diff options
| -rw-r--r-- | Problem Set 2.page | 16 |
1 files changed, 6 insertions, 10 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index c114851..5ba4b65 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -38,12 +38,10 @@ $\int_0^{2\pi} |\sin^2(x)|^2 dx = \sum |a_n|^2.$ # Solutions -2. Since -$$\sin x = \frac{e^{ix}-e^{-ix}}{2}$$, +2. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, $$\begin{array}{ccl} \sin^4 x &=& \frac{{( e^{ix}-e^{-ix} )}^4}{16} \\ - &=& \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}$. \end{array} $$ - + &=& \frac{e^{i 4x}+e^{-i 4x}-4 e^{i 2x} -4 e^{-i 2x}+6}{16}. \end{array} $$ If we express any periodic function $f(x)$ as @@ -58,14 +56,12 @@ $$\begin{array}{rcl} a_{-4} = a_{4} &=& \sqrt{2\pi} \times 1/16, \\ Since $a_m = < f_m, f >$ and setting $f(x) = \sin^4(x)$, -$\int_0^{2\pi} \sin^4(x) dx = <1, f> = \sqrt{2\pi} \times < f_0, f >$ - -$= \sqrt{2\pi} \times a_0 = \frac{3 \pi}{4}$ - +$$ \begin{array}{ccl} \int_0^{2\pi} \sin^4(x) dx = <1, f> &=& \sqrt{2\pi} \times < f_0, f > \\ +&=& \sqrt{2\pi} \times a_0 \\ + &=& \frac{3 \pi}{4} \end{array} $$ -3. Since -$\sin x = \frac{e^{ix}-e^{-ix}}{2}$, +3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, $\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$ |