posted solutions of 2 and 3 in pset2

1 files changed, 4 insertions, 2 deletions

diff --git a/Problem Set 2.page b/Problem Set 2.page
index 5788847..cc84acb 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -47,15 +47,17 @@ If we express any periodic function $f(x)$ as
 
 $f(x) = \sum a_n f_n(x)$, where $f_n(x) = \frac{e^{inx}}{\sqrt{2\pi}}$, $f_0(x) = \frac{1}{\sqrt{2\pi}}$,
 
+
 The Fourier coefficients for the above functions are:
 
 $a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \times 4/16$, $a_0 = \sqrt{2\pi} \times 6/16$
 
+
 Since $a_m = < f_m, f >$,
 
-$\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times $<f_0, f>$ 
+$\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times $<f_0, f>$$
 
-= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
+$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
 
 
 ## Cardinality