more formatting

1 files changed, 1 insertions, 1 deletions

diff --git a/ClassJuly5.page b/ClassJuly5.page
index fa72182..ff7cfb3 100644
--- a/ClassJuly5.page
+++ b/ClassJuly5.page
@@ -58,7 +58,7 @@ $$ \lambda = \frac{\pi n}{L} $$
 So, the most general solution we can generate in this manner is:
 $$ u(x,t) = \sum_{n = 1}^{\infty} c_n e^{-(\frac{\pi n}{L})^2t} \sin(\frac{\pi n x}{L}) $$
 
-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)'':
+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)'':
 $$ f(x) = \sum_{n = 1}^{\infty} c_n \sin(\frac{\pi n x}{L}) $$
 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.