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authorOpheliar99 <>2010-07-04 04:56:32 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:56:32 +0000
commit27194cc662f73877a413955c4c5f01761e2c8b08 (patch)
tree30f5f408005f1415cfe62df6716c0cac0b29a2cf
parent465bf280a3f16938d7423d7cd1cc4fafe8a6cec8 (diff)
downloadafterklein-wiki-27194cc662f73877a413955c4c5f01761e2c8b08.tar.gz
afterklein-wiki-27194cc662f73877a413955c4c5f01761e2c8b08.zip
posted solutions of 2 and 3 in pset2
-rw-r--r--Problem Set 2.page2
1 files changed, 2 insertions, 0 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page
index 523ce40..1cbcce1 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -62,6 +62,7 @@ $\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times < f_0, f >$
$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$
+
3. Since
$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
@@ -86,6 +87,7 @@ Then,
$\sum |a_n|^2 = {\sqrt{2\pi}/4}^2 + {- \sqrt{2\pi}/2}^2 + {\sqrt{2\pi}/4}^2 = \frac{3 \pi}{4}$
And, it was shown in Prob 2 that $\int_0^{2\pi} \sin^4(x) dx = \frac{3 \pi}{4}$.
+
Therefore,
$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$