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3) 35 36 LEGENDRE FUNCTIONS Just to motivate this formula, let’s consider the first terms in the Taylor series expansion about t = 0. Using the binomial expansion formula, which is convergent for t2 − 2xt < 1 (for |x| < 1 we can ensure that this holds by making |t| small enough), {1 + (−2xt + t2 )}−1/2 = 1+ − 1 2 (−2xt + t2 ) + − 12 − 32 (−2xt + t2 )2 + · · · 2! 1 = 1 + xt + (3x2 − 1)t2 + · · · = P0 (x) + P1 (x)t + P2 (x)t2 + · · · , 2 as expected. 3). 4) n=0 where, using the binomial expansion, we know that Zn (x) is a polynomial of degree n.

Y=A n=1 What is the radius of convergence of this series? Obtain the second solution of the equation in the form u2 (x) = u1 (x) log x + u1 (x) (−4x + · · · ) . 8 (a) The points x = ±1 are singular points of the differential equation d 2y dy − y = 0. + (x + 1) 2 dx dx Show that one of them is a regular singular point and that the other is an irregular singular point. (b) Find two linearly independent Frobenius solutions of (x2 − 1)2 dy d 2y − xy = 0, +4 2 dx dx which are valid for x > 0. 10 d 2y dy − ky = 0 + (1 + x) dx2 dx (where k is a real constant) in power series form.

22 1 ·2 Using n = 3 gives a3 = −a2 4 4·3·2 = −a0 2 2 2 , 2 3 1 ·2 ·3 and we conclude that an = (−1)n (n + 1)! n+1 a0 . )2 n! One solution is therefore ∞ (−1)n y = a0 n=0 (n + 1) n x , n! which can also be written as ∞ y = a0 x ∞ (−1)n xn−1 (−1)n xn + (n − 1)! n! n=1 n=0 ∞ = a0 −x (−1)m xm + e−x m! m=0 = a0 (1 − x)e−x . This solution is one that we could not have readily determined simply by inspection. We could now use the method of reduction of order to find the second solution, but we will proceed with the method of Frobenius so that we can see how it works in this case.

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