auto-update(nvim): 2025-02-19 03:00:27
This commit is contained in:
parent
c3278a4e6a
commit
133d84b018
GPG key ID:
865658ED1FE61EC3
3 changed files with 453 additions and 0 deletions
|
|
@ -835,3 +835,65 @@ To find gamma, we can simply use inverse trig functions.
|
|||
|
||||
Which makes $cos(5t + pi - arctan(1/2))$ negative.
|
||||
]
|
||||
|
||||
= Lecture #datetime(day: 18, year: 2025, month: 2).display()
|
||||
|
||||
== Linear systems of differential equations
|
||||
|
||||
Consider a matrix $A$ and solution vector $x$.
|
||||
$
|
||||
x'(t) = A(t) x(t)
|
||||
$
|
||||
|
||||
#fact[Superposition principle for linear system][
|
||||
If $x^((1)) (t)$ and $x^((2)) (t)$ are solutions, then
|
||||
$
|
||||
c_1 x^((1)) (t) + c_2 x^((2)) (t)
|
||||
$
|
||||
are also solutions.
|
||||
]
|
||||
|
||||
=== Homogenous case
|
||||
|
||||
This is when $g(t) = 0$. I want to solve for $arrow(x)' = A arrow(x)$, where
|
||||
$
|
||||
arrow(x) = vec(mu_1 (t), mu_2 (t), dots.v, mu_n (t))
|
||||
$
|
||||
Similar to the scalar case, we look for solutions in terms of exponential
|
||||
functions. Guessing
|
||||
|
||||
$
|
||||
arrow(x) (t) = e^(r t) arrow(v) \
|
||||
arrow(x)'(t) = r e^(r t) arrow(v) = A e^(r t) arrow(v) \
|
||||
r arrow(v) = A arrow(v) \
|
||||
(A - r I) arrow(v) = 0, arrow(v) != 0
|
||||
$
|
||||
|
||||
If there are non-trivial solutions then $A$ must have determinant 0.
|
||||
|
||||
$
|
||||
det (A - r I) = 0
|
||||
$
|
||||
|
||||
This is the eigenvalue equation for $A$!
|
||||
|
||||
#example[
|
||||
Find a fundamental set of solutions. FIrst find the eigenvalues and
|
||||
corresponding eigenvectors of $A$.
|
||||
|
||||
$
|
||||
x'(t) = A x(t), A = mat(2,1;1,2)
|
||||
$
|
||||
|
||||
Find eigenvalues of $A$.
|
||||
|
||||
$
|
||||
A - lambda I = mat(2 - lambda, 1; 1, 2- lambda) \
|
||||
|a - lambda I| = (2 - lambda)^2 - 1 = 0 \
|
||||
2 - lambda = plus.minus 1 \
|
||||
lambda = cases(1,3)
|
||||
$
|
||||
|
||||
Now we want the eigenvectors for our eigenvalues. Find an eigenvector
|
||||
corresponding to $lambda_1 = 1$.
|
||||
]
|
||||
|
|
|
|||
|
|
@ -201,3 +201,75 @@ $
|
|||
(diff F) / (diff x) = diff / (diff z) (3x^2 + 5 y z + z^3) = diff / (diff z) z^3 = 0 + (5 (diff y) / (diff x) z + 5y) + 3z^2 \
|
||||
(diff y) / (diff z) = - (5y + 3z^2) / (5z)
|
||||
$
|
||||
|
||||
= Lecture #datetime(day: 18, year: 2025, month: 2).display()
|
||||
|
||||
== Critical points
|
||||
|
||||
When optimizing in 2D, the strategy depends on whether we're
|
||||
|
||||
- optimizing for all of $RR^2$ (or a region in $RR^2$)
|
||||
- optimizing on a constraint (like a curve through $RR^n$)
|
||||
|
||||
We find critical points where the tangent plane is "flat": $m_x = 0$ and $m_y =
|
||||
0$.
|
||||
|
||||
We classify critical points using the determinant of the gradient.
|
||||
|
||||
$
|
||||
D = f_(x x) f_(y y) = f_(x y)^2
|
||||
$
|
||||
|
||||
- if $D >0$ and $f_(x x) (x_0, y_0) > 0$, then $f(x_0, y_0)$ is a relative minimum.
|
||||
- if $D > 0$ and $f_(x x) (x_0, y_0) < 0$, $f(x_0, y_0)$ is a relative maximum.
|
||||
- if $D < 0$ then $f(x_0, y_0)$ is neither and we call it a saddle point.
|
||||
- if $D = 0$ then we don't know
|
||||
|
||||
== Lagrange multipliers
|
||||
|
||||
Optimizing constrained curves. Idea: navigate along the curve and look for where
|
||||
the directional derivative is zero.
|
||||
|
||||
#example[
|
||||
Find the highest and lowest points on $f(x,y) = 81x^2 + y^2$ with the
|
||||
constraint $x^2 + y^2 = 0$.
|
||||
|
||||
For notational purposes, we'll call $g(x,y) = 4x^2 + y^2$ and keep in mind
|
||||
we're looking for $g(x,y) = 9$.
|
||||
|
||||
1. Find the gradients of $f$ and $g$.
|
||||
$
|
||||
vec(f_x,f_g) &= vec(162x,2y) \
|
||||
vec(g_x,g_y) &= vec(8x,2y)
|
||||
$
|
||||
2. We want to find the points where the gradients "align", i.e. we want these
|
||||
vectors to be parallel, that is:
|
||||
$
|
||||
arrow(F) = lambda arrow(G) \
|
||||
162x = 8x dot lambda \
|
||||
2y = 2y dot lambda
|
||||
$
|
||||
Remember to keep the constraint!
|
||||
$
|
||||
4x^2 + y^2 = 9
|
||||
$
|
||||
Breaking it down into cases,
|
||||
$
|
||||
162 = 8lambda => lambda = 81 / 4 "for" x != 0
|
||||
$
|
||||
which implies
|
||||
$
|
||||
2y (lambda - 1) 0 \
|
||||
y = 0
|
||||
$
|
||||
since $lambda - 1$ is nonzero. So if $x$ is nonzero, $y$ must be zero.
|
||||
|
||||
$
|
||||
4x^2 = 9 \
|
||||
x = plus.minus 3 / 2
|
||||
$
|
||||
Now consider when $x = 0$. Then $y = plus.minus 3$. So our critical points
|
||||
are $(plus.minus 3/2, plus.minus 3)$. Finally, just plug in these 4
|
||||
critical points into $f$ and find the biggest/smallest.
|
||||
|
||||
]
|
||||
|
|
|
|||
|
|
@ -688,3 +688,322 @@ $ z = x + (b c - a d) / (b d + 1) $
|
|||
$ z = (x + y) / 2 $
|
||||
Then $(x + y)/2$ is obviously rational and strictly between $x$ and $y$.
|
||||
]
|
||||
|
||||
= Induction
|
||||
|
||||
Induction is a way to prove that a statement is true for all $NN$. Formally, it
|
||||
is introduced like this: say you have a set $S subset.eq NN$ such that $1 in
|
||||
S$. Then, if $n in S => n + 1 in S$, $S = NN$. Usually we plug $n$ into a
|
||||
proposition $P$ to make it more useful for proving statements. If $(forall n in
|
||||
S)(P(n))$, and we can show $S = NN$ by induction, then $P(n)$ is true for all
|
||||
$n in NN$.
|
||||
|
||||
Induction consists of two main parts. First, we have to show that $1 in S$.
|
||||
This is the _base case_. Then, we assume the _inductive hypothesis_, $n in S$.
|
||||
If we can prove, using the inductive hypothesis, that $n in S$ implies that $n
|
||||
+ 1$ in $S$, then we'll have sufficient justification for $S = N$. When proving
|
||||
a statement in practice, if $S$ is the set of all inputs to $P(n)$ that make it
|
||||
true, we show that $P(1)$ is true ($1 in S$) and then show that $P(n) =>
|
||||
P(n+1)$, which implies $(forall n in NN) P(n)$.
|
||||
|
||||
There are two equivalent formulations of induction, the principle of
|
||||
mathematical induction (PMI, also called _weak induction_), and the principle
|
||||
of complete induction (PCI, also called _strong induction_). They are
|
||||
equivalent, but in some cases one is much more straightforward to show than the
|
||||
other.
|
||||
|
||||
The previous strategy we discussed was _weak induction_, named such because we
|
||||
need only show $P(1) and P(n) => P(n+1)$. _Strong induction_ requires us to
|
||||
show a much stronger result, but also allows us to make a stronger assumption.
|
||||
|
||||
#definition[Strong induction][
|
||||
If ${1,2,3,...,n-1} subset.eq S => n in S$, then $S = NN$.
|
||||
]
|
||||
|
||||
Essentially in strong induction we assume that every natural up to $n$ is in
|
||||
$S$, and if we can show that this implies $n$ is also in $S$, we have $S = NN$.
|
||||
Note that in this case we do not have to show $1 in S$, but this is implied.
|
||||
When proving a proposition $P$, we have to prove $P(1) and P(2) and ... and
|
||||
P(n-1) => P(n)$, but we can assume the strong inductive hypothesis $P(1) and
|
||||
P(2) and ... and P(n-1)$.
|
||||
|
||||
#example[Game theory of Nim][
|
||||
In the game of _Nim_, two players take turns taking coins from two piles of
|
||||
$n$ coins each. If there are $m$ coins in a pile, a player may choose to take
|
||||
$1,2,3,...,m$ coins, *except* $m - 1$ coins. That is, a player may take any
|
||||
nonzero amount, but may not leave exactly one coin left in the pile. The
|
||||
player to take the last coin in play wins.
|
||||
|
||||
Example game: if there are 6 coins on the left and 8 coins on the right,
|
||||
player 1 goes first and takes 4 coins from the left. Player 2 takes the
|
||||
remaining 2 coins on the left. Player 1 takes all 8 coins from the right and
|
||||
wins.
|
||||
|
||||
Prove that Player 2 has a winning strategy in all games of _Nim_.
|
||||
|
||||
This is actually a somewhat unexpected way in which strong induction can be
|
||||
very useful. Let $S$ be a set of the natural numbers where Player 2 has a
|
||||
winning strategy when there are $n in S$ coins left in both piles.
|
||||
|
||||
Suppose $m$ is a natural number and ${1,2,3,...,m-1} subset.eq S$. This is
|
||||
our inductive hypothesis, we assume that Player 2 has a winning strategy for
|
||||
when the game starts with $1,2,3,...,m-1$ coins in each pile. Now we seek to
|
||||
show that this implies that Player 2 has a winning strategy when there are
|
||||
$m$ coins in the pile. For the special case when $m = 1$, Player 1 takes 1
|
||||
coin from either pile and Player 2 takes both, winning the game. Otherwise,
|
||||
for $m > 1$, if Player 1 takes $m$ coins from a pile, Player 2 takes all the
|
||||
coins in the other pile, winning. If Player 1 takes $j$ coins from either
|
||||
pile where $1 <= j < m - 1$, leaving $m - j > 1$ coins in the pile, Player 2
|
||||
can take $j$ coins from the other pile, leaving both piles with $m - j$
|
||||
coins. This is essentially a new game of Nim with $m - j$ coins in each pile.
|
||||
Then $m - j in {1,2,3,...,m-1}$ since $j <= 1 < m-1$, so by our inductive
|
||||
hypothesis Player 2 has a winning strategy from this point on. Therefore
|
||||
Player 2 has a winning strategy when $m$ coins are in each pile.
|
||||
Since ${1,2,3,...,m-1} subset.eq S => m in S$, by strong induction, $S = NN$
|
||||
and Player 2 has a winning strategy for all possible games of _Nim_.
|
||||
]
|
||||
|
||||
#example[A hint for the fundamental theorem of arithmetic][
|
||||
Prove that all natural numbers can be expressed as the product of primes.
|
||||
|
||||
#proof[
|
||||
Let $m$ be a natural number. Note that $m = 2$ is prime so its the product
|
||||
of itself and 1. Now assume that all natural numbers $n$ such that $1 < n <
|
||||
m$ can be written as the product of primes. Then $m$ is either prime, in
|
||||
which case its prime factors are $1 dot m$, or $(exists s,t in
|
||||
{1,2,...,m-1})(m = s t)$. Then by our inductive hypothesis both $s$ and $t$
|
||||
can be written as the product of primes, so $m$ is also the product of
|
||||
primes. By strong induction, we conclude all natural numbers can be written
|
||||
as the product of primes.
|
||||
]
|
||||
]<fundamental-thm-arithmetic-hint>
|
||||
|
||||
== Well ordering principle
|
||||
|
||||
This is another property that characterizes $NN$.
|
||||
|
||||
#fact[Well ordering principle][
|
||||
Every nonempty subset of $NN$ has a smallest element.
|
||||
]
|
||||
|
||||
We demonstrate the WOP in an alternative proof of
|
||||
@fundamental-thm-arithmetic-hint.
|
||||
|
||||
#example[
|
||||
All natural numbers can be expressed as the product of primes.
|
||||
|
||||
#proof[
|
||||
Suppose there exists some non-empty set $S subset.eq NN$ that contains all
|
||||
of the natural numbers which cannot be expressed as a product of primes.
|
||||
By the WOP, there is a least number in this set, $T in S$. $T$ cannot be
|
||||
prime because it would be the product of $1 dot T$, so it must be
|
||||
composite. Then $T = s t$ for some natural numbers $s,t$, which are less
|
||||
than $T$. But $T$ is the least element of $S$ so neither $s$ nor $t$ are in
|
||||
$S$, and therefore they both have prime factorizations. But this implies
|
||||
that $T$ also has a prime factorization, a contradiction. Therefore no such
|
||||
set $S$ can exist, and all natural numbers have a prime factorization.
|
||||
]
|
||||
]
|
||||
|
||||
#theorem[The division algorithm][
|
||||
For all integers $a$ and $b$, with $a != 0$, there exist unique integers $q$
|
||||
and $r$ such that $b = a q + r$ and $0 <= r < |a|$.
|
||||
|
||||
#proof[
|
||||
Let us consider the case where $a > 0$. Then we have a set
|
||||
|
||||
$
|
||||
S = {b - a k, k in ZZ, b - a k >= 0}
|
||||
$
|
||||
|
||||
It follows that we should exclude 0 from $S$ because otherwise if $0 in S$,
|
||||
$b - a k = 0$ for some integer $k$, and $a | b$, so $b = a dot b/a$.
|
||||
|
||||
In our restricted $S$, there is a least element, call it $r$. Then for some
|
||||
integer $q$, $a q + r = b$. We know that $r > 0$ so let's show $r < |a|$.
|
||||
Since we're considering only $a > 0$ for simplicity, we just show $r < a$.
|
||||
If $r >= a$, then either $0 in S$ when $r = a$ because $a q + a = b$
|
||||
implies that $a | b$, or, when $r > a$, there exists another element in
|
||||
$S$, $b - a (q + 1)$. This is a contradiction since we assumed $b - a q$
|
||||
was the smallest element in $S$ and clearly $b - a q - a$ is smaller.
|
||||
Therefore $r < |a|$.
|
||||
|
||||
Now we show the uniqueness of $q, r$. Suppose there were other integers
|
||||
satisfying the properties of $q$ and $r$, $q'$ and $r'$. Then we have
|
||||
|
||||
$
|
||||
r' >= r
|
||||
$
|
||||
without loss of generality (otherwise just relabel). Then $a q + r = a q' +
|
||||
r'$, so $a(q - q') = r' - r$. Then $a$ divides $r' - r$.
|
||||
]
|
||||
]
|
||||
|
||||
== Relations, partitions
|
||||
|
||||
#definition[
|
||||
A relation on a set $A$ is a set of ordered pairs $(a,b)$ where $a,b in A$. A
|
||||
relation from a set $A$ to a set $B$ is set of ordered pairs $(a,b)$ where $a
|
||||
in A$ and $b in B$.
|
||||
]
|
||||
|
||||
#abuse[
|
||||
If $R$ is a relation from $A$ to $B$, for $a in A$ and $b in B$ where $(a,b)
|
||||
in R$, we can abbreviate it $a R b$.
|
||||
]
|
||||
|
||||
#definition[
|
||||
For any set $A$, the identity relation on $A$ is the set
|
||||
$
|
||||
I_A = {(a,a) : a in A}
|
||||
$
|
||||
]
|
||||
|
||||
#definition[
|
||||
The domain of the relation $R$ from $A$ to $B$ is the set
|
||||
$
|
||||
"Dom"(R) = {x in A : (exists y in B)(x R y)} \
|
||||
$
|
||||
The range is
|
||||
$
|
||||
"Rng"(R) = {y in B : (exists x in A)(x R y)}
|
||||
$
|
||||
]
|
||||
|
||||
So the domain of $R$ is the set of all first coordinates and the range is the
|
||||
set of all second coordinates.
|
||||
|
||||
#theorem[
|
||||
1. $"Dom"(R^(-1)) = "Rng"(R)$
|
||||
2. $"Rng"(R^(-1)) = "Dom"(R)$
|
||||
]
|
||||
|
||||
#definition[
|
||||
Let $R$ be a relation from $A$ to $B$, let $S$ be a relation from $B$ to $C$.
|
||||
The composite of $R$ and $S$ is
|
||||
|
||||
$
|
||||
S compose R = {(a,c) : (exists b in B)((a,b) in R and (b,c) in S)}
|
||||
$
|
||||
]
|
||||
|
||||
== Equivalence relations
|
||||
|
||||
Let $A$ be a set and $R$ a relation on $A$.
|
||||
|
||||
We say $R$ is reflexive on $A$ if $forall x in A, x R x$. $R$ is symmetric if
|
||||
$forall x,y in A$, if $x R y$, then $y R x$. $R$ is transitive if $forall x,y,z
|
||||
in A$, if $x R y$ and $y R z$, then $x R z$.
|
||||
|
||||
#theorem[
|
||||
Let $A$ be a set. For the power set $cal(P)(A)$, the relation "is a subset of"
|
||||
is reflexive on $cal(P)(A)$ and transitive but not symmetric.
|
||||
]
|
||||
|
||||
#definition[
|
||||
A relation $R$ on a set $A$ is an equivalence relation on $A$ if $R$ is
|
||||
reflexive on $A$, symmetric, and transitive.
|
||||
]
|
||||
|
||||
We make an equivalence relation when we think of objects as being related by
|
||||
having the same property.
|
||||
|
||||
An equivalence relation on a set divides the set into subsets of related
|
||||
elements.
|
||||
|
||||
#definition[
|
||||
Let $R$ be an equivalence class on a set $A$. For $x in A$, the equivalence
|
||||
class of modulo $R$ (or $x mod R$) is the set
|
||||
|
||||
$
|
||||
overline(x) = {y in A : x R y}
|
||||
$
|
||||
|
||||
Each element of $overline(x)$ is a *representative* of the class. The set
|
||||
$
|
||||
A \/ R = {overline(x) : x in A}
|
||||
$
|
||||
of all equivalence classes is called $A "modulo" R$.
|
||||
]
|
||||
|
||||
#theorem[
|
||||
Let $R$ be an equivalence relation on a nonempty set $A$. For all $x$ and $y$
|
||||
in $A$,
|
||||
|
||||
1. $x in overline(x)$ and $overline(x) subset.eq A$
|
||||
2. $x R y$ if and only if $overline(x) = overline(y)$
|
||||
3. $x cancel(R) y$ if and only if $overline(x) sect overline(y) = emptyset$
|
||||
]
|
||||
|
||||
== Congruence relations
|
||||
|
||||
Now we define congruence relations on $ZZ$ and show that congruence mod $m$ is
|
||||
an equivalence relation. We describe its equivalence classes.
|
||||
|
||||
#definition[
|
||||
Let $m$ be a fixed positive integer. For $x,y in ZZ$, we say $x$ is congruent
|
||||
to $y$ modulo $m$ and write $x = y (mod m)$ if $m$ divides $(x-y)$. The number
|
||||
$m$ is called the modulus of the congruence.
|
||||
]
|
||||
|
||||
#theorem[
|
||||
For every fixed positive integer $m$, congruence modulo $m$ is an equivalence
|
||||
relation on $ZZ$.
|
||||
]
|
||||
|
||||
#proof[
|
||||
1. Reflexivity. Let $x$ be an integer. We show $x = x (mod m)$. $m dot 0 = 0 = x - x$, $m | x - x$.
|
||||
2. For symmetry, suppose $x = y (mod m)$. Then $m | x - y$. Thus $(exists k)(x - y = k m)$. But $-(x-y) = -(k m)$, or $y - x = (-k) m$. So $m | y - x$ so $y = x (mod m)$
|
||||
3. Suppose $x = y (mod m)$ and $y = z (mod m)$. Thus $m | x - y$ and $m | y - z$. Thus $m | x - y + y - z <=> m | x - z$ so $x = z (mod m)$. So congruence modulo $m$ is transitive.
|
||||
]
|
||||
|
||||
#definition[
|
||||
The set of equivalence classes for the relation congruence modulo $m$ is
|
||||
denoted $ZZ_m$.
|
||||
]
|
||||
|
||||
#theorem[
|
||||
Let $m$ be a fixed positive integer. Then
|
||||
|
||||
1. for integers $x$ and $y$, $x = y (mod m)$ if and only if the remainder when $x$ is divided by $m$ equals the remainder when $y$ is divided by $m$
|
||||
2. $ZZ_m$ consists of $m$ distinct equivalence classes.
|
||||
]
|
||||
|
||||
The equivalence classes of $ZZ_m$ $overline(0), overline(1), ...,
|
||||
overline(m-1)$ which are exactly all of the possible remainders when integers
|
||||
are divided by $m$. For this reason the elements of $ZZ_m$ are sometimes called
|
||||
the residue classes modulo $m$.
|
||||
|
||||
== Partitions
|
||||
|
||||
#definition[
|
||||
Let $A$ be a nonempty set. $cal(P)$ is a partition of $A$ if $cal(P)$ is a set of subsets of $A$ such that
|
||||
1. if $X in cal(P)$, then $X != emptyset$
|
||||
2. if $X in cal(P)$ and $Y in cal(P)$, then $X = Y$ or $X sect Y = emptyset$
|
||||
3. $union.big_(X in cal(P)) X = A$
|
||||
]
|
||||
|
||||
In other words a partition of a set $A$ is a pairwise disjoint collection of
|
||||
nonempty subsets of $A$ whose union is $A$.
|
||||
|
||||
#theorem[
|
||||
If $R$ is an equivalence relation on a nonempty set $A$, then $A\/R$ is a
|
||||
partition of $A$.
|
||||
]
|
||||
|
||||
#proof[
|
||||
Every equivalence class $overline(x)$ is a subset of $A$ and is nonempty
|
||||
because it contains $x$. Any two equivalence classes are either equal or
|
||||
disjoint. Also,
|
||||
$
|
||||
union.big_(x in A) overline(x) subset.eq A
|
||||
$
|
||||
because each $overline(x) subset.eq A$. To prove $A subset.eq union.big _(x in A) overline(x)$, suppose $y in A$. Because $y in overline(y)$, then
|
||||
$
|
||||
y = union.big_(x in A) overline(x)
|
||||
$
|
||||
Thus,
|
||||
$
|
||||
A = union.big_(x in A) overline(x)
|
||||
$
|
||||
]
|
||||
|
|
|
|||
Loading…
Reference in a new issue