Formatting Fix

docs/math/abstract-algebra/16-rings.md

@@ -1,4 +1,5 @@
 # Chapter 16 - Rings
+
 ## 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:
@@ -139,4 +140,4 @@ $$
 
 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.
 
-**Theorem**. 16.40: In a commutative ring with identity, every maximal ideal is also a prime ideal.
+**Theorem**. 16.40: In a commutative ring with identity, every maximal ideal is also a prime ideal.

docs/math/abstract-algebra/17-polynomial-rings.md

@@ -1,4 +1,5 @@
 # Chapter 17 - Polynomial Rings
+
 ## Section 17.1 - Polynomial Rings
 
 Throughout this chapter, we will assume that $R$ is a commutative ring with identity.
@@ -63,4 +64,4 @@ Then, if $p | a_i$ for $0 \leq i < n$, but $p \nmid a_n$ and $p^2 \nmid a_0$, th
 
 **Theorem**. If $F$ is a field, then every ideal in $F[x]$ is a principal ideal.
 
-**Theorem**. Let $F$ be a field, and suppose $p(x) \in F[x]$. Then, the ideal $<p(x)>$ is maximal if and only if $p(x)$ is irreducible.
+**Theorem**. Let $F$ be a field, and suppose $p(x) \in F[x]$. Then, the ideal $<p(x)>$ is maximal if and only if $p(x)$ is irreducible.

docs/math/abstract-algebra/18-integral-domains.md

@@ -98,4 +98,4 @@ As a direct consequence, we see the following.
 
 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.
-3. Given $D$ is a UFD, $D[x]$ is a UFD. Thus, $D[x_1, x_2]$ is a UFD, and by induction, $D[x_1, \ldots, x_n]$ is a UFD.
+3. Given $D$ is a UFD, $D[x]$ is a UFD. Thus, $D[x_1, x_2]$ is a UFD, and by induction, $D[x_1, \ldots, x_n]$ is a UFD.

docs/math/diffeq/1-intro.md

@@ -36,4 +36,4 @@ Note that $a_n(t)$ does not depeond on any derivative of $y$, so the presence of
 
 This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/DirectionFields.aspx).
 
-**Definition**. A directional field is the graph of a $t$ vs. $y(t)$, with vectors drawn at each point with a slope corresponding to $y'(t)$. Notably, each arrow will be pointed right (towards increasing $t$).
+**Definition**. A directional field is the graph of a $t$ vs. $y(t)$, with vectors drawn at each point with a slope corresponding to $y'(t)$. Notably, each arrow will be pointed right (towards increasing $t$).

docs/math/diffeq/2-1st-order.md

@@ -144,4 +144,4 @@ $$
 \frac{1}{1-n}v' + p(x)v = q(x)
 $$
 
-After solving, be sure to rewrite in terms of $y$.
+After solving, be sure to rewrite in terms of $y$.

docs/math/diffeq/3-2nd-order.md

@@ -32,7 +32,7 @@ Thus, we allow the *charactaristic equation* of the differential equation to be
 
 $$ ar^2 + br + c = 0 $$
 
-# Section 3.2 - Real & Distinct Roots
+## Section 3.2 - Real & Distinct Roots
 
 This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/RealRoots.aspx).
 
@@ -42,11 +42,11 @@ $$ y_1(t) = e^{r_1 t} $$
 
 $$ y_2(t) = e^{r_2 t} $$
 
-Thus, 
+Thus,
 
 $$ y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t} $$
 
-# Section 3.3 - Complex Roots
+## Section 3.3 - Complex Roots
 
 This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/ComplexRoots.aspx).
 
@@ -120,7 +120,7 @@ Assume we have the differential equation as follows:
 
 $$ y'' + p(t) y' + q(t) y = g(t) $$
 
-The equivilent homogenous differential equation is 
+The equivilent homogenous differential equation is
 
 $$ y'' + p(t) y' + q(t) y = 0 $$
 
@@ -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 equivilent homogenous differential equation is
 
 $$ y'' + p(t) y' + q(t) y = 0 $$
 
@@ -164,4 +164,4 @@ For this method, we must have $y_1(t)$ and $y_2(t)$ known. Through a lot of math
 
 $$
 y_p = -y_1 \int \frac{y_2(t)g(t)}{W(y_1, y_2)} dt + y_2 \int \frac{y_1(t)g(t)}{W(y_1, y_2)} dt
-$$
+$$

docs/math/diffeq/4-laplace.md

@@ -1,4 +1,5 @@
 # Section 4 - Laplace Transformations
+
 ## Section 4.1 - Definition
 
 This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx).
@@ -78,4 +79,4 @@ We can take the Laplace transformation of an IVP, solve for $Y(s)$, then take th
 
 ## Section 4.6 - Nonconstant Coefficient IVPs
 
-This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithNonConstantCoefficient.aspx).
+This section is from [Paul's Online Math Notes](https://tutorial.math.lamar.edu/Classes/DE/IVPWithNonConstantCoefficient.aspx).

docs/physics/electrostatics/1-math.md

@@ -1,4 +1,5 @@
 # Chapter 1 - Mathematics
+
 ## 1.5 - Dyads and Tensors
 
 **Definition**. A *dyadic* is a representation of two-ish vectors.

docs/physics/electrostatics/2-coulomb.md

@@ -81,7 +81,6 @@ $$
 
 The curl of an electrostatic or magnetostatic is relatively simple.
 
-
 $$
 \begin{align}
     \curl{E(\vb{r})} &= \frac{1}{4 \pi \epsilon_0} \int_V \rho_e(\vb{r'}) \curl{(\frac{\vb{r}-\vb{r'}}{\abs{\vb{r}-\vb{r'}}^3})} \dd V' \\
@@ -132,4 +131,4 @@ $$
 
 **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 symnetry, the integral factor becomes simply $E(r)*A$, where $A$ is the area of the surface at $r$.

docs/physics/electrostatics/3-electro-magnetic-potentials.md

@@ -25,7 +25,7 @@ The units of magnetostatic potential is Joule/Weber, also known as an Ampere. Th
 With this, we can calculate work. Moving a charge $q$ from $A$ to $B$, we see that
 
 $$
-\delta W = \int_A^B \vb{F} \vdot \dd{\vb{l}} = q_e \int_A^B \vb{E} \vdot \dd{\vb{l}} = -q_e \int_A^B \grad{\vb{V}} \vdot \dd{\vb{l}} = -q_e \delta V_e 
+\delta W = \int_A^B \vb{F} \vdot \dd{\vb{l}} = q_e \int_A^B \vb{E} \vdot \dd{\vb{l}} = -q_e \int_A^B \grad{\vb{V}} \vdot \dd{\vb{l}} = -q_e \delta V_e
 $$
 
 Strictly speaking, this is a potential difference. To find the absolute potential, assume a point charge $Q$ at the origin, and a charge $q$. We take the work as $q$ moves from $\vb{r'} = \vb{\infty}$ to $\vb{r'} = \vb{r}$. Thus,
@@ -62,7 +62,6 @@ $$
 V_e(\vb{r}) = \frac{1}{4 \pi \epsilon_0} \int_{V'} \frac{\rho_e(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
 $$
 
-
 $$
 V_m(\vb{r}) = \frac{1}{4 \pi \mu_0} \int_{V'} \frac{\rho_m(\vb{r'})}{\abs{\vb{r}-\vb{r'}}} \dd{V'}
 $$
@@ -75,7 +74,7 @@ $$
 W_2 = W_{21} = \frac{1}{4 \pi \epsilon_0} \frac{Q_{e1} Q_{e2}}{\abs{\vb{r_2} - \vb{r_1}}}
 $$
 
-Superposition applies here. The energy to create $N$ charges is 
+Superposition applies here. The energy to create $N$ charges is
 
 $$
 W_n = \frac{1}{2} \frac{4 \pi \epsilon_0} \sum_{i = 1}^{N} \sum_{j > i}^{N} \frac{Q_{ei}Q_{ej}}{\abs{\vb{r_i}-\vb{r_j}}}
@@ -177,7 +176,7 @@ $$
 Taking the divergence, we find that
 
 $$
-- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\epsilon_0} = \frac{Q_e}{\epsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}}) 
+- \laplacian{V(\vb{r})} = G(\vb{r}, \vb{r'}) \frac{Q_e}{\epsilon_0} = \frac{Q_e}{\epsilon_0} \laplacian({\frac{-1}{4\pi \abs{\vb{r} - \vb{r'}}}})
 =  \frac{Q_e}{\epsilon_0} \div \frac{\vb{r} - \vb{r'}}{\abs{\vb{r} - \vb{r'}}^3} = \frac{Q_e}{\epsilon_0} \delta(\vb{r} - \vb{r'})
 $$
 
@@ -274,4 +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 volate or other equations, however, this becomes very messy.

docs/physics/electrostatics/4-conductors.md

@@ -16,4 +16,4 @@ Consider any two points internal to the conductor. The voltage between said poin
 
 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.
 
-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 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, $\sigma_e = \epsilon_0 E$.

docs/recipes/bread.md

@@ -1,9 +1,9 @@
 # Breads
 
-# Pumpkin Bread
+## Pumpkin Bread
 
 - Preheat oven to $350 \degree$ F.
 - Combine $1 \frac{2}{3}$ cups flour, $1 \frac{1}{2}$ cups sugar, 1 tsp. baking soda, 1 tsp cinnamon, $\frac{3}{4}$ tsp. salt, $\frac{1}{2}$ tsp. baking powder, $\frac{1}{2}$ tsp. nutmeg, $\frac{1}{4}$ tsp cloves.
 - In a separate bowl, combine 2 eggs, 1 can of pumpkin, $\frac{1}{2}$ cup canola oil, and $\frac{1}{2}$ cups water.
 - Combine. Mix in $\frac{1}{2}$ cups of walnuts.
-- Add to a greased 9x5 pan. Bake at $350 \degree$ F for 65-80 minutes.
+- Add to a greased 9x5 pan. Bake at $350 \degree$ F for 65-80 minutes.

docs/recipes/pies.md

@@ -43,10 +43,10 @@ From: [Link](https://www.allrecipes.com/recipe/234374/apple-hand-pies/)
 - Mix in apples, add sugar mixture.
 - Wait until apples are softened (approx. 5 minutes).
 
-### Pies
+### Hand Pies
 
 - Preheat oven to $400 \degree$ F.
 - Split pie crust into 4. Place fillin 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.
+- Bake at $400 \degree$ F. for 25-30 minutes.