Since the solution of the Legendre’s equations are the Legendre polynomials $P_n(\mu)$, write the general solution for $u(r, \theta)$. I am working on the problem

Consider the steady-state of the heat equation in a ball of radius a centred at the origin. In spherical coordinates, the ball occupied the region $0 \le r \le a$, $0 \le \theta \le \pi$ and $0 \le \phi < 2\pi$. It has a given temperature $g(\theta)$ imposed along its boundary, which is the sphere of radius $a$. Since the boundary condition is independent of $\phi$, we can assume that the temperature at the point $(r, \theta, \phi)$ in the ball is given as $u(r, \theta)$, which is given by the solution of the following boundary value problem,
$$\dfrac{1}{r^2} \dfrac{\partial}{\partial{r}} \left( r^2 \dfrac{\partial{u}}{\partial{r}} \right) + \dfrac{1}{r^2 \sin(\theta)} \dfrac{\partial}{\partial{\theta}} \left( \sin(\theta) \dfrac{\partial{u}}{\partial{\theta}} \right) = 0,$$
subject to boundary conditions
$u(a, \theta) = g(\theta)$ for $0 ≤ \theta ≤ \pi$.
(i) Show that the separation of variables $u(r, \theta) = R(r)S(\theta)$ leads to the equations
$$\dfrac{1}{\sin(\theta)} \dfrac{d}{d \theta} \left( \sin(\theta) \dfrac{dS}{d \theta} \right) + \lambda S = 0$$
and
$$(r^2 R')' - \lambda R = 0$$
(ii) Now let $\lambda = n(n + 1)$ for $n = 0,1,2,3, \dots$ and let $\mu = \cos(\theta)$, transform the ODE for $S(\theta)$ to the following Legendre’s equation:
$$(1 - \mu^2) \dfrac{\partial^2{S}}{\partial{\mu}^2} - 2\mu \dfrac{dS}{d \mu} + n(n + 1)S = 0$$
(iii) Solve the differential equation for $R$ for each eigenvalue $\lambda n = n(n + 1)$. (Hint: Try $R = Ar^m$.)
(iv) Given the solution of the Legendre’s equations are the Legendre polynomials $P_n(\mu) = P_n(\cos(θ))$, write the general solution for $u(r, \theta)$ as an infinite series.

I'm stuck on (iv) and just don't understand how to do this. I don't have very much experience with Legendre polynomials, so this is probably why. My textbook also doesn't have any solutions, so I am totally stuck. I would be very thankful if someone could please take the time to explain what (iv) is asking and show how (iv) is done. Thank you very much for your help!
 A: You have slightly different notation than my book. I am just going to write here. The steady state of the heat equation is Laplace's equation. 
We should get this equation.
$$ \nabla^{2}u = 0 \tag{1} $$
with boundary conditions
$$ u(a,\theta, \phi) = F(\theta,\phi) \tag{2} $$
which corresponds to what you were saying.  You should get an infinite series like this
$$ u(r,\theta, \phi) =\sum_{m=0}^{\infty} \sum_{n=m}^{\infty} \rho^{n}\big[ A_{mn} \cos(m\theta) + B_{mn} \sin(m\theta) \big] P_{n}^{m}(\cos(\phi)) \tag{3}$$
the non-homogeneous boundary condition gives 
$$ F(\theta,\phi) =\sum_{m=0}^{\infty} \sum_{n=m}^{\infty} a^{n}\big[ A_{mn} \cos(m\theta) + B_{mn} \sin(m\theta) \big] P_{n}^{m}(\cos(\phi)) \tag{4}$$
in order to find the coefficients, we use orthogonality. 
$$ a^{n}B_{mn} = \frac{\iint F(\theta,\phi)\sin(m\theta) P_{n}^{m}(\cos\phi) \sin(\phi)d \phi d\theta }{\iint \sin^{2}(m\theta)[P_{n}^{m}(\cos(\phi))]^{2} \sin(\phi) d\phi d\theta 
} \tag{5} $$
by the same method we find $A_{mn}$
$$ a^{n}A_{mn} = \frac{\iint F(\theta,\phi)\cos(m\theta) P_{n}^{m}(\cos\phi) \cos(\phi)d \phi d\theta }{\iint \cos^{2}(m\theta)[P_{n}^{m}(\cos(\phi))]^{2} \cos(\phi) d\phi d\theta 
} \tag{6} $$
in your case, you would simply have
$$ u(a,\theta ) = g(\theta) \tag{7} $$
Note that $g(\theta)$ isn't a function of $\phi$ so 
$$ g(\theta) =\sum_{m=0}^{\infty} \sum_{n=m}^{\infty} a^{n}\big[ A_{mn} \cos(m\theta) + B_{mn} \sin(m\theta) \big] \tag{8}$$
So when we solve for $A_{mn}, B_{mn}$
$$ a^{n}B_{mn} = \frac{\int_{0}^{\pi} g(\theta)\sin(m\theta)  d\theta }{\int_{0}^{\pi} \sin^{2}(m\theta) d\theta 
} \tag{9} $$
$$ a^{n}A_{mn} = \frac{\int_{0}^{\pi} g(\theta)\cos(m\theta) d\theta }{\int_{0}^{\pi} \cos^{2}(m\theta) d\theta 
} \tag{10} $$
There should have been boundary conditions on the integrals to figure out the normalization. I.e 
$$ B_{00} = \frac{\int_{0}^{\pi} g(\theta) \cdot  0 d\theta }{\int_{0}^{\pi} 0  d\theta 
} \tag{11} $$ 
Which means $ B_{00} $ can be anything. However, this shouldn't matter for $A_{00}$ Test it. Ok. Now you should be actually able to get a real normalization part since you have definite integral boundaries. 
$$ A_{00} = \frac{\int_{0}^{\pi} g(\theta) \cdot 1 d\theta }{\int_{0}^{\pi} \cdot 1  d\theta 
} \tag{12} $$
I am attempting to say this coefficient on the bottom is a function of $m$ and at $m=0$ it equals $\pi$ 
$$ A_{00} = \frac{\int_{0}^{\pi} g(\theta) \cdot 1d\theta } {\pi
} \tag{13} $$
It is "normalizing" the top integral 
