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The temperature in a rod of unit length in which there is heat transfer from its right boundary into a surrounding medium kept at a constant temperature zero is determined from

$\displaystyle {\rm k}\frac{\partial^2u}{\partial x^2} = \frac{\partial u}{\partial t}$
$\displaystyle u\left(0,t\right) = 0, \hskip 15pt \frac{\partial u}{\partial x} \Bigg\vert_{x=1} = -0.2 u\left(1,t\right),\hskip 25pt t > 0$
$\displaystyle u\left(x,0\right) = 0, \hskip 25pt 0 < x < 1$

Therefore

$u(x,t)=0.4\sum\limits_{n=1}^\infty\frac{1-\cos(\alpha_n)}{\alpha_n\left(0.2+\cos^2(\alpha_n)\right)}e^{-k\alpha_n^2t}\sin(\alpha_n x)$

where $\tan(\alpha_n)=-\frac{\alpha_n}{0.2}$ . Fill in the table below with the first four values of $\alpha_n$ :

n	$\alpha_n$
1
2
3
4

Hint: solve tan(x) = -x/h from 0 to a in wolframalpha works, you need to supply h and a.

You can earn partial credit on this problem.