fix some more math formatting

src/8_differential_equations_2.md

@@ -76,7 +76,7 @@ $$y(x) = c_1 f_1 (x) + c_2 f_2 (x) + \cdots + c_n f_{n}(x). $$
 
 To check that the $n$ solutions form a basis, it is sufficient to verify
 
-$$ det \begin{bmatrix} 
+$$ \det \begin{bmatrix} 
 f_1(x) & \cdots & f_{n}(x) \\
 f_1 ' (x) & \cdots & f_{n}'(x) \\
 \vdots & \vdots & \vdots \\
@@ -98,7 +98,7 @@ $$A = \begin{bmatrix}
 
 It is possible to show that 
 
-$$det(A - \lambda I) = -P(\lambda),$$
+$$\det(A - \lambda I) = -P(\lambda),$$
 
 in which $P(\lambda)$ is the characteristic polynomial of the system matrix $A$,
 
@@ -107,37 +107,37 @@ $$P(\lambda) = \lambda^n + a_{n-1} \lambda^{n-1} + \cdots + a_0.$$
 As we demonstrate below, the proof relies on the co-factor expansion technique 
 for calculating a determinant. 
 
-$$- det(A - \lambda I) = \begin{bmatrix} 
+$$- \det(A - \lambda I) = \begin{bmatrix} 
 \lambda & -1 & 0 & \cdots & 0 \\
 0 & \lambda & -1 & \cdots & 0 \\
 \vdots & \vdots & \vdots & \cdots & \vdots \\
 a_0 & a_1 & a_2 & \cdots & a_{n-1} + \lambda \\
 \end{bmatrix} $$
-$$- det(A - \lambda I) =  \lambda det \begin{bmatrix}
+$$- \det(A - \lambda I) =  \lambda \det \begin{bmatrix}
 \lambda & -1 & 0 & \cdots & 0 \\
 0 & \lambda & -1 & \cdots & 0 \\
 \vdots & \vdots & \vdots & \cdots & \vdots \\
 a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
-\end{bmatrix} + (-1)^{n+1}a_0 det \begin{bmatrix} 
+\end{bmatrix} + (-1)^{n+1}a_0 \det \begin{bmatrix} 
 -1 & 0 & 0 & \cdots & 0 \\
 \lambda & -1 & 0 & cdots & 0 \\
 \vdots & \vdots & \vdots & \cdots & \vdots \\
 0 & 0 & \cdots & \lambda & -1 \\
 \end{bmatrix}$$
-$$- det(A - \lambda I) = \lambda det \begin{bmatrix}
+$$- \det(A - \lambda I) = \lambda \det \begin{bmatrix}
 \lambda & -1 & 0 & \cdots & 0 \\
 0 & \lambda & -1 & \cdots & 0 \\
 \vdots & \vdots & \vdots & \cdots & \vdots \\
 a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
 \end{bmatrix} + (-1)^{n+1} a_0 (-1)^{n-1}$$
-$$- det(A - \lambda I) = \lambda det \begin{bmatrix}
+$$- \det(A - \lambda I) = \lambda \det \begin{bmatrix}
 \lambda & -1 & 0 & \cdots & 0 \\
 0 & \lambda & -1 & \cdots & 0 \\
 \vdots & \vdots & \vdots & \cdots & \vdots \\
 a_1 & a_2 & a_3 & \cdots & a_{n-1} + \lambda \\
 \end{bmatrix} + a_0$$
-$$- det(A - \lambda I) = \lambda (\lambda (\lambda \cdots + a_2) + a_1) + a_0$$
-$$- det(A - \lambda I) = P(\lambda).$$
+$$- \det(A - \lambda I) = \lambda (\lambda (\lambda \cdots + a_2) + a_1) + a_0$$
+$$- \det(A - \lambda I) = P(\lambda).$$
 
 In the second last line of the proof we indicated that the method of co-factor 
 expansion demonstrated is repeated an additional $n-2$ times. This completes the
@@ -198,7 +198,7 @@ $$f(x) = e^{\lambda_1 x}, \ x e^{\lambda_1 x} , \ \cdots, \ x^{m_{1}-1} e^{\lamb
     since they are linear combinations of $f_1$ and $f_2$ which remain linearly
     independent,
     
-    $$\tilde{f_1}(x)=cos(kx), \tilde{f_2}(x)=sin{kx}.$$
+    $$\tilde{f_1}(x)=\cos(kx), \tilde{f_2}(x)=sin{kx}.$$
     
     **Case 2: $E<0$**
     This time, define $E=-k^2$, for constant $k$. The characteristic polynomial 
@@ -369,7 +369,7 @@ $$\phi''(x)+k^2 \phi(x)=0.$$
 
 Two linearly independent solutions to this equation are 
 
-$$\phi_{1}(x)=sin(k x), \ \phi_{2}(x) = cos(k x).$$
+$$\phi_{1}(x)=\sin(k x), \ \phi_{2}(x) = \cos(k x).$$
 
 The solution to this homogeneous equation is then 
 
@@ -378,11 +378,11 @@ $$\phi(x)=c_1 \phi_1(x)+c_2 \phi_2(x).$$
 The eigenvalue, $\lambda$ as well as one of the constant coefficients can be 
 determined using the boundary conditions. 
 
-$$\phi(0)=0 \ \Rightarrow \ \phi(x)=c_1 sin(k x), \ c_2=0.$$
+$$\phi(0)=0 \ \Rightarrow \ \phi(x)=c_1 \sin(k x), \ c_2=0.$$
 
-$$\phi(L)=0 \ \Rightarrow \ 0=c_1 sin(k L) .$$
+$$\phi(L)=0 \ \Rightarrow \ 0=c_1 \sin(k L) .$$
 
-In turn, using the properties of the $sin(\cdot)$ function, it is now possible
+In turn, using the properties of the $\sin(\cdot)$ function, it is now possible
 to find the allowed values of $k$ and hence also $\lambda$. The previous 
 equation implies, 
 
@@ -404,7 +404,7 @@ PDE was supplied with an initial condition.
 Putting the solutions to the two ODEs together and redefining 
 $\tilde{A}=A \cdot c_1$, we arrive at the solutions for theb PDE,
 
-$\psi_n(x,t) = \tilde{A}_n e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}).$
+$\psi_n(x,t) = \tilde{A}_n e^{-i \frac{\lambda_n t}{\hbar}} \sin(\frac{n \pi x}{L}).$
 
 Notice that there is one solution $\psi_{n}(x,t)$ for each natural number $n$. 
 These are still very special solutions. We will begin discussing next how to 
@@ -488,18 +488,18 @@ the eigenfunctions of $L$.
     In terms of hermitian operators and their eigenfunctions, the eigenfunctions
     play the role of the orthonormal basis. In reference to our running example,
     the 1D Schrödinger equation of a free particle, the eigenfunctions 
-    $sin(\frac{n \pi x}{L})$ play the role of the basis functions $\ket{u_n}$.
+    $\sin(\frac{n \pi x}{L})$ play the role of the basis functions $\ket{u_n}$.
     
 To close our running example, consider the initial condition 
-$\psi(x,o) = \psi_{0}(x)$. Since the eigenfunctions $sin(\frac{n \pi x}{L})$ 
+$\psi(x,o) = \psi_{0}(x)$. Since the eigenfunctions $\sin(\frac{n \pi x}{L})$ 
 form a basis, we can now write the general solution to the problem as 
 
-$$\psi(x,t)  = \overset{\infinity}{\underset{n}{\Sigma}} c_n e^{-i \frac{\lambda_n t}{\hbar}} sin(\frac{n \pi x}{L}),$$
+$$\psi(x,t)  = \overset{\infinity}{\underset{n}{\Sigma}} c_n e^{-i \frac{\lambda_n t}{\hbar}} \sin(\frac{n \pi x}{L}),$$
 
 where in the above we have defined the coefficients using the Fourier 
 coefficient,
 
-$$c_n:= \int^{L}_{0} dx sin(\frac{n \pi x}{L}) \psi_{0}(x). $$
+$$c_n:= \int^{L}_{0} dx \sin(\frac{n \pi x}{L}) \psi_{0}(x). $$
     
 ### General recipie for seperable PDEs
 
@@ -530,9 +530,9 @@ necessary to work with numerical methods of solution.
 
 1.  [:grinning:] Which of the following equations for $y(x)$ is linear?
 
-    (a) y''' - y'' + x cos(x) y' + y - 1 = 0
+    (a) y''' - y'' + x \cos(x) y' + y - 1 = 0
 
-    (b) y''' + 4 x y' - cos(x) y = 0
+    (b) y''' + 4 x y' - \cos(x) y = 0
 
     (c) y'' + y y' = 0