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48) implies that the only contribution to the integral comes through this term, namely, 1 −1 [Pnm (t)]2 1 dt = −1 (n + m)! an Pn (t)tn dt. (n − m)! 50) to finally obtain 1 −1 [Pnm (t)]2 dt = 2(n + m)! (2n + 1)(n − m)! 19 Let BR be the open ball with a radius R and a center at the origin. For x ∈ BR , denote by R2 x˜ := 2 x |x| 53 the inverse point of x with respect to the sphere ∂BR . It is convenient to define the ideal point ∞ as the inverse of the origin. Fix y ∈ BR . Recall that as a function of x the function Γ(|x − y|) is harmonic for all x = y and satisfies −∆Γ(x; y) = δ(x − y) .

In this case we get UN −1,n = UN −2,n . Unfortunately this is a first order approximation and the error due to it might spoil the entire (second order) scheme. Therefore, it is beneficial to add an artificial point UN,n , and to approximate the Neumann condition at x = 1 by UN,n = UN −2,n . Notice that now UN −1,n is an internal point. 7 The analytic solution: It is easy to see that u(x, t) = t5 satisfies all the problem’s conditions, and thus is the unique solution. 12 N= 1 n i + 1 = 11, tn = , xi = .

4(N − 2) be the set of boundary point. For each i define the harmonic function Ti , such that Ti (pi ) = 1, while Ti (pj ) = 0 if j = i. Clearly the set {Ti } spans all solutions to the Laplace equation in the grid. It also follows directly from the construction that the set {Ti } is linearly independent.