auto-update(nvim): 2025-01-24 23:00:23

documents/by-course/math-6a/course-notes/main.typ

@@ -1,7 +1,7 @@
-#import "./dvd.typ": *
+#import "@youwen/zen:0.1.0": *
 #import "@preview/cetz:0.3.1"
 
-#show: dvdtyp.with(
+#show: zen.with(
   title: "Math 6A Course Notes",
   author: "Youwen Wu",
   date: "Winter 2025",
@@ -72,3 +72,30 @@ point or vector.
 
   $ arrow(v) dot arrow(w) = ||arrow(v)|| dot ||arrow(w)|| cos theta $
 ]
+
+= Lecture #datetime(day: 23, month: 1, year: 2025).display()
+
+Midterm is next Thursday in class!
+
+== Arclength and curvature
+
+Easy way of finding curvature: reparameterize curve with speed 1, then
+curvature is acceleration. If we can't do that then we need some other
+technique.
+
+Given $arrow(c)(t) = <2t^(-1), 6, 2t>$, find the curvature $kappa(t)$.
+$
+  kappa (t) = (||arrow(c)'(t) times arrow(c)''(t)||) / (||arrow(c)'(t)||^3)
+$
+
+== Arclength parameterization
+
+Find an arc-length parameterization of $arrow(c)(t) = <e^t sin(t), e^t cos(t), 5e^t>$.
+
+Let $s = 0$ when $t = 0$ and let $s$ be the arc-length that has traveled along
+the curve after $t$ seconds, then we can find $s$ by integrating the curve's
+speed over $t$.
+
+$
+  s(t) = integral^t_0 ||arrow(c)'(u)|| dif u
+$

documents/by-course/math-8/pset-2/main.typ

@@ -33,7 +33,7 @@ $ ((forall x)H(x)) or ((forall x)(not H(x))) $
 
 Again let $H(p)$ be true if a person is honest and false otherwise.
 
-$ (exists x)(exists y)(H(x) and not H(y)) $
+$ (exists x)(H(x)) and ((exists y) not H(y)) $
 
 1j.
 
@@ -41,25 +41,25 @@ $ (forall x)(exists y)(x > y) $
 
 1k.
 
-$ (exists.not x)(forall y)(x > y) $
+$ (forall x)(exists y)(not (x > y)) $
 
 1L.
 
-$ (x in ZZ)(y in ZZ)(y > x)(exists z in RR)(x < z < y) $
+$ (forall x in ZZ)(forall y in ZZ)(y > x)(exists z in RR)(x < z < y) $
 
 1m.
 
-$ (exists x in ZZ^+)(exists.not y in ZZ^+)(y < x) $
+$ (exists x in ZZ^+)(forall y in ZZ^+)(x <= y) $
 
 1p.
 
-$ (forall x)(x > 0)(exists y)(2^y = x) $
+$ (forall x)(x > 0)(exists! y)(2^y = x) $
 
 2f.
 
 Let $H(p)$ be true if a person is honest and false otherwise.
 
-$ (exists x)(exists y)(H(x) and not H(y)) $
+$ (exists x)(H(x) and ((exists y) not H(y)) $
 
 In English: Some people are honest and some people are not honest.
 
@@ -73,25 +73,36 @@ In English: all people are honest or no one is honest.
 
 2j.
 
+$ (exists x)(forall y) (x <= y) $
 
-$ (exists x)(forall y) not (x > y) $
+There is an integer such that it is smaller than every other integer y.
 
 2k.
 
-$ (forall x)(exists y)(y > x) $
+$ (exists y)(forall x)(y >= x) $
+
+There is an integer that is greater than all other integers.
 
 2L.
 
-$ (exists x in ZZ)(exists y in ZZ)(forall z)((z > x) and (z > y)) $
+$ (exists x in ZZ)(exists y in ZZ)(x < y)(forall z)(not (x < z < y)) $
+
+There exists an integer and a larger integer such that there is no real number
+between them.
 
 2m.
 
 $ (forall x in ZZ^+)(exists y in ZZ^+)(y < x) $
 
+Any positive integer has an integer less than itself.
+
 2p.
 
 $ (exists x)(x > 0)(forall y) not (2^y = x) $
 
+There is a positive real number such that there is no real number $y$ that
+satisfies $2^y = x$.
+
 6a.
 
 $T$, $U$, $V$.
@@ -182,7 +193,7 @@ one.
     x y = 4 j k
   $
 
-  Clearly $x y$ has $4$ in its factors and so $x y | 4$.
+  $x y$ has $4$ in its factors and so $4 | x y $.
 ]
 
 5d.
@@ -263,7 +274,7 @@ one.
   $
 
   Clearly for any $b$ the left side is strictly lower than the right. Repeat
-  this exact for $a$ is negative.
+  this exact reasoning for when $a$ is negative.
 ]
 
 7d.
@@ -284,15 +295,15 @@ one.
 7e.
 
 #proof[
-  If $1 | a$, then we can find some $k in ZZ$ such that $1k = a$. Such a $k$ is
-  $a$, because $1 dot a = a$. Therefore $1 | a$.
+  $1 | a$ if and only if we can find some $k in ZZ$ such that $1k = a$. We can
+  always find such a $k$, which is $a$, because $1 dot a = a$. Therefore $1 | a$ is always true.
 ]
 
 7f.
 
 #proof[
-  If $a | a$, we can find some $k in ZZ$ such that $a k = a$. Such a $k$ is $1$,
-  because $a dot 1 = a$. Therefore $a | a$.
+  $a | a$ if and only if we can find some $k in ZZ$ such that $a k = a$. Such a
+  $k$ is $1$, because $a dot 1 = a$. Therefore $a | a$.
 ]
 
 7g.
@@ -366,8 +377,8 @@ one.
 
   $ k j a c = b d $
 
-  If $exists n in ZZ, n a c = b d$, then $a c | b d$. We see that $n = k j$.
-  Therefore $a c$ indeed divides $b d$.
+  If $exists n in ZZ, n a c = b d$, then $a c | b d$. We have such an $n = k
+  j$. Therefore $a c$ indeed divides $b d$.
 ]
 
 8a.
@@ -399,9 +410,9 @@ one.
     n^2 + n + 3 = n(n + 1) + 3
   $
 
-  By 7(d), we know that $n(n+1)$ is even $forall n$. Then by 5(h) we can take
-  $x := n(n+1)$ and $y := 3$. Since $x$ is even, $y$ is odd, $x + y$ is odd. So
-  $n^2 + n + 3$ is odd.
+  By the proof in 7(d), we know that $forall n, n(n+1)$ is even. Then by 5(h)
+  we can take $x := n(n+1)$ and $y := 3$. Since $x$ is even, $y$ is odd, $x +
+  y$ is odd. So $n^2 + n + 3$ is odd.
 ]
 
 9a.
@@ -410,18 +421,14 @@ one.
   In the case of $x=0$ and $y=0$ we trivially have $0 >= 0$. Otherwise,
 
   $
-    (x+y) / 2 &>= sqrt(x y) \
-    x+y &>= 2sqrt(x y) \
-    (x+y)^2 &>= 4x y \
-    x^2 + 2 x y + y^2 &>= 4 x y \
-    x^2 - 2 x y + y^2 &>= 0
+    &(x+y) / 2 >= sqrt(x y) \
+    &<=> x+y >= 2sqrt(x y) \
+    &<=> (x+y)^2 >= 4x y \
+    &<=> x^2 + 2 x y + y^2 >= 4 x y \
+    &<=> x^2 - 2 x y + y^2 >= 0 \
+    &<=> (x - y)^2 >= 0
   $
-
-  THIS IS WRONG FIX IT!!!
-
-  You can think of $x^2 - 2 x y + y^2$ as quadratic with $y$ as a constant. For
-  all possible values of $y$ the equation is nonnegative (since the absolute
-  minimum occurs at the vertex). Therefore the inequality holds true.
+  We know that $forall x in RR, x^2 >= 0$, so this statement is always true, and thus the original is always true as well.
 ]
 
 11b.
@@ -431,7 +438,7 @@ Grade: C.
 The assertions that $exists q, b = a q$ and $exists q, c = a q$ are essentially
 correct but these $q$ are not the same. This can be corrected fairly
 straightforwardly by replacing one of the $q$ with another variable serving the
-same purpose, then proceeding.
+same purpose, then proceeding in a similar fashion.
 
 11c.