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authorOpheliar99 <>2010-07-04 04:12:05 +0000
committerbnewbold <bnewbold@adelie.robocracy.org>2010-07-04 04:12:05 +0000
commit14244b4506f772541be6b2289b2ace8651d0e7f1 (patch)
tree7f850e8089b6db44b6c54681580a63c7d217553f
parentf5cb3cc89c0eb29347cdcb95dc6543f46af095d9 (diff)
downloadafterklein-wiki-14244b4506f772541be6b2289b2ace8651d0e7f1.tar.gz
afterklein-wiki-14244b4506f772541be6b2289b2ace8651d0e7f1.zip
posted solutions of 2 and 3 in pset2
-rw-r--r--Problem Set 2.page4
1 files changed, 3 insertions, 1 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page
index be65cd5..2d5a503 100644
--- a/Problem Set 2.page
+++ b/Problem Set 2.page
@@ -40,7 +40,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$
2. Since
$\sin x = \frac{e^{ix}-e^{-ix}}{2}$,
-$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$,
+
+
+$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$
$= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$.
If we express any periodic function $f(x)$ as