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| -rw-r--r-- | Problem Set 2.page | 7 |
1 files changed, 6 insertions, 1 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 5365ceb..5788847 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -44,13 +44,18 @@ $\int_0^{2\pi} \sin^4(x) dx = \frac{{e^{ix}-e^{-ix}}^4}{16}$ $ = \frac{e^{i4x}+e^{-4ix}-4 e^{i2x} -4 e^{-i2x}+6}{16}$ If we express any periodic function $f(x)$ as + $f(x) = \sum a_n f_n(x)$, where $f_n(x) = \frac{e^{inx}}{\sqrt{2\pi}}$, $f_0(x) = \frac{1}{\sqrt{2\pi}}$, The Fourier coefficients for the above functions are: + $a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \times 4/16$, $a_0 = \sqrt{2\pi} \times 6/16$ Since $a_m = < f_m, f >$, -$\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times <f_0, f> = \frac{3 \pi}{4}$ + +$\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}$ ## Cardinality |