha^2})} = 1 \qquad \ and \ \qquad \lim_{z \to +\infty} (1 - F_Z(z)) \frac{z}{\phi (z)} = 0 \] \[ \lim_{z \to +\infty} \frac{1}{2} (1 - F_Z(z)) \frac{z}{\phi (z)} = 1 \] \[ log(\sqrt{\frac{2}{\pi}}) - \frac{1}{2} \alpha^2 z^2 - log (1 + \alpha^2) - log (\vert \alpha \vert z) < 0 \] \[ log(\sqrt{\frac{2}{\pi}}) - \frac{1}{2} \alpha^2 z^2 - log (1 + \alpha^2) - log (\vert \alpha \vert z) ? 0 \] \[ z = 0.7134{exp(-\alpha) - 1}^{0.979} \] \[ log \begin{Bmatrix} \frac{P_i^u (z; \alpha) - \ + F_Z(z)}{1 - F_Z(z)} \end{Bmatrix} \qquad \ for \ \qquad i = 1, 2 \] \[ 1 - F_Z(z) < 2 \frac{\phi (z)}{z} - g (z; \alpha) \qquad \ where \ \qquad \] \[ g(z; \alpha) = \sqrt{\frac{2}{\pi}} \frac{1}{\vert \alpha \vert} \frac{ \phi (z \sqrt{1 + \alpha^2})}{(1 + \alpha^2) z^2} - \sqrt{\frac{2}{\pi}} \frac{1}{\vert \alpha \vert} (2 + \frac{1 + \alpha^2}{\alpha^2}) \frac{\phi (z \sqrt{1 + \alpha^2})}{(1 + \alpha^2)^2 z^4} \] \end{document} `
- Devi essere connesso per rispondere a questo topic.