dollar signs

1 files changed, 4 insertions, 4 deletions

diff --git a/ClassJuly5.page b/ClassJuly5.page
index dde84c7..632ecec 100644
--- a/ClassJuly5.page
+++ b/ClassJuly5.page
@@ -11,13 +11,13 @@ $$ g(x,t) = u_0 + \frac{u_L- u_0}{L}x $$
 This satisfies the heat equation for trivial reasons since it is time-independent and its second spatial derivative is zero, hence both sides of the heat equation are zero independently of one another.
 
 Now suppose that $h(x,t)$ is another solution satisfying the same boundary conditions.  Then the function
-$$ u(x,t) = g(x,t) - h(x,t)$
+$$ u(x,t) = g(x,t) - h(x,t)$$
 also satisfies the heat equation, but it is zero at both endpoints.
 $$ u(x,0) = u(x,L) = 0 $$
 To solve for $h$, it is clear that we only need to solve for $u$.  First we'll use what's called the ``separation of variables'' trick to generate a lot of nice solutions, then hope and pray that any other solution can be expressed as a linear combination of these.
 
 Here's how the separation of variables trick works.  We seek a solution of the form
-$$u(x,t) = a(t)b(x)$
+$$u(x,t) = a(t)b(x)$$
 for some functions $a$ and $b$.  Then the heat equation tells us:
 $$ \frac{da}{dt}b = a \frac{d^2b}{dx^2} $$
 Rearranging, we get:
@@ -42,10 +42,10 @@ So, the most general solution we can generate in this manner is:
 $$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-(\frac{\pi n}{L})^2t} \sin(\frac{\pi n x}{L}) $$
 
 We would like to assert that any solution takes this form.  One way to prove this assertion would be to show that any function $f:[0,L] \to \R$ satisfying $f(0) = f(L) = 0$ has a unique ``[Fourier sine expansion](http://mathworld.wolfram.com/FourierSineSeries.html)'':
-$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L})
+$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$
 One could then allow the coefficients $c_n$ to vary with $t$ and apply the same method of solution that we used in the case of periodic boundary conditions.
 
-In fact, every function of the kind described above does have a Fourier sine expansion.  The link above contains a hint of how to do it.  First you extend the function $f(x)$ to an odd, periodic function $\tilde{f}(x)$ defined on the interval $[-L,L]$.  Then you can use convergence of the usual Fourier series for $\tilde{f}(x)$.
+In fact, every function of the kind described above does have a Fourier sine expansion.  The link above contains a hint of how to do it.  First you extend the function $f(x)$ to a certain odd, periodic function $\tilde{f}(x)$ defined on the interval $[-L,L]$.  Then you can use convergence of the usual Fourier series for $\tilde{f}(x)$.
 
 ## Convergence for not-so-nice Fourier series.