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| -rw-r--r-- | Problem Set 2.page | 7 |
1 files changed, 3 insertions, 4 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 6fec599..329265b 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -39,10 +39,9 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ # Solutions 2. Since -$sin(x) = \frac{e^{ix}-e^{-ix}}{2}$, +$\sin(x) = \frac{e^{ix}-e^{-ix}}{2}$, $\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^4}{16}$, - -$ = \frac{e^{i4x}+e^{-4ix}-4 e^{i2x} -4 e^{-i2x}+6}{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 @@ -57,7 +56,7 @@ $a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $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 <f_0, f>$ -$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$ +$ = \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$ ## Cardinality |