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- #1

$$

\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}A_{nm}\sin\frac{n\pi x}{L}\sin\frac{m\pi y}{H} = -\frac{4}{\pi}\sum_{k = 1}^{\infty}\frac{1}{(2k-1)\sinh\frac{\pi(2k-1)H}{L}}\sin\frac{\pi(2k-1)x}{L}\sinh\frac{\pi(2k-1)y}{L}

$$

If I start with x on the left, can I then end up with:

$$

\frac{L}{2}\sum_{m = 1}^{\infty}A_{nm}\sin\frac{m\pi y}{H} = -\frac{4L}{2\pi}\Rightarrow\sum_{m = 1}^{\infty}A_{nm}\sin\frac{m\pi y}{H} = -\frac{4}{\pi}

$$

So

$$

A_{nm} = -\frac{8}{H\pi}\int_0^H\sin\frac{m\pi y}{H}dy = \begin{cases}

0, & \text{if m is even}\\

-\frac{16}{m\pi^2}, & \text{if m is odd}

\end{cases}

$$

$$

\frac{16}{\pi^2}\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\sin\frac{n\pi x}{L}\sin\frac{(2m-1)\pi y}{H} = \frac{4}{\pi}\sum_{k = 1}^{\infty}\frac{1}{(2k-1)\sinh\frac{\pi(2k-1)H}{L}}\sin\frac{\pi(2k-1)x}{L}\sinh\frac{\pi(2k-1)y}{L}

$$

Is this true though?

\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}A_{nm}\sin\frac{n\pi x}{L}\sin\frac{m\pi y}{H} = -\frac{4}{\pi}\sum_{k = 1}^{\infty}\frac{1}{(2k-1)\sinh\frac{\pi(2k-1)H}{L}}\sin\frac{\pi(2k-1)x}{L}\sinh\frac{\pi(2k-1)y}{L}

$$

If I start with x on the left, can I then end up with:

$$

\frac{L}{2}\sum_{m = 1}^{\infty}A_{nm}\sin\frac{m\pi y}{H} = -\frac{4L}{2\pi}\Rightarrow\sum_{m = 1}^{\infty}A_{nm}\sin\frac{m\pi y}{H} = -\frac{4}{\pi}

$$

So

$$

A_{nm} = -\frac{8}{H\pi}\int_0^H\sin\frac{m\pi y}{H}dy = \begin{cases}

0, & \text{if m is even}\\

-\frac{16}{m\pi^2}, & \text{if m is odd}

\end{cases}

$$

$$

\frac{16}{\pi^2}\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\sin\frac{n\pi x}{L}\sin\frac{(2m-1)\pi y}{H} = \frac{4}{\pi}\sum_{k = 1}^{\infty}\frac{1}{(2k-1)\sinh\frac{\pi(2k-1)H}{L}}\sin\frac{\pi(2k-1)x}{L}\sinh\frac{\pi(2k-1)y}{L}

$$

Is this true though?

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