From 290e2e626bd69b36dd7f383395c75ccd3757398e Mon Sep 17 00:00:00 2001 From: Youwen Wu Date: Wed, 5 Mar 2025 02:11:45 -0800 Subject: [PATCH] auto-update(nvim): 2025-03-05 02:11:45 --- .../by-course/math-4b/course-notes/main.typ | 59 +++++++++ .../by-course/math-8/course-notes/main.typ | 115 ++++++++++++++++ documents/by-course/math-8/pset-8/main.typ | 125 ++++++++++++++++++ documents/by-course/math-8/pset-8/package.nix | 37 ++++++ 4 files changed, 336 insertions(+) create mode 100644 documents/by-course/math-8/pset-8/main.typ create mode 100644 documents/by-course/math-8/pset-8/package.nix diff --git a/documents/by-course/math-4b/course-notes/main.typ b/documents/by-course/math-4b/course-notes/main.typ index 00bfa4a..7a14db4 100644 --- a/documents/by-course/math-4b/course-notes/main.typ +++ b/documents/by-course/math-4b/course-notes/main.typ @@ -1146,6 +1146,65 @@ $ arrow(x)_"Im" (t) = e^(lambda t) (sin(mu t) arrow(a) + cos(mu t) arrow(b)) \ $ +== Visualizing complex eigenvalue solutions + +Consider the system +$ + arrow(x)' = mat(2,-1;1,2) arrow(x) +$ + +The phase portrait is a spiral going out as $t -> infinity$, and facing +counterclockwise. + +How can we determine the direction of motion? Test it at a simple point, say, +$(1,0)$. + +$ + arrow(x)' = mat(2,-1;1,2) vec(1,0) = vec(2,1) +$ + +This vector points up, so the spiral is counterclockwise. In general we can +determine it by checking the first column vector of $A$. + +== Second order linear DE as linear system + +Consider a damped, forced harmonic motion: +$ + x'' + 3x' + 2x = cos(t) +$ + +Let $y(t) = x'(t)$. Then: +$ + x'(t) &= y \ + y'(t) &= -2x - 3y + cos(t) +$ + +The system is written in matrix form as +$ + arrow(x)'(t) = mat(0,1;-2,-3) arrow(x) + vec(0,cos(t)) +$ +This is a system of first order nonhomogenous differential equations. + +Consider the associated homogenous equations +$ + x'' + 3x' + 2x = 0 +$ +corresponding to a system +$ + arrow(x)(t) = mat(0,1;-2,-3) arrow(x) +$ +A fundamental set is given by +$ + r_1 &= -1 <-> vec(1,-1) \ + r_2 &= -2 <-> vec(1,-2) +$ +Any vector valued function of the following form is a solution +$ + arrow(x)(t) = c_1 e^(-t) vec(1,-1) + c_2 e^(-2t) vec(1,-2) +$ +Solutions tend to 0 as $t -> infinity$. The equilibrium $x=0$ is asymptotically +stable. Hence it is called an *asymptotically stable node*. + = Repeated eigenvalues, nonhomogenous systems == Classification of equilibria $n=2$ diff --git a/documents/by-course/math-8/course-notes/main.typ b/documents/by-course/math-8/course-notes/main.typ index 0233bad..a3f1dc1 100644 --- a/documents/by-course/math-8/course-notes/main.typ +++ b/documents/by-course/math-8/course-notes/main.typ @@ -1197,3 +1197,118 @@ $ z = x + (b c - a d) / (b d + 1) $ Then $(x + y)/2$ is obviously rational and strictly between $x$ and $y$. ] += Constructions of functions + +#definition[ + For functions $f : A -> B$ and $g : B -> C$, the *inverse* of $f$ is the + relation from $B -> A$: + $ + f^(-1) = {(x,y) | (y,x) in f} + $ + and the *composite* of $f$ and $g$ is the relation from $A$ to $C$ + $ + g compose f = {(x,z) | (exists y in B)((x,y) in f and (y,z) in g)} + $ +] + +These are _relations_, nothing in their definitions guarantee that either is a +function. + +#theorem[ + Let $A$, $B$, and $C$ be sets, let $f : A -> B$ and $g : B -> C$. Then $g + compose f$ is a function from $A$ to $C$, and $"Dom"(g compose f) = A$. +] + +#proof[ + First let us show that $g compose f$ is a relation from $A$ to $C$. + + Suppose $(a,b) in f$ such that $a in A$ and $b in B$. Then there exists a + pair $(b,c) in g$ with some $c in C$. Then by the definition of function + composition, we can form a pair $(a,c)$ from these, still with $a in A$ and + $c in C$. By definition, $g compose f$ is the set of all these $(a,c)$ we + form and is a relation from $A$ to $C$. Hence $g compose f$ is a relation + from $A$ to $C$. + + Now we show that $g compose f$ is indeed a function. The key idea is to show + that any $(x,y)$ and $(x,z)$ in $g compose f$ must have $y = z$, hence + mapping each input to exactly one output. + + Suppose that $(x,y) in g + compose f$ and $(x,z) in g compose f$. Because $(x,y) in g compose f$, there + exists $u in B$ such that $(x,u) in f$ and $(u,y) in g$. Likewise, there + exists $v in B$ such that $(x,v) in f$ and $(v,y) in g$. Because $f$ is a + function and $(x,u)$ and $(x,v)$ are in $f$, we have $u = v$. Because $g$ is + a function, $(u,y)$ and $(v,z)$ are in $g$, and $u = v$, we have $y = z$. + + Now we show the domain of $g compose f$. Suppose $(x,y) in g compose f$. Then + $(x,u) in f$ for some $u in B$ such that $(u, y) in g$. Hence the domain of + $g compose f$ is the same as the domain of $f$. So $"Dom"(g compose f)$ is + $A$. +] + +#theorem[ + Let $A$, $B$, $C$, and $D$ be sets. Let $f : A -> B$, $g : B -> C$, and $h : + C -> D$. Then $(h compose g) compose f = h compose (g compose f)$. +] + +#proof[ + By @compose-domain, the domain of both functions are $A$. Let $x in A$. Then + $((h compose g) compose f)(x) = (h compose g)(f(x)) = h(g(f(x))) = h((g + compose f)(x)) = (h compose (g compose f))(x)$. The functions are equal. +] + +#theorem[ + Let $f : A -> B$. Then $f compose I_A = f$ and $I_B compose f = f$. +] + +#proof[ + $"Dom"(f compose I_A) = "Dom"(I_A) = A = "Dom"(f)$. Suppose $x in A$, then + $(f compose I_A)(x) = f(I_A (x)) = f(x)$. Therefore $f compose I_A = f$. + + $"Dom"(I_B compose f) = "Dom"(f) = A$. Now suppose $x in A$, then $(I_B + compose f)(x) = I_B (f(x)) = f(x)$. Therefore $I_B compose f = f$. +] + +#theorem[ + Let $f : A -> B$ with $"Rng"(f) = C$. If $f^(-1)$ is a function, then $f^(-1) compose f = I_A$ and $f compose f^(-1) = I_C$. +] + +#proof[ + Suppose that $f : A -> B$ and $f^(-1)$ is a function. Then $"Dom"(f^(-1) compose f) = "Dom"(f)$. Thus $"Dom"(f^(-1) compose f) = A = "Dom"(I_A)$. Additionally, $"Dom"(f compose f^(-1)) = "Dom"(f^(-1))$. By a known theorem, $"Dom"(f^(-1)) = "Rng"(f) = C$. + + Suppose that $x in A$. Because $(x,f(x)) in f$, we have $(f(x), x) in f^(-1)$. Therefore $(f^(-1) compose f)(x) = f^(-1) (f(x)) = x = I_A (x)$. Therefore $f^(-1) compose f = I_A$. + + Suppose that $y in C$. Because $(y,f^(-1) (y)) in f^(-1)$, $(f^(-1) (y),y) in f$. Then $f compose f^(-1) = f(f^(-1) (y)) = y = I_C$. +] + +#theorem[ + Let $h$ and $g$ be functions with $"Dom"(h) = A$ and $"Dom"(g) = B$. If $A sect B = emptyset$, then $h union g$ is a function with domain $A union B$. Furthermore, + $ + (h union g)(x) = cases(h(x) &"if" x in A, g(x) &"if" x in B) + $ +] + +#proof[ + Suppose $(x,y) in (h union g)$. Then either $x in A$ or $x in B$. So $"Dom"(h + union g) = A union B$. Suppose that $(x,u)$ and $(x,v)$ are in $h union g$. + Then either $(x,u) in h$ or $(x,u) in g$, but not both, since $"Dom"(h)$ is + disjoint with $"Dom"(g)$. The same goes for $(x,v)$. Then if both $(x,u)$ and + $(x,v)$ are in $f$, because it is a function, $u = v$. Repeat the same + reasoning if they are in $g$. Therefore $h union g$ is indeed a function with + domain $A union B$. + + Now we show the second part. Suppose $(x,y) in h union g$. Then either $x in + A$ or $x in B$, but not both. If $x in A$, then $(x,y) in h$. Hence $(h union + g)(x) = h(x)$. If $x in B$, then $(x,y) in g$. Hence $(h union g)(x) = g(x)$. +] + +#definition[ + Let $f : A -> B$, and $D subset.eq A$. The *restriction* of $f$ to $D$ is the function + $ + lr(f|)_D = {(x,y) | y = f(x) "and" x in D} + $ +] + +#definition[ + Let $f$ be a real function defined on an interval $I$. Then we say $f$ is *increasing on $I$* if $x < y => f(x) < f(y)$ for all $x,y in I$. Likewise, $f$ is *decreasing on $I$* if $x < y => f(x) > f(y)$ for all $x,y in I$. +] diff --git a/documents/by-course/math-8/pset-8/main.typ b/documents/by-course/math-8/pset-8/main.typ new file mode 100644 index 0000000..c7bb6bf --- /dev/null +++ b/documents/by-course/math-8/pset-8/main.typ @@ -0,0 +1,125 @@ +#import "@youwen/zen:0.1.0": * +#import "@preview/cetz:0.3.2" + +#show: zen.with( + title: "Homework 8", + author: "Youwen Wu", +) + +#set heading(numbering: none) +#show heading.where(level: 2): it => [#it.body.] +#show heading.where(level: 3): it => [#it.body.] +#set par(first-line-indent: 0pt) +#show figure: it => { + pad( + y: 10pt, + it, + ) +} + +Problems: + +- 4.2: \#1g, 2b, 3b, 6, 7, 8, 14bd + +- 4.3: \#1bdeh, 2bdeh, 3ab, 5, 6, 7, 8 + +#outline() + += 4.2 + +== 1 + +=== g + +$f(x) = 1 / (1-x)$ + +$f^(-1) (x) = 1 - 1 / x$ + +== 2 + +=== b + +$f(x) = x^2 + 2x$, $g(x) = 2x + 1$. + +$ + f compose g &= (2x+1)^2 + 2(2x+1) = 4x^2 + 8x + 3 \ + g compose f &= 2(x^2 + 2x) + 1 = 2x^2 + 4x + 1 +$ + +== 3 + +=== b + +$"Dom"(f compose g) = RR$, $"Dom"(g compose f = RR$. + +$"Rng"(f compose g) = [3,infinity]$, $"Rng"(g compose f) = [1, infinity]$. + +== 6 + +#theorem[ + If $f : A -> B$, then $I_B compose f = f$. +] + +#proof[ + $"Dom"(I_B compose f) = "Dom"(f) = A$. Now suppose $x in A$, then $(I_B +compose f)(x) = I_B (f(x)) = f(x)$. Therefore $I_B compose f = f$. +] + +== 7 + +#theorem[ + Let $f : A -> B$ with $"Rng"(f) = C$. If $f^(-1)$ is a function, $f compose f^(-1) = I_C$. +] + +#lemma[ + $"Dom"(f^(-1)) = "Rng"(f)$. + + #proof[ + Suppose $(x,y) in f$. Then $(y,x) in f^(-1)$. Therefore for any $x$ in $"Dom"(f)$, $x$ is also in $"Rng"(f^(-1))$, so $"Dom"(f) subset.eq "Rng"(f^(-1))$. + + Suppose $(x,y) in f^(-1)$. Then $(y,x) in f$. So for any $y in "Rng"(f^(-1))$, $y$ is also in $"Dom"(f)$. So $"Rng"(f^(-1)) subset.eq "Dom"(f)$. Therefore $"Dom"(f) = "Rng"(f^(-1))$. + ] +] + +#proof[ + Suppose that $f : A -> B$ and $f^(-1)$ is a function. By @dom-inverse-rng, $"Dom"(f^(-1)) = "Rng"(f) = "Dom"(I_C) = C$. + + Suppose that $y in C$. Because $(y,f^(-1) (y)) in f^(-1)$, $(f^(-1) (y),y) in f$. Then $f compose f^(-1) = f(f^(-1) (y)) = y = I_C$. +] + +== 8 + +The graph of $lr(f|)_A$ is simply the collection of points ${(1,1),(2,-2),(3,-5),(4,-8)}$ + +The graph of $lr(f|)_[-1,3]$ is simply the line given by $f(x) = 4 - 3x$ +restricted between the $x$-coordinates $[-1,3]$. + +The graph of $lr(f|)_(2,4]$ is the line given by $f(x) = 4 - 3x$ between $x = +2$ to $x = 4$, with a hole at $x = 2$. + +The graph of $lr(f|)_{6}$ is simply the point at $(6,-14)$. + +== 14 + +=== b + +$ + &h : [-1, infinity) -> RR, &&h(x) = x^2 + 1 \ + &g : (-infinity, -1] -> RR, &&g(x) = x + 3 +$ + +$h union g$ is a function only if $h(x) = g(x)$ where their domain overlaps at +-1. We check $h(-1) = g(-1) = 2$ is indeed true. Everywhere else $h union g$ is +either $g(x)$ or $h(x)$ which are both functions. Therefore it is a function. + +=== d + +$ + &h : (-infinity,2] -> RR, &&h(x) = cos x \ + &g : [2, infinity) -> RR, &&g(x) = x^2 +$ + +We check that $h$ and $g$ agree at their overlapping domain. $h(2) = cos 2$, +but $g(2) = 4$. Without further computation note that $cos(x)$ gives values +only in $[-1,1]$ so $h(2) != g(2)$. $h union g$ would contain both $(2,4)$ and +$(2,cos 2)$ so it's not a function. diff --git a/documents/by-course/math-8/pset-8/package.nix b/documents/by-course/math-8/pset-8/package.nix new file mode 100644 index 0000000..7879e7e --- /dev/null +++ b/documents/by-course/math-8/pset-8/package.nix @@ -0,0 +1,37 @@ +{ + pkgs, + typstPackagesCache, + typixLib, + cleanTypstSource, + flakeSelf, + ... +}: +let + src = cleanTypstSource ./.; + commonArgs = { + typstSource = "main.typ"; + + fontPaths = [ + # Add paths to fonts here + # "${pkgs.roboto}/share/fonts/truetype" + ]; + + virtualPaths = [ + # Add paths that must be locally accessible to typst here + # { + # dest = "icons"; + # src = "${inputs.font-awesome}/svgs/regular"; + # } + ]; + + XDG_CACHE_HOME = typstPackagesCache; + SOURCE_DATE_EPOCH = builtins.toString flakeSelf.lastModified; + }; + +in +typixLib.buildTypstProject ( + commonArgs + // { + inherit src; + } +)