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| -rw-r--r-- | Problem Set 2.page | 12 |
1 files changed, 5 insertions, 7 deletions
diff --git a/Problem Set 2.page b/Problem Set 2.page index 4fe7a1f..2637ab2 100644 --- a/Problem Set 2.page +++ b/Problem Set 2.page @@ -61,17 +61,15 @@ $$ \begin{array}{ccl} \int_0^{2\pi} \sin^4(x) dx = <1, f> &=& \sqrt{2\pi} \times &=& \frac{3 \pi}{4} \end{array} $$ -3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, $\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$, and therefore, +3. Since $\sin x = \frac{e^{ix}-e^{-ix}}{2}$, we have $\sin^2 x = \frac{e^{i 2x}+e^{-i 2x}-2}{4}$. Therefore, -$$\begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx \\ - -&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx \\ - -&=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx. \end{array} $$ +$$ \begin{array}{ccl} a_m &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \sin^2(x) e^{-imx} dx \\ + &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{i 2x}+e^{-i 2x}-2}{4} e^{-imx} dx \\ + &=& \frac{1}{\sqrt 2\pi} \int_0^{2\pi} \frac{e^{-i (m-2)x}+e^{-i (m+2)x}-2e^{-imx}}{4} dx. \end{array} $$ Because $\int_0^{2\pi} e^{inx} dx = 2\pi$ for $n = 0$ and $\int_0^{2\pi} e^{inx} dx = 0$ for $n \neq 0$, we have -$$a_2 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\pi}/4,$$ +$$a_2 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi = \sqrt{2\ pi}/4,$$ $$a_0 = \frac{1}{4 \sqrt{2\pi}} \times 2\pi \times (-2) = - \frac{\sqrt{2\pi}}{2},$$ |