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In summary, we have $u(\rho,\phi) = \frac{A_0}{2} + \sum_{n=1}^\infty \rho^n \left(A_n\cos n\phi+B_n\sin n \phi \right).$ The one remaining boundary condition can now be used to determine the coefficients $$A_n$$ and $$B_n$$, \begin{aligned} U(c,\phi) &= \frac{A_0}{2} + \sum_{n=1}^\infty c^n \left(A_n\cos n\phi+B_n\sin n \phi \right)\nonumber\\ &= \begin{cases} 100 & \text{if 0 < \phi < \pi} \\ 0 & \text{if \pi < \phi < 2\pi} \end{cases}\quad.\end{aligned} We find \begin{aligned} A_0 &= \frac{1}{\pi} \int_0^\pi 100\, d\phi = 100, \nonumber\\ c^n A_n &= \frac{1}{\pi}\int_0^\pi 100\cos n\phi\, d\phi = \frac{100}{n\pi} \sin(n\phi)|^\pi_0=0,\nonumber\\ c^n B_n &= \frac{1}{\pi}\int_0^\pi 100\sin n\phi\, d\phi = -\frac{100}{n\pi} \cos(n\phi)|^\pi_0 \nonumber\\&= \begin{cases} 200/(n\pi) & \text{if n is odd}\\ 0 & \text{if n is even} \end{cases}\quad.\end{aligned} In summary $u(\rho,\phi) = 50 + \frac{200}{\pi} \sum_{n~\rm odd} \left(\frac{\rho}{c}\right)^n \frac{\sin n \phi}{n}. \label{eq:T0-100}$ We clearly see the dependence of $$u$$ on the pure number $$r/c$$, rather than $$\rho$$. A three dimensional plot of the temperature is given in Fig. $$\PageIndex{1}$$.