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| -rw-r--r-- | Problem Set 2.page | 15 |
1 files changed, 8 insertions, 7 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 218287f..c114851 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -39,20 +39,21 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ # Solutions 2. Since -$\sin x = \frac{e^{ix}-e^{-ix}}{2}$, +$$\sin x = \frac{e^{ix}-e^{-ix}}{2}$$, -$\sin^4 x = \frac{{( e^{ix}-e^{-ix} )}^4}{16}$ -$= \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. +$$\begin{array}{ccl} \sin^4 x &=& \frac{{( e^{ix}-e^{-ix} )}^4}{16} \\ + &=& \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. \end{array} $$ If we express any periodic function $f(x)$ as -$f(x) = \sum a_n f_n(x)$, where $f_n(x) = \frac{e^{inx}}{\sqrt{2\pi}}$ and $f_0(x) = \frac{1}{\sqrt{2\pi}}$, - +$$f(x) = \sum a_n f_n(x),$$ -The Fourier coefficients for the above functions are: +where $f_n(x) = \frac{e^{inx}}{\sqrt{2\pi}}$ and $f_0(x) = \frac{1}{\sqrt{2\pi}}$, then the Fourier coefficients for the above functions are: -$a_{-4} = a_{4} = \sqrt{2\pi} \times 1/16$, $a_{-2} = a_{2} = - \sqrt{2\pi} \times 4/16$, $a_0 = \sqrt{2\pi} \times 6/16$ +$$\begin{array}{rcl} a_{-4} = a_{4} &=& \sqrt{2\pi} \times 1/16, \\ + a_{-2} = a_{2} &=& - \sqrt{2\pi} \times 4/16 \\ + a_0 &=& \sqrt{2\pi} \times 6/16 \end{array} $$ Since $a_m = < f_m, f >$ and setting $f(x) = \sin^4(x)$, |