more more formatting

1 files changed, 2 insertions, 2 deletions

diff --git a/ClassJuly5.page b/ClassJuly5.page
index ff7cfb3..664c440 100644
--- a/ClassJuly5.page
+++ b/ClassJuly5.page
@@ -56,10 +56,10 @@ and to satisfy $b(L) = 0$, we must impose a constraint on $\lambda$:
 $$ \lambda = \frac{\pi n}{L} $$
 
 So, the most general solution we can generate in this manner is:
-$$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-(\frac{\pi n}{L})^2t} \sin(\frac{\pi n x}{L}) $$
+$$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-\frac{\pi^2 n^2 t}{L^2}} \sin \left(\frac{\pi n x}{L} \right) $$
 
 We would like to assert that any solution takes this form.  One way to prove this assertion would be to show that any function $f:[0,L] \to \mathbb{R}$ satisfying $f(0) = f(L) = 0$ has a unique ``[Fourier sine expansion](http://mathworld.wolfram.com/FourierSineSeries.html)'':
-$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$
+$$ f(x) = \sum_{n = 1}^{\infty} c_n \sin\left(\frac{\pi n x}{L}\right) $$
 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.