I have acquired the sum below through Fourier, and was wondering if there was anyway to simplify it, since it is large and ugly.
$$\sum \limits_{n=1}^\infty \frac{-2K_1}{n\pi} \left(\cos\left(\frac{3n\pi}{4}\right)-\cos\left(\frac{n\pi}{4}\right)\right) e^{-\left(\frac{n\pi K_2}{L}\right)^2 t} \sin\left(\frac{n\pi x}{L} \right)$$
$K_1,K_2$ are constants(that can't be found).
I can see that $\left(\cos\left(\frac{3n\pi}{4}\right)-\cos\left(\frac{n\pi}{4}\right)\right)$ is:
$$n=1\Longrightarrow-\sqrt{2}$$ $$n=2\Longrightarrow0$$ $$n=3\Longrightarrow\sqrt{2}$$ $$n=4\Longrightarrow0$$ $$n=5\Longrightarrow\sqrt{2}$$ $$n=6\Longrightarrow0$$ $$n=7\Longrightarrow-\sqrt{2}$$ $$n=8\Longrightarrow0$$
But I don't see how I can use this to simplify the problem, or any other way. Any suggestions would be pleasant.
It's not that complicated. When you get to solving the Laplace equation in a rectangle, the formulas will get uglier. The only thing that could be simplified here is the difference of cosines. You can use the identity $$\cos a-\cos b = -2\sin\frac{a+b}{2}\sin\frac{a-b}{2}$$ to get $$\cos\left(\frac{3n\pi}{4}\right)-\cos\left(\frac{n\pi}{4}\right) =-2 \sin\frac{n\pi}{2}\sin \frac{n\pi}{4}$$ and consider various cases. Or just deal with the difference of cosines in the original form: for $n=8k\pm 1$ there is one value, for $n=8k\pm 3$ there is another.
But I don't think these manipulations win you anything, because instead of straightforward summation over $n$ you get complicated sums with modular arithmetics. I would leave the sum as is.