However we might suspect that the solutions to

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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| < 1. If we substitute u = x2 , we can conclude that 1 = 1 + x2 + x4 + x6 + · · · , 1 − x2 the power series converging when |x| < 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| < 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...
View Full Document

Ask a homework question - tutors are online