more formatting

1 files changed, 2 insertions, 1 deletions

diff --git a/ClassJuly5.page b/ClassJuly5.page
index 19b3d47..fa72182 100644
--- a/ClassJuly5.page
+++ b/ClassJuly5.page
@@ -35,7 +35,7 @@ Rearranging, we get:
 
 $$ \frac{da}{dt}/a = \frac{d^2b}{dx^2}/b $$
 
-The left hand side only depends on $t$ and the right hand side depends only on $x$.  Therefore, neither side depends on either $x$ or $t$.  We conclude that both sides are equal to a constant, which for convenience we'll write as $-\lambda^2$.  
+The left hand side only depends on $t$ and the right hand side depends only on $x$.  Therefore, neither side depends on either $x$ or $t$.  We conclude that both sides are equal to a constant, which for convenience we'll write as $-\lambda^2$.
 
 First we solve for $a$.  It satisfies the equation
 
@@ -62,6 +62,7 @@ We would like to assert that any solution takes this form.  One way to prove thi
 $$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$
 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 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.