auto-update(nvim): 2025-01-23 14:21:33
This commit is contained in:
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commit
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GPG key ID:
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3 changed files with 116 additions and 342 deletions
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@ -250,7 +250,7 @@ When $F(t,y)$ (RHS) is not nice (differentiable) at the initial value $(t_0,
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y_0)$, existence and uniqueness (E/U) could fail. We do have E/U when $F(t,y)$
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y_0)$, existence and uniqueness (E/U) could fail. We do have E/U when $F(t,y)$
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is nice.
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is nice.
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== Existence of solutions to IVT - linear case
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== Existence of solutions to IVP - linear case
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The first order linear ODE is
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The first order linear ODE is
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@ -339,3 +339,52 @@ $ y' = 1 / 2 cos(y) + t, y(0) = -1 $
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The righthand side is nice, by @eutheorem, it has a unique solution. We can't
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The righthand side is nice, by @eutheorem, it has a unique solution. We can't
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find an explicit solution.
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find an explicit solution.
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= Lecture #datetime(day: 23, month: 1, year: 2025).display()
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Second order linear differential equation.
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$
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y'' = p(t)y' + q(t)y = g(t)
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$
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We say it is homogenous if $g(t) = 0$.
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== Initial value problem and uniqueness on 2nd order linear ODEs
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Suppose $p(t)$, $q(t)$, and $g(t)$ are given continuous functions defined on
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$I$.
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$
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y'' = p(t)y' + q(t)y = g(t) \
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y(t_0) = y_0, y'(t_0) = y'_0
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$
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has a unique solution $y(t)$ defined on $I$. Same as first order case. Always write in standard form first before identifying.
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#example[
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What is the largest interval on which the IVP
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$
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(t+1)y'' + y = 3, y(0) = 1, y'(0) = 0
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$
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is certain to have a solution?
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$
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y'' + 1 / (t+1) y = 3 / (t+1)
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$
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So the functions are nice except at -1. Our intervals are
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$
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(-infinity, -1), (-1, infinity)
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$
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But we have initial values at 0, so we know for certain that a solution (but not for certain it's the only solution) exists
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$
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(-1, infinity)
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$
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]
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== 2nd order linear homogenous ODEs
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@ -1,341 +0,0 @@
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#import "@preview/ctheorems:1.1.3": *
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#import "@preview/showybox:2.0.3": showybox
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#let colors = (
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rgb("#9E9E9E"),
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rgb("#F44336"),
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rgb("#E91E63"),
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rgb("#9C27B0"),
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rgb("#673AB7"),
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rgb("#3F51B5"),
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rgb("#2196F3"),
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rgb("#03A9F4"),
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rgb("#00BCD4"),
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rgb("#009688"),
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rgb("#4CAF50"),
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rgb("#8BC34A"),
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rgb("#CDDC39"),
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rgb("#FFEB3B"),
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rgb("#FFC107"),
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rgb("#FF9800"),
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rgb("#FF5722"),
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rgb("#795548"),
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rgb("#9E9E9E"),
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)
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#let dvdtyp(
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title: "",
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subtitle: "",
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author: "",
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abstract: none,
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bibliography: none,
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paper-size: "a4",
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date: "today",
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body,
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) = {
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set document(title: title, author: author)
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set std.bibliography(style: "springer-mathphys", title: [References])
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show: thmrules
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set page(
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numbering: "1",
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number-align: center,
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header: locate(loc => {
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if loc.page() == 1 {
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return
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}
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box(stroke: (bottom: 0.7pt), inset: 0.4em)[#text(
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font: "New Computer Modern",
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)[
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*#author* --- #datetime.today().display("[day] [month repr:long] [year]")
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#h(1fr)
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*#title*
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]]
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}),
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paper: paper-size,
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// The margins depend on the paper size.
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margin: (
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left: (86pt / 216mm) * 100%,
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right: (86pt / 216mm) * 100%,
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),
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)
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set heading(numbering: "1.")
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show heading: it => {
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set text(font: "Libertinus Serif")
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block[
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#if it.numbering != none {
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text(rgb("#2196F3"), weight: 500)[#sym.section]
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text(rgb("#2196F3"))[#counter(heading).display() ]
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}
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#it.body
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#v(0.5em)
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]
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}
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set text(font: "New Computer Modern", lang: "en")
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show math.equation: set text(weight: 400)
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// Title row.
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align(center)[
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#set text(font: "Libertinus Serif")
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#block(text(weight: 700, 26pt, title))
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#if subtitle != none [#text(12pt, weight: 500)[#(
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subtitle
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)]]
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#if author != none [#text(16pt)[#smallcaps(author)]]
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#v(1.2em, weak: true)
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#if date == "today" {
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datetime.today().display("[day] [month repr:long] [year]")
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} else {
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date
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}
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]
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if abstract != none [
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#v(2.2em)
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#set text(font: "Libertinus Serif")
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#pad(x: 14%, abstract)
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#v(1em)
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]
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set outline(fill: repeat[~.], indent: 1em)
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show outline: set heading(numbering: none)
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show outline: set par(first-line-indent: 0em)
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show outline.entry.where(level: 1): it => {
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text(font: "Libertinus Serif", rgb("#2196F3"))[#strong[#it]]
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}
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show outline.entry: it => {
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h(1em)
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text(font: "Libertinus Serif", rgb("#2196F3"))[#it]
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}
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// Main body.
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set par(
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justify: true,
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spacing: 0.65em,
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first-line-indent: 2em,
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)
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body
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// Display the bibliography, if any is given.
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if bibliography != none {
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show std.bibliography: set text(footnote-size)
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show std.bibliography: set block(above: 11pt)
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show std.bibliography: pad.with(x: 0.5pt)
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bibliography
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}
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}
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#let thmtitle(t, color: rgb("#000000")) = {
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return text(
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font: "Libertinus Serif",
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weight: "semibold",
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fill: color,
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)[#t]
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}
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#let thmname(t, color: rgb("#000000")) = {
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return text(font: "Libertinus Serif", fill: color)[(#t)]
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}
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#let thmtext(t, color: rgb("#000000")) = {
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let a = t.children
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if (a.at(0) == [ ]) {
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a.remove(0)
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}
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t = a.join()
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return text(font: "New Computer Modern", fill: color)[#t]
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}
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#let thmbase(
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identifier,
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head,
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..blockargs,
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supplement: auto,
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padding: (top: 0.5em, bottom: 0.5em),
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namefmt: x => [(#x)],
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titlefmt: strong,
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bodyfmt: x => x,
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separator: [. \ ],
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base: "heading",
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base_level: none,
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) = {
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if supplement == auto {
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supplement = head
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}
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let boxfmt(name, number, body, title: auto, ..blockargs_individual) = {
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if not name == none {
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name = [ #namefmt(name)]
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} else {
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name = []
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}
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if title == auto {
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title = head
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}
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if not number == none {
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title += " " + number
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}
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title = titlefmt(title)
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body = [#pad(top: 2pt, bodyfmt(body))]
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pad(
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..padding,
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showybox(
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width: 100%,
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radius: 0.3em,
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breakable: true,
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padding: (top: 0em, bottom: 0em),
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..blockargs.named(),
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..blockargs_individual.named(),
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[
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#title#name#titlefmt(separator)#body
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],
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),
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)
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}
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let auxthmenv = thmenv(
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identifier,
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base,
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base_level,
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boxfmt,
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).with(supplement: supplement)
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return auxthmenv.with(numbering: "1.1")
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}
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#let styled-thmbase = thmbase.with(
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titlefmt: thmtitle,
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namefmt: thmname,
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bodyfmt: thmtext,
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)
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#let builder-thmbox(color: rgb("#000000"), ..builderargs) = styled-thmbase.with(
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titlefmt: thmtitle.with(color: color.darken(30%)),
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bodyfmt: thmtext.with(color: color.darken(70%)),
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namefmt: thmname.with(color: color.darken(30%)),
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frame: (
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body-color: color.lighten(92%),
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border-color: color.darken(10%),
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thickness: 1.5pt,
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inset: 1.2em,
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radius: 0.3em,
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),
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..builderargs,
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)
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#let builder-thmline(
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color: rgb("#000000"),
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..builderargs,
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) = styled-thmbase.with(
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titlefmt: thmtitle.with(color: color.darken(30%)),
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bodyfmt: thmtext.with(color: color.darken(70%)),
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namefmt: thmname.with(color: color.darken(30%)),
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frame: (
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body-color: color.lighten(92%),
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border-color: color.darken(10%),
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thickness: (left: 2pt),
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inset: 1.2em,
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radius: 0em,
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),
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..builderargs,
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)
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#let problem-style = builder-thmbox(
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color: colors.at(11),
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shadow: (offset: (x: 2pt, y: 2pt), color: luma(70%)),
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)
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#let exercise = problem-style("item", "Exercise")
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#let problem = exercise
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#let theorem-style = builder-thmbox(
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color: colors.at(6),
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shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
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)
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#let example-style = builder-thmbox(
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color: colors.at(16),
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shadow: (offset: (x: 3pt, y: 3pt), color: luma(70%)),
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)
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#let theorem = theorem-style("item", "Theorem")
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#let lemma = theorem-style("item", "Lemma")
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#let corollary = theorem-style("item", "Corollary")
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#let definition-style = builder-thmline(color: colors.at(8))
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// #let definition = definition-style("definition", "Definition")
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#let proposition = definition-style("item", "Proposition")
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#let remark = definition-style("item", "Remark")
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#let observation = definition-style("item", "Observation")
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// #let example-style = builder-thmline(color: colors.at(16))
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#let example = example-style("item", "Example")
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#let proof(body, name: none) = {
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v(0.5em)
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[_Proof_]
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if name != none {
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[ #thmname[#name]]
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}
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[.]
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body
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h(1fr)
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// Add a word-joiner so that the proof square and the last word before the
|
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// 1fr spacing are kept together.
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sym.wj
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// Add a non-breaking space to ensure a minimum amount of space between the
|
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// text and the proof square.
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sym.space.nobreak
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$square.stroked$
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v(0.5em)
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}
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#let fact = thmplain(
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"item",
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"Fact",
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titlefmt: content => [*#content.*],
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namefmt: content => [_(#content)._],
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separator: [],
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inset: 0pt,
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padding: (bottom: 0.5em, top: 0.5em),
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)
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#let abuse = thmplain(
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"item",
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"Abuse of Notation",
|
|
||||||
titlefmt: content => [*#content.*],
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||||||
namefmt: content => [_(#content)._],
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||||||
separator: [],
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||||||
inset: 0pt,
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||||||
padding: (bottom: 0.5em, top: 0.5em),
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||||||
)
|
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||||||
#let definition = thmplain(
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|
||||||
"item",
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|
||||||
"Definition",
|
|
||||||
titlefmt: content => [*#content.*],
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|
||||||
namefmt: content => [_(#content)._],
|
|
||||||
separator: [],
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|
||||||
inset: 0pt,
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|
||||||
padding: (bottom: 0.5em, top: 0.5em),
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|
||||||
)
|
|
||||||
|
|
@ -774,4 +774,70 @@ us generalize to more than two colors.
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||||||
|
|
||||||
= Discussion section #datetime(day: 22, month: 1, year: 2025).display()
|
= Discussion section #datetime(day: 22, month: 1, year: 2025).display()
|
||||||
|
|
||||||
|
= Lecture #datetime(day: 23, month: 1, year: 2025).display()
|
||||||
|
|
||||||
|
== Independence
|
||||||
|
#definition("Independence")[
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||||||
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Two events $A subset Omega$ and $B subset Omega$ are independent if and only if
|
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$ P(B sect A) = P(B)P(A) $
|
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|
"Joint probability is equal to product of their marginal probabilities."
|
||||||
|
]
|
||||||
|
|
||||||
|
#fact[This definition must be used to show the independence of two events.]
|
||||||
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|
||||||
|
#fact[
|
||||||
|
If $A$ and $B$ are independent, then,
|
||||||
|
$
|
||||||
|
P(A | B) = underbrace((P(A sect B)) / P(B), "conditional probability") = (P(A) P(B)) / P(B) = P(A)
|
||||||
|
$
|
||||||
|
]
|
||||||
|
|
||||||
|
#example[
|
||||||
|
Flip a fair coin 3 times. Let the events:
|
||||||
|
|
||||||
|
- $A$ = we have exactly one tails among the first 2 flips
|
||||||
|
- $B$ = we have exactly one tails among the last 2 flips
|
||||||
|
- $D$ = we get exactly one tails among all 3 flip
|
||||||
|
|
||||||
|
Show that $A$ and $B$ are independent.
|
||||||
|
What about $B$ and $D$?
|
||||||
|
|
||||||
|
Compute all of the possible events, then we see that
|
||||||
|
|
||||||
|
$
|
||||||
|
P(A sect B) = (hash (A sect B)) / (hash Omega) = 2 / 8 = 4 / 8 dot 4 / 8 = P(A) P(B)
|
||||||
|
$
|
||||||
|
|
||||||
|
So they are independent.
|
||||||
|
|
||||||
|
Repeat the same reasoning for $B$ and $D$, we see that they are not independent.
|
||||||
|
]
|
||||||
|
|
||||||
|
#example[
|
||||||
|
Suppose we have 4 red and 7 green balls in an urn. We choose two balls with replacement. Let
|
||||||
|
|
||||||
|
- $A$ = the first ball is red
|
||||||
|
- $B$ = the second ball is greeen
|
||||||
|
|
||||||
|
Are $A$ and $B$ independent?
|
||||||
|
|
||||||
|
$
|
||||||
|
hash Omega = 11 times 11 = 121 \
|
||||||
|
hash A = 4 dot 11 = 44 \
|
||||||
|
hash B = 11 dot 7 = 77 \
|
||||||
|
hash (A sect B) = 4 dot 7 = 28
|
||||||
|
$
|
||||||
|
]
|
||||||
|
|
||||||
|
#definition[
|
||||||
|
Events $A_1, ..., A_n$ are independent (mutually independent) if for every collection $A_i_1, ..., A_i_k$, where $2 <= k <= n$ and $1 <= i_1 < i_2 < dots.c < i_k <= n$,
|
||||||
|
|
||||||
|
$
|
||||||
|
P(A_i_1 sect A_i_2 sect dots.c sect A_i_k) = P(A_i_1) P(A_i_2) dots.c P(A_i_k)
|
||||||
|
$
|
||||||
|
]
|
||||||
|
|
||||||
|
#definition[
|
||||||
|
We say that the events $A_1, ..., A_n$ are *pairwise independent* if any two
|
||||||
|
different events $A_i$ and $A_j$ are independent for any $i != j$.
|
||||||
|
]
|
||||||
|
|
|
||||||
Loading…
Reference in a new issue