# However we might suspect that the solutions to

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Unformatted text preview: x2 ) d2 y dy + p(p + 1)y = 0, − 2x dx2 dx (1.11) where p is a parameter. This equation is very useful for treating spherically symmetric potentials in the theories of Newtonian gravitation and in electricity and magnetism. To apply our theorem, we need to divide by 1 − x2 to obtain d2 y p(p + 1) 2x dy + − y = 0. 2 2 dx dx 1−x 1 − x2 Thus we have P (x) = − 2x , 1 − x2 Q(x) = 12 p(p + 1) . 1 − x2 Now from the preceding section, we know that the power series 1 + u + u2 + u3 + · · · converges to 1 1−u for |u| &lt; 1. If we substitute u = x2 , we can conclude that 1 = 1 + x2 + x4 + x6 + · · · , 1 − x2 the power series converging when |x| &lt; 1. It follows quickly that P (x) = − 2x = −2x − 2x3 − 2x5 − · · · 1 − x2 and p(p + 1) = p(p + 1) + p(p + 1)x2 + p(p + 1)x4 + · · · . 1 − x2 Both of these functions have power series expansions about x0 = 0 which converge for |x| &lt; 1. Hence our theorem implies that any solution to Legendre’s equation will be expressible as a power series about x0 = 0 which converges for...
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