一道计算题求解

  1. 2月前

    \(\frac{4V_0}{\pi} \sum_{n=1,3,5,\cdots}^{\infty}\quad \frac{1}{n} e^{-\frac{n\pi x}{a}}\sin(\frac{n\pi y}{a}) = \frac{2V_0}{\pi}\arctan(\frac{\sin(\frac{\pi y}{a})}{\sinh(\frac{\pi x}{a})})\)
    这是我在看格里菲斯的电动力学导论遇到的问题。

    因为\[\arctan x = \sum_{n = 0}^{\infty} \dfrac{\left( -1 \right) ^ n x ^ {2 n + 1}}{2 n + 1},\]所以\[\arctan{\mathrm{j} x} = \mathrm{j} \sum_{n = 0}^{\infty} \dfrac{x ^ {2 n + 1}}{2 n + 1} = \mathrm{j} \sum_{n = 1, 3, 5, \cdots}^{\infty} \dfrac{x ^ n}{n}.\]即\[ \sum_{n = 1, 3, 5, \cdots}^{\infty} \dfrac{x ^ n}{n} = - \mathrm{j} \arctan {\mathrm{j} x}. \]设\[ f \left( x \right) = \sum_{n = 1, 3, 5, \cdots}^{\infty} \dfrac{1}{n} \mathrm{e}^{- \frac{n \pi x} {a}} \sin \left( \frac{n \pi y} {a} \right), \]且令\[g \left( x \right) = \sum_{n = 1, 3, 5, \cdots}^{\infty} \dfrac{1}{n} \mathrm{e}^{- \frac{n \pi x} {a}} \cos \left( \frac{n \pi y} {a} \right),\]那么可以算出来\[\begin{array}{ll}g \left( x \right) + \mathrm{j} f \left( x \right) &= - \mathrm{j} \arctan \left[ \mathrm{j} \exp \left( - \dfrac{\pi x}{a} + \dfrac{\mathrm{j} \pi y}{a} \right) \right], \\ g \left( x \right) - \mathrm{j} f \left( x \right) &= - \mathrm{j} \arctan \left[ \mathrm{j} \exp \left( - \dfrac{\pi x}{a} - \dfrac{\mathrm{j} \pi y}{a} \right) \right].\end{array}\] 解得\[\begin{array}{ll} f \left( x \right) &= - \dfrac{1}{2} \arctan \left[ \mathrm{j} \exp \left( - \dfrac{\pi x}{a} + \dfrac{\mathrm{j} \pi y}{a} \right) \right] + \dfrac{1}{2} \arctan \left[ \mathrm{j} \exp \left( - \dfrac{\pi x}{a} - \dfrac{\mathrm{j} \pi y}{a} \right) \right] \\ &= \dfrac{1}{2} \arctan \dfrac{ \sin \dfrac{\pi y}{a}} {\sinh \dfrac{\pi x}{a}}. \end{array}\]
    附:由\[\tan \left( \arctan A - \arctan B \right) = \dfrac{A - B}{1 + A B} \]可得\[ \arctan A - \arctan B = \arctan \dfrac{A - B}{1 + A B} .\]

  2. 廣夏原

    2楼 11月25日 被采纳
    2月前廣夏原 重新编辑

    因为\[\arctan x = \sum_{n = 0}^{\infty} \dfrac{\left( -1 \right) ^ n x ^ {2 n + 1}}{2 n + 1},\]所以\[\arctan{\mathrm{j} x} = \mathrm{j} \sum_{n = 0}^{\infty} \dfrac{x ^ {2 n + 1}}{2 n + 1} = \mathrm{j} \sum_{n = 1, 3, 5, \cdots}^{\infty} \dfrac{x ^ n}{n}.\]即\[ \sum_{n = 1, 3, 5, \cdots}^{\infty} \dfrac{x ^ n}{n} = - \mathrm{j} \arctan {\mathrm{j} x}. \]设\[ f \left( x \right) = \sum_{n = 1, 3, 5, \cdots}^{\infty} \dfrac{1}{n} \mathrm{e}^{- \frac{n \pi x} {a}} \sin \left( \frac{n \pi y} {a} \right), \]且令\[g \left( x \right) = \sum_{n = 1, 3, 5, \cdots}^{\infty} \dfrac{1}{n} \mathrm{e}^{- \frac{n \pi x} {a}} \cos \left( \frac{n \pi y} {a} \right),\]那么可以算出来\[\begin{array}{ll}g \left( x \right) + \mathrm{j} f \left( x \right) &= - \mathrm{j} \arctan \left[ \mathrm{j} \exp \left( - \dfrac{\pi x}{a} + \dfrac{\mathrm{j} \pi y}{a} \right) \right], \\ g \left( x \right) - \mathrm{j} f \left( x \right) &= - \mathrm{j} \arctan \left[ \mathrm{j} \exp \left( - \dfrac{\pi x}{a} - \dfrac{\mathrm{j} \pi y}{a} \right) \right].\end{array}\] 解得\[\begin{array}{ll} f \left( x \right) &= - \dfrac{1}{2} \arctan \left[ \mathrm{j} \exp \left( - \dfrac{\pi x}{a} + \dfrac{\mathrm{j} \pi y}{a} \right) \right] + \dfrac{1}{2} \arctan \left[ \mathrm{j} \exp \left( - \dfrac{\pi x}{a} - \dfrac{\mathrm{j} \pi y}{a} \right) \right] \\ &= \dfrac{1}{2} \arctan \dfrac{ \sin \dfrac{\pi y}{a}} {\sinh \dfrac{\pi x}{a}}. \end{array}\]
    附:由\[\tan \left( \arctan A - \arctan B \right) = \dfrac{A - B}{1 + A B} \]可得\[ \arctan A - \arctan B = \arctan \dfrac{A - B}{1 + A B} .\]

  3. 谢谢二楼详细的解答,之后按照我喜欢的方式抄了一遍。这种问题我不会处理,主要是复数的运算没怎么接触过。
    \begin{align*}
    &= 2\sum\limits_{n=1,3,5,\cdots}^{\infty }\frac{1}{n} \exp(-\frac{n\pi x}{a})\cdot \sin(\frac{n\pi y}{a})\\
    &= -j\sum\limits_{n=1,3,5,\cdots}^{\infty }\frac{1}{n} \exp(-\frac{n\pi x}{a})\cdot {\color{Blue} (\exp(j\frac{n\pi y}{a})-\exp(-j\frac{n\pi y}{a}))}\\
    &=-j\sum\limits_{n=1,3,5,\cdots}^{\infty }\frac{1}{n} \cdot (\exp(-\frac{n\pi x}{a}+j\frac{n\pi y}{a})-\exp(-\frac{n\pi x}{a}-j\frac{n\pi y}{a}))\\
    &=-j[\sum\limits_{n=1,3,5,\cdots}^{\infty }{\color{Purple} \frac{1}{n} [\exp(-\frac{\pi x}{a}+j\frac{\pi y}{a})]^n}-\sum\limits_{n=1,3,5,\cdots}^{\infty }{\color{DarkGreen}\frac{1}{n}[ \exp(-\frac{\pi x}{a}-j\frac{\pi y}{a})]^{n}}]\\
    &=-{\color{Purple} \arctan(j\exp(-\frac{\pi x}{a}+j\frac{\pi y}{a}))}+{\color{DarkGreen} \arctan(j\exp(-\frac{\pi x}{a}-j\frac{\pi y}{a}))}\\
    &=\arctan(\frac{2\exp(-\frac{\pi x}{a})\cdot \sin(\frac{\pi y}{a})}{1-\exp(-\frac{2\pi x}{a})})\\
    &=\arctan(\frac{\sin(\frac{\pi y}{a})}{\sinh(\frac{\pi x}{a})})
    \end{align*}

 

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