kruby/notes-archive
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

Spellchecker

Browse Source
This commit is contained in:
Nathan Nguyen
2024-12-06 12:48:11 -06:00
parent ac4e2030c1
commit bb7fc5f6e8
24 changed files with 139 additions and 113 deletions
Download Patch File Download Diff File Expand all files Collapse all files

25
.vscode/settings.json vendored Normal file
View File

@@ -0,0 +1,25 @@
{
"cSpell.words": [
// TeX
"infty",
"mathbb",
"mathcal",
"stackrel",
"vmatrix",
// Vocab - Math
"subring",
"subrings",
"monic",
"Noetherian",
"abelian",
"indeterminates",
"infimum",
"supremum",
"Wronskian",
// Vocab - Physics
"Magnetostatic",
"Magnetostatics"
]
}

1
docs/index.md
View File

@@ -1,4 +1,5 @@
# Personal Notes Collection
## Textbook Reference
Abstract Algebra: [Abstract Algebra, Theory and Applications](http://abstract.ups.edu/download/aata-20220728.pdf)

18
docs/math/abstract-algebra/16-rings.md
View File

@@ -2,7 +2,7 @@
## Section 16.1 - Rings
**Definition**. A nonempty set $S$ is a *ring* if, with two binary operations called addition and multipllication, the following are satisfied:
**Definition**. A nonempty set $S$ is a *ring* if, with two binary operations called addition and multiplication, the following are satisfied:
1. Addition is commutative. $a + b = b + a$ for $a, b \in R$
2. Addition is associative. $(a + b) + c = a + (b + c)$ for $a, b, c \in R$
@@ -46,17 +46,17 @@ Note that some books impose the condition that $1 \neq 0$. If $1 = 0$, we can sh
**Definition**. For any non-negative integer $n \in \mathbb{N}$ and $r \in R$, we say that $nr = r + \ldots + r \text{(n times)}$.
**Definition**. The *charactaristic* of a ring is the leat possible $n \in \mathbb{N}$ such that $nr = 0$ for all $r \in R$.
**Definition**. The *characteristic* of a ring is the least possible $n \in \mathbb{N}$ such that $nr = 0$ for all $r \in R$.
**Example**. For every prime number $p$, $\mathbb{N}_p$ is a field of charactaristic $p$.
**Example**. For every prime number $p$, $\mathbb{N}_p$ is a field of characteristic $p$.
**Lemma**. 16.18: Given $R$ is a ring with identity, the charactaristic of $1$ is the charactartistic of the field.
**Lemma**. 16.18: Given $R$ is a ring with identity, the characteristic of $1$ is the characteristic of the field.
**Theorem**. 16.19: The charactaristic of an integral domain is prime or zero.
**Theorem**. 16.19: The characteristic of an integral domain is prime or zero.
## Section 16.3 - Ring Homomorphisms and Ideals
**Definition** Given rins $R$ and $S$, and a mapping $\varphi: R \rightarrow S$, we say that $\varphi$ is a *ring homomorphism* if the following are satisfied for all elements of $R$:
**Definition** Given rings $R$ and $S$, and a mapping $\varphi: R \rightarrow S$, we say that $\varphi$ is a *ring homomorphism* if the following are satisfied for all elements of $R$:
$$
\begin{align}
@@ -92,7 +92,7 @@ is an ideal in $R$. Specifically, $\langle a \rangle$ is a *principal ideal*.
**Example**. Theorem 16.25. Every ideal in $\mathbb{Z}$ is a principal ideal.
**Examplee**. With $\varphi: R \rightarrow S$, $\ker \varphi$ is an ideal of $R$.
**Example**. With $\varphi: R \rightarrow S$, $\ker \varphi$ is an ideal of $R$.
**Remark**. 16.28: We are working with *two-sided ideals*. If rings are not commutative, we may deal with *left ideals* and *right ideals*.
@@ -118,7 +118,7 @@ $$
R/I \cong \frac{R/J}{I/J}
$$
**Theorem**. 16.34, *Correspondence Theorem*. Let $I$ be an ideal of $R$. Then, $S \mapsto S/I$ is a one-to-one correspeondence between the set of subrings $S$ containing $I$ (that is, $I \in S$) and the set of subrings of $R/I$. Furthermore, the ideals of $R$ containing $I$ correspond to the ideals of $R/I$.
**Theorem**. 16.34, *Correspondence Theorem*. Let $I$ be an ideal of $R$. Then, $S \mapsto S/I$ is a one-to-one correspondence between the set of subrings $S$ containing $I$ (that is, $I \in S$) and the set of subrings of $R/I$. Furthermore, the ideals of $R$ containing $I$ correspond to the ideals of $R/I$.
## Section 16.4 - Maximal and Prime Ideals
@@ -140,6 +140,6 @@ $$
(a + P)(b + P) = ab + P = 0 + P = P
$$
Thus, $ab \in P$. By symnetry, assume $a \notin P$. Thus, $b \in P$ by the devinition of a prime ideal, so $b + P = 0 + P$, meaning $R/P$ is an integral domain.
Thus, $ab \in P$. By symnetry, assume $a \notin P$. Thus, $b \in P$ by the definition of a prime ideal, so $b + P = 0 + P$, meaning $R/P$ is an integral domain.
**Theorem**. 16.40: In a commutative ring with identity, every maximal ideal is also a prime ideal.

4
docs/math/abstract-algebra/17-polynomial-rings.md
View File

@@ -14,7 +14,7 @@ where $a_i \in R$ and $a_n \neq 0$ is called a *polynomial over $R$* with *indet
**Definition**. A polynomial is known as *monic* if the leading coefficient is equal to $1$.
**Definition**. The *degree* of $f$ is the largest nonnegative number such that $a_n \neq 0$, written as $\deg f(x) = n$. If no such mumber exists, that is, $f(x) = 0$, we say the degree of $f$ is $-\infty%.
**Definition**. The *degree* of $f$ is the largest nonnegative number such that $a_n \neq 0$, written as $\deg f(x) = n$. If no such number exists, that is, $f(x) = 0$, we say the degree of $f$ is $-\infty%.
**Definition**. We denote the set of all polynomials with coefficients in $R$ as $R[x]$.
@@ -46,7 +46,7 @@ where either $\deg r(x) < \deg g(x)$ or $r(x)$ is the zero polynomial.
## Section 17.3 Irreducible Polynomials
**Definition** A nonconstant polynomial $f(x) \ in F[x]$ is *irreducible* over a field $F$ if it cannot be expressed as the product of two non-identity polynomials $g(x)$ and $h(x)$ in $F[x]$, with the degree of both polynomials strictly less than the degree of $f(x)$.
**Definition** A non-constant polynomial $f(x) \ in F[x]$ is *irreducible* over a field $F$ if it cannot be expressed as the product of two non-identity polynomials $g(x)$ and $h(x)$ in $F[x]$, with the degree of both polynomials strictly less than the degree of $f(x)$.
**Lemma**. Let $p(x) \in \mathbb{Q}[x]$. Then, with $r, s \in \mathbb{Z}, a(x) \in \mathbb{N}[x]$, we can write $p(x) = \frac{r}{s} a(x)$.

26
docs/math/abstract-algebra/18-integral-domains.md
View File

@@ -2,7 +2,7 @@
## Section 18.1 - Fields of Fractions
**Definition**. Given an integral domain $D$, we can construct a field $F$ containing $D$ by stating that any $p/q \in F$, annd that any two elements $a/b = c/d$ if and only if $ad = bc$. We can consider this akin o a set of ordered pairs
**Definition**. Given an integral domain $D$, we can construct a field $F$ containing $D$ by stating that any $p/q \in F$, and that any two elements $a/b = c/d$ if and only if $ad = bc$. We can consider this akin o a set of ordered pairs
$$
S = \{(a, b) : a, b \in D \text{ and } b \neq 0 \}
@@ -18,13 +18,13 @@ $$
Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$, there exists a map $\psi: F_D \rightarrow D$ giving an isomorphism such that $\psi(a) = a$ for all $a \in D$.
**Corollary**. 18.6: Let $F$ be a field of charactaristic $0$. Then, $F$ contains a subfield isomorphic to $\mathbb{Q}$.
**Corollary**. 18.6: Let $F$ be a field of characteristic $0$. Then, $F$ contains a subfield isomorphic to $\mathbb{Q}$.
**Corollary**. 18.6: Let $F$ be a field of charactaristic $p$. Then, $F$ contains a subfield isomorphic to $\mathbb{Z}_p$.
**Corollary**. 18.6: Let $F$ be a field of characteristic $p$. Then, $F$ contains a subfield isomorphic to $\mathbb{Z}_p$.
## Section 18.2 - Factorization in Integral Domains
**Definition**. Let $R$ be a commutative ring with identity, and $a, b \in R$. We say that $a$ *divideds* $b$, that is, $a | b$, if there exists some $c \in R$ such that $b = ac$.
**Definition**. Let $R$ be a commutative ring with identity, and $a, b \in R$. We say that $a$ *divides* $b$, that is, $a | b$, if there exists some $c \in R$ such that $b = ac$.
**Definition**. A *unit* element is any element that has a multiplicative inverse.
@@ -37,7 +37,7 @@ Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$,
**Definition**. Given integral domain $D$, we say that $D$ is a *Unique Factorization Domain (UFD)* if it satisfies the following criteria:
1. Given $a \in D, a \neq 0$, and $a$ is not a unit, $a$ can be written as a product of irreducible elements in $D$.
2. Let $a = p_1 \ldots p_r = q_1 \ldots q_s$, where $p_i$ and $q_i$ are all irreducible. Then, $r = s$, and there exists some fuction $\pi \in S_r$ such that $p_i$ and $q_{\pi(j)}$ are associates for $j = 1, \ldots, r$.
2. Let $a = p_1 \ldots p_r = q_1 \ldots q_s$, where $p_i$ and $q_i$ are all irreducible. Then, $r = s$, and there exists some function $\pi \in S_r$ such that $p_i$ and $q_{\pi(j)}$ are associates for $j = 1, \ldots, r$.
**Definition**. A ring $R$ is a *principal ideal domain (PID)* if every ideal of $R$ is principal.
@@ -61,24 +61,24 @@ Additionally, $F_D$ is unique. That is, given field $E$ such that $E \supset D$,
---
**Definition**. Any integral domain $D$ is a *euclidian domain* with a *euclidian function* $nu: D \\ \{0\} \rightarrow \mathbb{N}$ that satisfies the following:
**Definition**. Any integral domain $D$ is a *Euclidean domain* with a *Euclidean function* $nu: D \\ \{0\} \rightarrow \mathbb{N}$ that satisfies the following:
1. Given $a, b \neq 0$, then $\nu(a) \leq \nu(ab)$.
2. Given, $a, b \in D$ and $b \neq 0$, there exists some $q, r \in D$ such that $a = bq + r$ and either $r = 0$ or $\nu(r) < \nu(b)$.
**Example**. Absolute value on $\mathbb{Z}$ is a Euclidian validation.
**Example**. Absolute value on $\mathbb{Z}$ is a Euclidean validation.
**Example**. Degree on $F[x]$ is a Euclidian validation.
**Example**. Degree on $F[x]$ is a Euclidean validation.
**Example**. $\nu(a + bi) = a^2 + b^2$ is a Euclidian validation over $\mathbb{Z}[i]$.
**Example**. $\nu(a + bi) = a^2 + b^2$ is a Euclidean validation over $\mathbb{Z}[i]$.
**Theorem**. 18.21: Every Euclidian domain is a PID.
**Theorem**. 18.21: Every Euclidean domain is a PID.
**Corollary**. Every Euclidian domain is a UFD.
**Corollary**. Every Euclidean domain is a UFD.
---
**Definition**. Given a polynomial $p(x) \in D$, with $D$ bein an integer domain, we say that the *content* of $p(x)$ is the greatest common divisor of its coefficients. Additionally, if the content is $1$, we say that $p(x)$ is *primitive*.
**Definition**. Given a polynomial $p(x) \in D$, with $D$ being an integer domain, we say that the *content* of $p(x)$ is the greatest common divisor of its coefficients. Additionally, if the content is $1$, we say that $p(x)$ is *primitive*.
**Theorem**. 18.24: Let $D$ be a UFD, and $f(x), g(x) \in D[x]$ be primitive. Then, $f(x)g(x)$ is primitive.
@@ -94,7 +94,7 @@ As a direct consequence, we see the following.
**Theorem**. If $D$ is as UFD, then $D[x]$ is a UFD.
**Corollary**. This theorem has several collaries:
**Corollary**. This theorem has several corollaries:
1. Given a field $F$, since $F$ is a PID, it is also a UFD. Thus, $F[x]$ is a UFD.
2. The ring of polynomials over integers, $\mathbb{Z}[x]$ is a UFD.

22
docs/math/abstract-algebra/DF-10-modules.md
View File

@@ -15,13 +15,13 @@
**Remark**. Modules over a field $F$ and vector spaces over $F$ are identical.
**Definition** An *R-submodule* is a subset$N \subseteq M$ which is closed under the action taken forall $r \in R$. That is, given $r \in R, n \in N$, tthen $rn \in N$. Every module has at least two submodules: itself and the trivial (empty) submodule.
**Definition** An *R-submodule* is a subset$N \subseteq M$ which is closed under the action taken forall $r \in R$. That is, given $r \in R, n \in N$, then $rn \in N$. Every module has at least two submodules: itself and the trivial (empty) submodule.
**Remark**. If $F$ is a field, submodules are equivilent to subspaces.
**Remark**. If $F$ is a field, submodules are equivalent to subspaces.
---
**Example**. Let $F$ be a field and $F[x]$ a polynomial ring. Then, let $V$ be a vector space of $F$, and $T$ be a linear transformaion from $V$ to itself. That is, $V: T \rightarrow T$. We know that $V$ is an $F$-module. We will want to show that $V$ can be written as an $F[x]$-module for some choice of $T$. That is, we want an action $F[x] \cross V \rightarrow V$.
**Example**. Let $F$ be a field and $F[x]$ a polynomial ring. Then, let $V$ be a vector space of $F$, and $T$ be a linear transformation from $V$ to itself. That is, $V: T \rightarrow T$. We know that $V$ is an $F$-module. We will want to show that $V$ can be written as an $F[x]$-module for some choice of $T$. That is, we want an action $F[x] \cross V \rightarrow V$.
Now, for a given linear transformation $T$, consider some polynomial $p(x) = a_n x^n + \ldots + a_0$ and some $v \in V$. We define $p(x) \cross v$ by$
@@ -42,17 +42,17 @@ with $T^n$ being defined as applying $T$ a total of $n$ times.
**Recall**. The center of a ring $A$ is the subring $A'$ such that for all $x, y \in R'$, then $xy = yx$. In other words, it is the commutative subring of $A$.
**Definition**. Given two $R$-algebras $A, B$, an *$R$-algebra homomorphhism$ is a ring homomorphism $\varphi: A \rightarrow B$ that maps $1_A \rightarrow 1_B$ such that $\varphi(ra) = r\varphi(a)$.
**Definition**. Given two $R$-algebras $A, B$, an *$R$-algebra homomorphism$ is a ring homomorphism $\varphi: A \rightarrow B$ that maps $1_A \rightarrow 1_B$ such that $\varphi(ra) = r\varphi(a)$.
## Section 10.2 - Quotient Modules and Module Homomorphisms
**Definition**. Let $R$ be a ring and $M, N$ be $R$-modules. then a ring homomorphhism $\varphi: M \rightarrow N$ is an *$R$-module homomorphism* if for all $r \in R$, $\varphi(rx) = r\varphi(x)$.
**Definition**. Let $R$ be a ring and $M, N$ be $R$-modules. then a ring homomorphism $\varphi: M \rightarrow N$ is an *$R$-module homomorphism* if for all $r \in R$, $\varphi(rx) = r\varphi(x)$.
**Theorem**. An $R$-module homomorphism is an *isomorphism* if it is 1-1 and onto, and said modules are *isomorphic*.
**Definition**. Let $M, N$ be $R$-modules. The set $\Hom_R(M, N)$ is the set of all homomorphisms from $M$ to $N$.
**Promposition**. Let $M$, $N$, and $L$ be $R$-modules. Then,
**Proposition**. Let $M$, $N$, and $L$ be $R$-modules. Then,
1. A function $\varphi: M \rightarrow N$ is an $R$-module homomorphism if and only if $\varphi(rx + y) = r\varphi(x) + \varphi(y)$ for all $x, y \in M$ and $r \in R$.
2. Let $\varphi, \psi \in \Hom_R(M, N)$. Then, define $\varphi + \psi$ as
@@ -85,13 +85,13 @@ $$
This is the smallest submodule that contains both $A$ and $B$.
**Theorem**. First Isomorphism Theorem. Let $M, N$ be $R$-modules, and $\varphi: M \rightarrow N$ be an $R$-module homomorphhiism. Then, $\ker \varphi$ is a submodule of $M$, and $M / \ker \varphi \cong \varphi(M)$.
**Theorem**. First Isomorphism Theorem. Let $M, N$ be $R$-modules, and $\varphi: M \rightarrow N$ be an $R$-module homomorphism. Then, $\ker \varphi$ is a submodule of $M$, and $M / \ker \varphi \cong \varphi(M)$.
**Theorem**. Second Isomorphism Theorem. Let $A, B$ be submodules of the $R$-module $M$. Then, $(A + B)/B \cong A/(A \cap B)$.
**Theorem**. Third Isomorphism Theorem. Let $M$ be an $R$-module, and $A \subseteq B$ be submodules of $M$. Then, $\frac{M/A}{B/A} \cong M/B$.
**Theorem**. Lattice Isomorphism Theorem. Let $N$ be a submodule of the $R$-module $M$. Then, there is a bijection between submoudles of $M$ containing $N$ and submodules of $M/N$. This is given by $A \leftrightarrow A/N$, for $A \supseteq N$.
**Theorem**. Lattice Isomorphism Theorem. Let $N$ be a submodule of the $R$-module $M$. Then, there is a bijection between submodules of $M$ containing $N$ and submodules of $M/N$. This is given by $A \leftrightarrow A/N$, for $A \supseteq N$.
## Section 10.3 - Generation of Modules, Direct Sums, and Free Modules
@@ -99,7 +99,7 @@ This is the smallest submodule that contains both $A$ and $B$.
1. The *sum* of $N_1, \ldots, N_n$ is the set of all finite sums of elements from the sets $N_i$. That is, $N_1, \ldots, N_n := \{a_1 + a_2 + \ldots + a_n | a_i \in N_i\}$
2. For any subset $A$ of $M$, let $RA = \{r_1 a_1 + r_2 a_2 + \ldots + r_m a_m | r_i \in R, a_i \in A\}$. If $N$ is a submodule of $M$ such that $N = RA$, then $A$ is called the *generating set* for $N$.
3. A submodule $N$ of $M$ is *finitely generaated* if there is some finite subset $A$ of $M$ such that $N = RA$. That is, $N$ is generated by some finite subset.
3. A submodule $N$ of $M$ is *finitely generated* if there is some finite subset $A$ of $M$ such that $N = RA$. That is, $N$ is generated by some finite subset.
4. A submodule of $M$ (up to equality) is $cyclic$ if there exists some element $a \in M$ such that $N = Ra = \{ra | r \in R\}$.
**Definition**. Let $M_1, \ldots, M_k$ be a collection of $R$-modules. Then, the *direct product* is defined as
@@ -118,13 +118,13 @@ This direct product is in itself an $R$-module.
**Definition**. An $R$-module $F$ is said to be *free* on the subset $A$ of $F$ if for every nonzero $x \in F$, there exists nonzero elements $r_1, \ldots, r_n$ of $R$ and unique $a_1, \ldots, a_n$ such that $x = r_1 a_1 + \ldots + r_n a_n$ for some $n \in \mathbb{Z}^+$. That is, $A$ is a *basis* or *set of free generators* of $F$.
**Theorem**. For any set $A$, there is a free $R$-module $F(A)$ on $A$ such that $F(A)$ satisfies the universal property: if $M$ is any $R$-module, and $\varphi: A \rightarrow M$ is a map of sets, there eixts a unique $R$-module homomorphism: $\Phi: F(A) \rightarrow M$ such that $\Phi(a) = \varphi(a)$ for all $a \in A$.
**Theorem**. For any set $A$, there is a free $R$-module $F(A)$ on $A$ such that $F(A)$ satisfies the universal property: if $M$ is any $R$-module, and $\varphi: A \rightarrow M$ is a map of sets, there exists a unique $R$-module homomorphism: $\Phi: F(A) \rightarrow M$ such that $\Phi(a) = \varphi(a)$ for all $a \in A$.
**Corollary**. If $F_1$ and $F_2$ are free modules on $A$, then there is a unique isomorphism between $F_1$ and $F_2$, which is the identity map on A.
**Corollary**. If $F$ is a free $R$-module with basis $A$, then $F \cong F(A)$.
**Definition** For a free module $F$ with basis $A$, if $R$ is commutative, tthen the *rank* of $F$ is the cardinality of $A$.
**Definition** For a free module $F$ with basis $A$, if $R$ is commutative, then the *rank* of $F$ is the cardinality of $A$.
## Section 10.4 - Tensor Products of Modules

8
docs/math/abstract-algebra/DF-12-modules-pids.md
View File

@@ -1,4 +1,4 @@
# Chapter 12 - Modules over Principal Ideal Domains
# Dummit & Foote Chapter 10 Chapter 12 - Modules over Principal Ideal Domains
## Section 12.1 The Basic Theory
@@ -8,7 +8,7 @@ $$
M_1 \subseteq M_2 \subseteq \ldots
$$
there exists some $k \in \mathbb{N}$ such thaht given any $n \in \mathbb{N}$ with $n \geq k$, then $M_n = M_k$.
there exists some $k \in \mathbb{N}$ such that given any $n \in \mathbb{N}$ with $n \geq k$, then $M_n = M_k$.
**Definition**. A ring $R$ is *Noetherian* if it is Noetherian when viewed as a left $R$-module over itself.
@@ -20,7 +20,7 @@ there exists some $k \in \mathbb{N}$ such thaht given any $n \in \mathbb{N}$ wit
**Corollary**. If $R$ is a principal ideal domain (PID), then all nonempty set of ideals of $R$ has a maximal element. Additionally, $R$ is as Noetherian ring.
**Proposition**. Let $R$ be an integral doman, and $M$ be a free $R$-module of rank $n < \infty$. Then, given $S$ is subset $M$ with $|S| > n$, the elements of $S$ are $R$-linearly dependent.
**Proposition**. Let $R$ be an integral domain, and $M$ be a free $R$-module of rank $n < \infty$. Then, given $S$ is subset $M$ with $|S| > n$, the elements of $S$ are $R$-linearly dependent.
**Definition**. Given $R$ an integral domain and $M$ an $R$-module,
@@ -49,5 +49,5 @@ Additionally, if $R$ is a PID, as $\Ann_R(N)$ is an ideal, $\Ann(N) = (n)R$ and
**Theorem**. Let $R$ be a principal ideal domain, and $M$ be a free $R$-module of finite rank $m$, and $N$ be a submodule of $M$. Then,
1. $N$ is a free submodule with rank $n \leq m$.
2. There exiss a basis $y_1, y_2, \ldots, y_m$ of $M$ so that $r_1 y_1, r_2 y_2, \ldots, r_m y_n$ is a basis of $N$ for some $r_i \in R$ and $r_1 | r_2 | \ldots | r_n$
2. There exists a basis $y_1, y_2, \ldots, y_m$ of $M$ so that $r_1 y_1, r_2 y_2, \ldots, r_m y_n$ is a basis of $N$ for some $r_i \in R$ and $r_1 | r_2 | \ldots | r_n$

6
docs/math/diffeq/1-intro.md
View File

@@ -16,15 +16,15 @@ This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/
\[ a_n(t)y^{(n)}(t) + a_{n-1}(t)+y^{n-1}(t) + \ldots + a_1(t)y'(t) + a_0(t)y(t) = g(t) \]
Note that $a_n(t)$ does not depeond on any derivative of $y$, so the presence of terms such as $e^y$ or $\sqrt{y'}$ signal that the equation is *nonlinear*.
Note that $a_n(t)$ does not depend on any derivative of $y$, so the presence of terms such as $e^y$ or $\sqrt{y'}$ signal that the equation is *nonlinear*.
**Definition**. The *solution(s)* to a differential equation over an inverval $\alpha < t < \beta$ are any funcion(s) $y(t)$ that satisfy the differential equation.
**Definition**. The *solution(s)* to a differential equation over an interval $\alpha < t < \beta$ are any function(s) $y(t)$ that satisfy the differential equation.
**Definition**. The *initial conditions* are a condition or set of conditions that constrain the possible solution sets.
**Definition**. An *Initial Value Problem* is a differential equation along with the appropriate boundary or initial conditions.
**Definition**. The *integral of validity* for a solution to a differential equation is the largest possible interval containing the initial coniditions for which the solution is valid.
**Definition**. The *integral of validity* for a solution to a differential equation is the largest possible interval containing the initial conditions for which the solution is valid.
**Definition**. The *general solution* to a differential equation is the most general form a solution to a differential equation can take without requiring the initial conditions.

4
docs/math/diffeq/2-1st-order.md
View File

@@ -24,7 +24,7 @@ $$
\mu(t)\frac{dy}{dt} + \mu'(t)y = \mu(t)g(t)
$$
The left of the preceeding equation is simply the product rule, so we can write $(\mu(t)y(t))' = \mu(t)g(t)$. Take the integral of both sides.
The left of the preceding equation is simply the product rule, so we can write $(\mu(t)y(t))' = \mu(t)g(t)$. Take the integral of both sides.
\begin{align}
\int (\mu(t)y(t))' dt &= \int \mu(t)g(t) \\
@@ -88,7 +88,7 @@ Let the following differential equation of the following forms be given.
\frac{dy}{dx} &= N(y)M(x) \\
\end{align}.
For the sake of simplicty, select the following form:
For the sake of simplicity, select the following form:
$$
N(y) \frac{dy}{dx} = M(x)

24
docs/math/diffeq/3-2nd-order.md
View File

@@ -14,21 +14,21 @@ $$ ay'' + by' + cy = g(t) $$
This is a second-order differential equation with constant coefficients.
**Definition**. In the event that $g(t) = 0$, we say the equation is *homogenous*. Otherwise, the equation is *nonhomogenous*.
**Definition**. In the event that $g(t) = 0$, we say the equation is *homogenous*. Otherwise, the equation is *nonhomogeneous*.
**Definition**. Principal of Superposition. Let $y_1(t)$ and $y_2(t)$ be solutions to a linear, homogenous differential equation. Then, any linear combination of said solutions is also a solution to the differential equation. In other words, with $c_1, c_2 \in \mathbb{R}$, the following is a solution to a differential equation.
$$ y(t) = c_1 y_1(t) + c_2 y_2(t) $$
Given a second-order homogenous differential equation with constant coeffictions, we assume solutions of the following form:
Given a second-order homogenous differential equation with constant coefficients, we assume solutions of the following form:
$$ y(t) = e^{rt} $$
Substituting this equation into the differential equationm, we see the following:
Substituting this equation into the differential equation, we see the following:
$$ e^{rt}(ar^2 + br + c) = 0 $$
Thus, we allow the *charactaristic equation* of the differential equation to be as follows:
Thus, we allow the *characteristic equation* of the differential equation to be as follows:
$$ ar^2 + br + c = 0 $$
@@ -36,7 +36,7 @@ $$ ar^2 + br + c = 0 $$
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx).
When the two roots to the charactaristic equation are discrete roots in the real numbers, we see the following solutions.
When the two roots to the characteristic equation are discrete roots in the real numbers, we see the following solutions.
$$ y_1(t) = e^{r_1 t} $$
@@ -50,7 +50,7 @@ $$ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} $$
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx).
Let the solutions to the charactaristic equation be of the following form:
Let the solutions to the characteristic equation be of the following form:
$$ r_{1,2} = \lambda \pm \mu i $$
@@ -64,7 +64,7 @@ Recall Euler's Formula:
$$ e^{i \theta} = \cos \theta + i \sin \theta $$
A colliloquy of Euler's formula is the following:
A corollary of Euler's formula is the following:
$$ e^{-i \theta} = \cos(-\theta) + i \sin(-\theta) = \cos \theta - i \sin \theta $$
@@ -83,7 +83,7 @@ $$ y(t) = c_1 e^{\lambda t} \cos(\mu t) + c_2 e^{\lambda t} \sin(\mu t) $$
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/RepeatedRoots.aspx).
Assume the solutions to the charactaristic equations are $r = r_1 = r_2$. Thus, the two equations $y_t(t)$ and $y_2(t)$ are not linearly independent.
Assume the solutions to the characteristic equations are $r = r_1 = r_2$. Thus, the two equations $y_t(t)$ and $y_2(t)$ are not linearly independent.
After a *lot* of algebra, we see that
@@ -112,7 +112,7 @@ $$
**Definition**. If $W(f, g) \neq 0$, then $f(t)$ and $g(t)$ are said to form a *fundamental set of solutions*, and can be superimposed to form the general solution.
## Section 3.8 - Nonhomogenous Differential Equations
## Section 3.8 - Nonhomogeneous Differential Equations
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx).
@@ -120,11 +120,11 @@ Assume we have the differential equation as follows:
$$ y'' + p(t) y' + q(t) y = g(t) $$
The equivilent homogenous differential equation is
The equivalent homogenous differential equation is
$$ y'' + p(t) y' + q(t) y = 0 $$
**Theorem**. Assume $Y_1(t)$, $Y_2(t)$ are solutions to the nonhomogenous differential equations. Then, $Y_1(t) - Y_2(t)$ is a solution to the homogenous differential equation. This can be proved by substitution.
**Theorem**. Assume $Y_1(t)$, $Y_2(t)$ are solutions to the nonhomogeneous differential equations. Then, $Y_1(t) - Y_2(t)$ is a solution to the homogenous differential equation. This can be proved by substitution.
Thus, with $y_h(t)$ the solution to the homogenous problem, and $y_p(t)$ the solution to this particular problem, we can say that the general form of the solution to this differential equation is
@@ -156,7 +156,7 @@ Assume we have the differential equation as follows:
$$ y'' + p(t) y' + q(t) y = g(t) $$
The equivilent homogenous differential equation is
The equivalent homogenous differential equation is
$$ y'' + p(t) y' + q(t) y = 0 $$

4
docs/math/diffeq/4-laplace.md
View File

@@ -60,7 +60,7 @@ $$
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithLaplace.aspx).
**Theorum**. Given a function $f(t)$ with $C^n$ continuity, then
**Theorem**. Given a function $f(t)$ with $C^n$ continuity, then
$$
\mathcal{L} \{ f^{(n)} (t) \} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \ldots - s f^{(n-2)} (0) - f^{(n-1)} (0)
@@ -77,6 +77,6 @@ $$
We can take the Laplace transformation of an IVP, solve for $Y(s)$, then take the inverse to find the solution.
## Section 4.6 - Nonconstant Coefficient IVPs
## Section 4.6 - Non-constant Coefficient IVPs
This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithNonConstantCoefficient.aspx).

4
docs/math/real-analysis/2-reals.md
View File

@@ -56,7 +56,7 @@
2. $|ab| = |a||b|$
3. $|a + b| \leq |a| + |b|$
**Corollary**. Givem $a, b \in \mathbb{R}$, then $\abs{\abs{a} - \abs{b}} \leq \abs{a - b}$.
**Corollary**. Given $a, b \in \mathbb{R}$, then $\abs{\abs{a} - \abs{b}} \leq \abs{a - b}$.
**Remark**. Every field has at least one absolute value function.
@@ -86,7 +86,7 @@ $$
---
**Decimals**. Let $x \in \mathbb{R}$ such that $x > 0$. By the archimedian property, there exists some $b_0 \in \mathbb{N} \cup {0}$ such that $b_0 < x < b_0 + 1$. We can repeat this to see
**Decimals**. Let $x \in \mathbb{R}$ such that $x > 0$. By the archimedean property, there exists some $b_0 \in \mathbb{N} \cup {0}$ such that $b_0 < x < b_0 + 1$. We can repeat this to see
$$
x = b_0 + \frac{b_1}{10} + \frac{b_2}{100} + \ldots + \frac{b_n}{100^n} + \ldots

6
docs/math/real-analysis/3-sequences-series.md
View File

@@ -16,7 +16,7 @@ $$
(x_n) = (a, ar, ar^2, ar^3, \ldots)
$$
**Example**. The *arithmatic sequence*, given base $a \in \mathbb{R}$ and distance $d \in \mathbb{R}$,
**Example**. The *arithmetic sequence*, given base $a \in \mathbb{R}$ and distance $d \in \mathbb{R}$,
$$
(x_n) = (a, a + d, a + 2d, a + 3d, \ldots)
@@ -52,7 +52,7 @@ $$
2. $x_n \cdot y_n \rightarrow xy$
3. If $x_n \neq 0$ for all $n$, then $\frac{1}{x_n} \rightarrow \frac{1}{x}$
**Theorem**. Suppose $(x_n)$ aand $(y_n)$ are convergent sequences and $N \in \mathbb{N}$. Then,
**Theorem**. Suppose $(x_n)$ and $(y_n)$ are convergent sequences and $N \in \mathbb{N}$. Then,
1. If $x_n \leq y_n$ for all $n \geq N$, then $\lim(x_n) \leq \lim(y_n)$
2. If $x_n \leq a$ for all $n \geq N$, then $\lim(x_n) \leq a$
@@ -110,7 +110,7 @@ $$
## Section 3.7 - Series
**Definition**. Let $(x_n)$ be a sequence in $\mathbb{R}$. Then, the *infinite series genearted by $X$* is the sequence $S = (s_n)$ with terms
**Definition**. Let $(x_n)$ be a sequence in $\mathbb{R}$. Then, the *infinite series generated by $X$* is the sequence $S = (s_n)$ with terms
$$
s_1 = x_1; \; s_{n+1} = s_n + x_{n+1}

14
docs/math/real-analysis/4-limits.md
View File

@@ -1,12 +1,12 @@
# Chapter 4 - Limits
## Secrion 4.1 - Limits of Functions
## Section 4.1 - Limits of Functions
**Definition**. Let $A \subseteq \mathbb{R}$. Then, a point $c \in \mathbb{R}$ is a *cluster point* of $A$ if for every $\delta > 0$, the $\delta$-neighborhood of $c$ contains a point $a \in A$ such thhat $a \neq c$. That is, there exists some $a$ such that $0 < |a - c| < \delta$.
**Definition**. Let $A \subseteq \mathbb{R}$. Then, a point $c \in \mathbb{R}$ is a *cluster point* of $A$ if for every $\delta > 0$, the $\delta$-neighborhood of $c$ contains a point $a \in A$ such that $a \neq c$. That is, there exists some $a$ such that $0 < |a - c| < \delta$.
**Theorem**. A real number $c$ is a cluster point for a set $A$ if and only if there exists a sequence $(a_n)$ in $A\\ \{c\}$ such that $a_n \rightarrow c$
**Corollary**. A real number $c$ is a cluster point of a set $A$ if and only if every $\delta$-neighborhood conttains infinitely many points of $A$.
**Corollary**. A real number $c$ is a cluster point of a set $A$ if and only if every $\delta$-neighborhood contains infinitely many points of $A$.
**Definition**. The set of every cluster point of $A$ is called the *derived set* of $A$, and denoted $A'$.
@@ -18,21 +18,21 @@
---
**Definition**. Suppose $f: A \rightarrow \mathbb{R}$ is a function with domain $A \subseteq \mathbb{R}$, and let $c \in A$ be a cluster point of $A$. then, a real number $L$ is a *limit of $f$ at $c$* if goven any $\epsilon > 0$, there exists some $\delta > 0$ such that
**Definition**. Suppose $f: A \rightarrow \mathbb{R}$ is a function with domain $A \subseteq \mathbb{R}$, and let $c \in A$ be a cluster point of $A$. then, a real number $L$ is a *limit of $f$ at $c$* if given any $\epsilon > 0$, there exists some $\delta > 0$ such that
$$
0 < |x-c| < \delta \Rightarrow |f(x) - L| < \epsilon
$$
**Therorem**. For a given function and cluster point, there can be at most one limit at said point.
**Theorem**. For a given function and cluster point, there can be at most one limit at said point.
**Theorem**. Let $A \subseteq \mathbb{R}$ and $f: A \rightarrow \mathbb{R}$. Then, to show that $lim_{x \rightarrow c} f(x) = L$, it suffices to show that for every sequence $(a_n)$ in $A\\ \{c\}$, the sequence $(f(a_n))$ converges tto $L$.
---
**Definition**. The *extended real numbers* are $\hat{\mathbb{R}} = \mathbb{R} \cup \{ \infty, -\infty \}$ are a totally-ordered set witth supremum and infimum. Note that this set is no longer a field.
**Definition**. The *extended real numbers* are $\hat{\mathbb{R}} = \mathbb{R} \cup \{ \infty, -\infty \}$ are a totally-ordered set with supremum and infimum. Note that this set is no longer a field.
**Definition**. At any point $c$, the limitt of $f$ at $c$ is infinite if given some $\alpha$, there exists some $V_\delta(c)$ such that forr all $x \in V_\epsilon(c)$, then $f(x) \in V_\alpha(\infty)$.
**Definition**. At any point $c$, the limit of $f$ at $c$ is infinite if given some $\alpha$, there exists some $V_\delta(c)$ such that for all $x \in V_\epsilon(c)$, then $f(x) \in V_\alpha(\infty)$.
**Definition**. The limit of a function at infinity is defined if for a given $\epsilon$, there exists some $\alpha$ so that there exists some $V_\delta(c)$ such that for all $x \in A$,

16
docs/math/real-analysis/5-continuity.md
View File

@@ -1,4 +1,4 @@
# Chapter 5 - Continuiy
# Chapter 5 - Continuity
## Section 5.1 - Continuous Functions
@@ -8,7 +8,7 @@ $$
|x - a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon
$$
Note that if $a$ is an *isolateed point* of $A$, that is, not a cluster point, then $a$ is automatically continuous.
Note that if $a$ is an *isolated point* of $A$, that is, not a cluster point, then $a$ is automatically continuous.
If $a$ is a cluster point of $A$, then this definition collapses to the definition of $\lim_{x \rightarrow a} f(x) = f(a)$.
@@ -20,15 +20,15 @@ Note that a function cannot be continuous at a point outside of its domain, even
---
**Definition**. Let $(S, d_S)$ and $(T, d_T)$ be metric spaces. A function $f: S \rightarrow T$ is continuous at a point $a \in S$ if given any $\epsilon > 0$, there exists some $\delta > 0$ such thaat for all $x \in S$,
**Definition**. Let $(S, d_S)$ and $(T, d_T)$ be metric spaces. A function $f: S \rightarrow T$ is continuous at a point $a \in S$ if given any $\epsilon > 0$, there exists some $\delta > 0$ such that for all $x \in S$,
$$
d_S(x, a) < \delta \Rightarrow d_T(f(x), f(a)) < \epsilon
$$
**Theorem**. A function $f: S \rightarrow T$ is continuous at a point $a \in A$ if and only if given some neighborhood $V(f(a)) \in B$, there xists some $U(a) \in A$ such that $f(U) \subseteq V$.
**Theorem**. A function $f: S \rightarrow T$ is continuous at a point $a \in A$ if and only if given some neighborhood $V(f(a)) \in B$, there exists some $U(a) \in A$ such that $f(U) \subseteq V$.
## Section 5.2 - Combinations of continuous Functions.
## Section 5.2 - Combinations of continuous Functions
**Theorem**. Let $f, g: A \rightarrow \mathbb{R}$ be continuous at $a \in A$. Then,
@@ -51,15 +51,15 @@ $$
**Corollary**. Let $f: A \rightarrow \mathbb{R}$ be a continuous function, with $A$ being a compact subset of metric space $S$. Then, $f(A)$ is closed and bounded. Moreover, there exists a $p, q \in A$ such that $f(p)$ and $f(q)$ are the supremum and infimum of $f(A)$.
**Corollary**. Maximum-Minimum Theorem. If $I = [a, b]$ is a closed and bounded interval and $f: I \rightarrow \mathbb{R}$ is continuous on $I$, then $f$ has an absolute minumum and maximum on $I$.
**Corollary**. Maximum-Minimum Theorem. If $I = [a, b]$ is a closed and bounded interval and $f: I \rightarrow \mathbb{R}$ is continuous on $I$, then $f$ has an absolute minimum and maximum on $I$.
---
**Theorem**. Let $S, T$ be metric spaces and $A \subseteq S$. Then, if $f: A \rightarrow T$ is continuous on $A$, and $A$ is a connected subset of $S$, then $f(A)$ is a connected subset of $T$.
**Corollary**. Suppose that $I$ is an interval. Let $f: I \rightarrow \mathbb{R}$ be continuous on $I$. Then, $f(I)$ is an intterval.
**Corollary**. Suppose that $I$ is an interval. Let $f: I \rightarrow \mathbb{R}$ be continuous on $I$. Then, $f(I)$ is an interval.
**Theorem**. (Bolzano's) Invermediate Value Theorem. Suppose $f: [a, b] \rightarrow \mathbb{R}$ is continuous on $[a, b]$ with $a \neq b$. Then, given some $k$ such that $f(a) < k < f(b)$, there exists some $c \in (a, b)$ such that $k = f(c)$.
**Theorem**. (Bolzano's) Intermediate Value Theorem. Suppose $f: [a, b] \rightarrow \mathbb{R}$ is continuous on $[a, b]$ with $a \neq b$. Then, given some $k$ such that $f(a) < k < f(b)$, there exists some $c \in (a, b)$ such that $k = f(c)$.
---

6
docs/physics/electrostatics/2-coulomb.md
View File

@@ -98,11 +98,11 @@ $$
We can verify that $\curl{(\vb{r}-\vb{r'})} = 0$, cancelling the first term. Additionally, $\curl{\frac{1}{\abs{\vb{r}-\vb{r'}}^3}} = -3 \frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^5}$, which when crossed with $\vb{r}-\vb{r'}$, will cancel. Thus, all terms in the curl cancel, so for a static field, the curl is zero.
## Section 2.4 - Eletric and Magnetic Flux Densities
## Section 2.4 - Electric and Magnetic Flux Densities
The electric and magnetic flux density vectors are given by $\epsilon_0 \vb{E}$ and $\mu_0 \vb{H}$.
Now, given $S$ is a surfance enclosing $Q_e$ or $Q_m$ total charge, we denotate flux as following:
Now, given $S$ is a surface enclosing $Q_e$ or $Q_m$ total charge, we denote flux as following:
$$
\Phi_e = \epsilon_0 \int_S \vb{E} \vdot \vu{n} \dd = Q_e S \text{ or } \Phi_m = \mu_0 \int_S \vb{H} \vdot \vu{n} \dd S = Q_m
@@ -131,4 +131,4 @@ $$
**Definition**. This is known as *Gauss' Law*.
With applicable symnetry, the integral factor becomes simply $E(r)*A$, where $A$ is the area of the surface at $r$.
With applicable symmetry, the integral factor becomes simply $E(r)*A$, where $A$ is the area of the surface at $r$.

14
docs/physics/electrostatics/3-electro-magnetic-potentials.md
View File

@@ -14,7 +14,7 @@ $$
\vb{E}(\vb{r}) = \frac{1}{q_e} \vb{F_e}(\vb{r}) = - \frac{1}{q_e} \grad{U_e(\vb{r})} = -\grad{V_e(\vb{r})}
$$
The units of electrostatic potential is Joule/Coublomb, also known as a Volt. Thus, the units of the electric field should be expressed in Volts/meter. Similarly,
The units of electrostatic potential is Joule/Coulomb, also known as a Volt. Thus, the units of the electric field should be expressed in Volts/meter. Similarly,
$$
\vb{H}(\vb{r}) = \frac{1}{q_m} \vb{F_m}(\vb{r}) = - \frac{1}{q_m} \grad{U_m(\vb{r})} = -\grad{V_m(\vb{r})}
@@ -80,7 +80,7 @@ $$
W_n = \frac{1}{2} \frac{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
$$
For the sake of symnetry, sum overall charges and divide by 2.
For the sake of symmetry, sum overall charges and divide by 2.
$$
W_n = \frac{1}{2} \frac{1}{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
@@ -93,7 +93,7 @@ W_n = \frac{1}{2} \sum_{i = 1}^{N}Q_{ei} \sum_{i \neq j}^{N} \frac{1}{4 \pi \eps
= \frac{1}{2}\sum_{i = 1}^{N} Q_{ei} V(\vb{r_i})
$$
We can rewrite this as a Reimann sum and convert to an integral.
We can rewrite this as a Riemann sum and convert to an integral.
$$
W_e = \frac{1}{2} \int_V p_e(\vb{r}) V_e(\vb{r}) \dd V ; \quad
@@ -129,7 +129,7 @@ $$
We know that $\vb{E}(\vb{r}) = -\div{V_e(\vb{r})}$ and $\vb{H}(\vb{r}) = -\div{V_m(\vb{r})}$
Combinind this, as well as the first of the Maxwell equations, we see that
Combined this, as well as the first of the Maxwell equations, we see that
$$
\div{\vb{E}} = -\div{\grad{V_e}} = - \laplacian{V_e} = \frac{\rho_e}{\epsilon_0}
@@ -139,7 +139,7 @@ $$
\div{\vb{H}} = -\div{\grad{V_m}} = - \laplacian{V_m} = \frac{\rho_m}{\mu_0}
$$
The last inequatlity is called the Poisson Equation, or the inhomogenous Laplace equation.
The last inequality is called the Poisson Equation, or the inhomogeneous Laplace equation.
To solve this equation, we define a Green function as follows:
@@ -147,7 +147,7 @@ $$
\laplacian G(\vb{r}, \vb{r'}) = \delta(\vb{r} - \vb{r'})
$$
Now, we can construct a potential function in terms of said green function that satisfies the lapalce equation.
Now, we can construct a potential function in terms of said green function that satisfies the Laplace equation.
$$
V_e(\vb{r}) = - \int_V G(\vb{r}, \vb{r'}) \frac{\rho_e(\vb{r'})}{\epsilon_0} \dd{V'}
@@ -273,4 +273,4 @@ $$
Note that as a quirk of the function, $P_n(1) = 1$ for all $n$.
We can apply these quadrupole and beyond terms to the volate or other equations, however, this becomes very messy.
We can apply these quadrupole and beyond terms to the violate or other equations, however, this becomes very messy.

12
docs/physics/electrostatics/4-conductors.md
View File

@@ -6,23 +6,23 @@ We will focus primarily on electric fields and charges. For the purposes for thi
## Section 4.2 - Electrostatic Properties of a Conductor
In a metal or conductor, there are plentiful charges not bound to a particular atom and are thus free to move throughought the material.
In a metal or conductor, there are plentiful charges not bound to a particular atom and are thus free to move throughout the material.
We note that there is no electric fiend inside a conductor, as charges internal to the material would move under the force it generates until they find a configuration that eliminates the field. This may happen, but not in electrostatics.
Additionally, as the field is zero, it follows from Maxwell's equations that there is no charge inside a conductor. However, charge may be present at the surface. For sufficiently symnetric charges, this charge may be calculated.
Additionally, as the field is zero, it follows from Maxwell's equations that there is no charge inside a conductor. However, charge may be present at the surface. For sufficiently symmetric charges, this charge may be calculated.
Consider any two points internal to the conductor. The voltage between said points is defined as $\int_A^B \vb{E} \vdot \dd{\vb{l}}$. Since $\vb{E} = 0$ inside the conductor, the volage difference must be zero. Thus, any two points in or on the surface (TODO: Why on the surface?) of a conductor must be at the same potential.
Consider any two points internal to the conductor. The voltage between said points is defined as $\int_A^B \vb{E} \vdot \dd{\vb{l}}$. Since $\vb{E} = 0$ inside the conductor, the voltage difference must be zero. Thus, any two points in or on the surface (TODO: Why on the surface?) of a conductor must be at the same potential.
The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $\dd{\vb{l}}$. Now, $\vb{E} \vdot \dd{\vb{l}} = \vb{E}_s \vdot \dd{\vb{l}}_s + \vb{E}_p \vdot \dd{\vb{l}}_p = \dd{V_s} + \dd{V_p}$, in terms of parallel and perpendicular components. The paralell voltage difference is zero, so the electric field must be zero.
The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $\dd{\vb{l}}$. Now, $\vb{E} \vdot \dd{\vb{l}} = \vb{E}_s \vdot \dd{\vb{l}}_s + \vb{E}_p \vdot \dd{\vb{l}}_p = \dd{V_s} + \dd{V_p}$, in terms of parallel and perpendicular components. The parallel voltage difference is zero, so the electric field must be zero.
Consider the surface of a conductor with surface charge density $\sigma_e$. A cyliner with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\epsilon_0} = \frac{A\sigma_e}{\epsilon_0}$. Thus, $\sigma_e = \epsilon_0 E$.
Consider the surface of a conductor with surface charge density $\sigma_e$. A cylinder with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\epsilon_0} = \frac{A\sigma_e}{\epsilon_0}$. Thus, $\sigma_e = \epsilon_0 E$.
## Section 4.3 - Exercises involving conductors at fixed potentials
Consider a square with left and right potentials $V(0, y) = V(l, y) = V_1$ and $V(x, 0) = V(x, l) = V_2$. Since we are uniform in $z$, we can say that $V(x, y) = X(x)Y(y)$ and apply separation of variables.
In spherical polar coordinates, we see that with azimuthal symnetry, $V(r, \theta) = \sum_{l=0}^\infty a_l r^l P_l(cos\theta)$ where $P_l(x)$ are Legendre polynomials.
In spherical polar coordinates, we see that with azimuthal symmetry, $V(r, \theta) = \sum_{l=0}^\infty a_l r^l P_l(cos\theta)$ where $P_l(x)$ are Legendre polynomials.
**Theorem**. 4.3.3: A Laplace equation's solution must be unique inside a volume $\Omega$ if $\int_{\dd{\Omega}}[\Phi(\vb{r})\grad{\Phi{\vb{r}}} \vdot \vu{n} \dd{S} = 0]$. With this, consider a surface $\dd{\Omega}$ that surrounds conductors. The integral vanishes if a) the potential is specified on each conductor or b) the total charge on each conductor is specified.

6
docs/physics/electrostatics/5-moving-charges.md
View File

@@ -1,6 +1,6 @@
# Chapter 5 - Electrodynamics with Moving Charges
## Section 5.1 - Currents in Steady-State Regine
## Section 5.1 - Currents in Steady-State Regime
We want to work in a steady-state system. Thus, we restrict ourselves to currents that do not change in time.
@@ -24,7 +24,7 @@ If we assume cylindrical coordinates and that $\vb{H}(vb{r}) = H_\varphi(s) \vu{
---
By Helmholtz Theorem, we know that $\vb{H}(\vb{r}) = \curl{\vb{A}(\vb{r})}$. For a current-carying wire, $\vb{A}(\vb{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{\dd{\vb{l'}}}{|\vb{r}-\vb{r'}|}$. Applying identities, we see the *Law of Biot and Savart$, where
By Helmholtz Theorem, we know that $\vb{H}(\vb{r}) = \curl{\vb{A}(\vb{r})}$. For a current-carrying wire, $\vb{A}(\vb{r}) = \frac{I_e}{4\pi} \int_{\text{wire}} \frac{\dd{\vb{l'}}}{|\vb{r}-\vb{r'}|}$. Applying identities, we see the *Law of Biot and Savart$, where
$$
\vb{H}(\vb{r}) = \int{I_e}{4\pi}\int_{\text{wire}} \frac{-(\vb{r}-\vb{r'}) \cross \dd{\vb{l'}}}{|\vb{r}-\vb{r'}|^3}
@@ -60,7 +60,7 @@ $$
Consider an electric charge moving with velocity $\vb{v}$ in a magnetic parallel plate capacitor with charge densities $\plusminus \sigma_m$. That is, $\mu_0 \vb{H} = \sigma_m \vu{z}$. Then, we can apply theorems to see the resulting force.
**Theorem**. *Lorentz Force Law* states that $\vb{F} = q_e \vb{v} \cross \u_0 \vb{H}$ in the presence of a magnetic field. In the presence of both an electic andmagnetic field, $\vb{F} = q_e (\vb{E} + \vb{v} \cross \u_0 \vb{H})$.
**Theorem**. *Lorentz Force Law* states that $\vb{F} = q_e \vb{v} \cross \u_0 \vb{H}$ in the presence of a magnetic field. In the presence of both an electric and magnetic field, $\vb{F} = q_e (\vb{E} + \vb{v} \cross \u_0 \vb{H})$.
**Theorem**. *Ampere's Force Law* states that generalizing the previous theorem, we can see that

4
docs/recipes/cookies.md
View File

@@ -9,7 +9,7 @@ Original: [Link](https://www.allrecipes.com/recipe/10813/best-chocolate-chip-coo
- In a medium bowl, mix together $\frac{1}{2}$ cup butter, $\frac{1}{2}$ cup sugar, $\frac{1}{2}$ cup brown sugar.
- To the wet ingredients, add 1 egg and 1 tsp. vanilla.
- To the wet ingredients, add a combination of 1 tsp. hot water and $\frac{1}{2}$ tsp. baking soda.
- Mix the wet and dry ingredients. Stir in 1 cup chocolate chips and optionallt $\frac{1}{2}$ cup walnuts.
- Mix the wet and dry ingredients. Stir in 1 cup chocolate chips and optionally $\frac{1}{2}$ cup walnuts.
- Bake for 10m at $350 \degree$ F.
## Peanut Butter Cookies
@@ -36,7 +36,7 @@ Original: [Link](https://www.allrecipes.com/recipe/10813/best-chocolate-chip-coo
## Snickerdoodle Cookies
- Preheat oven to $400 \degree$ F.
- Blend together $\frac{3}{4}$ cup sugar, $\frac{1}{4}$ cup butter (softened), $\frac{1}{4} cup shortening.
- Blend together $\frac{3}{4}$ cup sugar, $\frac{1}{4}$ cup butter (softened), $\frac{1}{4}$ cup shortening.
- Add 1 egg, 1 tsp. vanilla
- In a separate bowl, combine $1 \frac{1}{4}$ cups flour, 1 tsp. cream of tartar, $\frac{1}{2}$ tsp. baking soda, and $\frac{1}{4}$ tsp. salt
- Combine

4
docs/recipes/cupcakes.md
View File

@@ -4,9 +4,9 @@
- Preheat oven to $350 \degree$ F
- Combine $1 \frac{1}{4}$ cups flour, $\frac{3}{4}$ tsp. baking powder, $\frac{1}{4}$ tsp. baking soda.
- In a separate bowl, combine $\frac{1}{4}$ cup butter, $\frac{3}{4}$ cup sugar, $\frac{1}{4}$ cup vegetable oil, $\frac{1}{4}$ tsp. vaninna
- In a separate bowl, combine $\frac{1}{4}$ cup butter, $\frac{3}{4}$ cup sugar, $\frac{1}{4}$ cup vegetable oil, $\frac{1}{4}$ tsp. vanilla
- Add 2 eggs.
- Combine with half of the dry mixture. Add 6 tbsp. milk, 1 lemon of zest and juce. Stir. Add the rest of the dry ingredients.
- Combine with half of the dry mixture. Add 6 tbsp. milk, 1 lemon of zest and juice. Stir. Add the rest of the dry ingredients.
- Bake for 15-18 minutes at $350 \degree$ F.
### Frosting

6
docs/recipes/pies.md
View File

@@ -4,13 +4,13 @@
- Whisk $1 \frac{1}{4}$ cups of flour with $\frac{1}{4}$ tsp. salt
- Cut in $\frac{1}{2}$ cups of cubed butter (chilled), $\frac{1}{4}$ cups cold water.
- Refridgerate
- Refrigerate
## Pie Crust 2
- Mix $\frac{1}{3}$ cup flour, $\frac{1}{3}$ tsp. salt.
- Cut in $\frac{1}{2}$ cups of shortening, 3 tbsp. cold water.
- Refridgerate
- Refrigerate
## Pumpkin Pie 1
@@ -46,7 +46,7 @@ From: [Link](https://www.allrecipes.com/recipe/234374/apple-hand-pies/)
### Hand Pies
- Preheat oven to $400 \degree$ F.
- Split pie crust into 4. Place fillin in crust, fold.
- Split pie crust into 4. Place filling in crust, fold.
- Sprinkle with $\frac{1}{4}$ tsp. white sugar.
- Whisk 2 tsp. milk, 1 egg. Brush pastries.
- Bake at $400 \degree$ F. for 25-30 minutes.

4
docs/recipes/snacks.md
View File

@@ -16,11 +16,11 @@
## Chocolate Fudge
- Melt 1 bag of chocolate chips on low. Stir in 1 can of sweeteneed condensed milk. Pour into buttered or oiled tray.
- Melt 1 bag of chocolate chips on low. Stir in 1 can of sweetened condensed milk. Pour into buttered or oiled tray.
## Peanut Butter Fudge
- Melt $\frac{1}{2}$ cup of butter on medium heat. Stir in a 16oz bag of brown sugar, $\frac{1}{2}$ cup of milk.
- Remove from heat. Stir in $\frac{3}{4}$ cups of peanut butter and 1 tsp banilla.
- Remove from heat. Stir in $\frac{3}{4}$ cups of peanut butter and 1 tsp bvnilla.
- Add to $3 \frac{1}{2}$ cups of powdered sugar
- Pour into buttered or oiled tray.