From 3f965d6b32b5672b350b63444dbf75d72123f14d Mon Sep 17 00:00:00 2001 From: Opheliar99 <> Date: Sun, 4 Jul 2010 04:10:01 +0000 Subject: posted solutions of 2 and 3 in pset2 --- Problem Set 2.page | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) (limited to 'Problem Set 2.page') diff --git a/Problem Set 2.page b/Problem Set 2.page index 3f1a214..078d5e1 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -40,8 +40,8 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ 2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, -$ \sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$, -$ = \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{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 @@ -54,9 +54,9 @@ $a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \tim Since $a_m = < f_m, f >$, -$ \int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times $ +$\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times $ -$ = \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$ +$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$ ## Cardinality -- cgit v1.2.3