auto-update(nvim): 2025-03-05 02:11:45

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 $n=2$

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 $"Dom"(g compose f) = A$.
+]<compose-domain>
+
+#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 $x,y in I$.
+]

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 $"Dom"(f) = "Rng"(f^(-1))$.
+  ]
+]<dom-inverse-rng>
+
+#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.

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;
+  }
+)