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@@ -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,
$\int_0^{2\pi} \sin^4(x) dx = \sum |a_n|^2 = \frac{3 \pi}{4}$