Mathematical Methods for Physics and Engineering - Matematica.NET

16.3 SERIES SOLUTIONS ABOUT A REGULAR SINGULAR POINTone solution in the form of a Frobenius series. We therefore substitute y = z ∑ σ ∞n=0 a nz ninto (16.21) and, using (16.13) and (16.14), we obtain∞∑(n + σ)(n + σ − 1)a n z n+σ−2 + 3 ∞∑(n + σ)a n z n+σ−1z − 1n=0which, on dividing through by z σ−2 ,gives∞∑n=0n=0[(n + σ)(n + σ − 1) + 3z (n + σ)+zz − 1+1z(z − 1)∞∑a n z n+σ =0,n=0]a n z n =0.z − 1Although we could use this expression to find the indicial equation and recurrence relations,the working is simpler if we now multiply through by z − 1togive∞∑[(z − 1)(n + σ)(n + σ − 1) + 3z(n + σ)+z] a n z n =0. (16.22)n=0If we set z = 0 then all terms in the sum with the exponent of z greater than zero vanish,and we obtain the indicial equationσ(σ − 1) = 0,which has the roots σ =1andσ = 0. Since the roots differ by an integer (unity), it may notbe possible to find two linearly independent solutions of (16.21) in the form of Frobeniusseries. We are guaranteed, however, to find one such solution corresponding to the largerroot, σ =1.Demanding that the coefficients of z n vanish separately in (16.22), we obtain therecurrence relation(n − 1+σ)(n − 2+σ)a n−1 − (n + σ)(n + σ − 1)a n +3(n − 1+σ)a n−1 + a n−1 =0,which can be simplified to give(n + σ − 1)a n =(n + σ)a n−1 . (16.23)On substituting σ = 1 into this expression, we obtain( ) n +1a n = a n−1 ,nandonsettinga 0 = 1 we find a n = n + 1; so one solution to (16.21) is given by∞∑y 1 (z) =z (n +1)z n = z(1+2z +3z 2 + ···)n=0z=(1 − z) . (16.24)2If we attempt to find a second solution (corresponding to the smaller root of the indicialequation) by setting σ = 0 in (16.23), we find( n)a n = a n−1 .n − 1But we require a 0 ≠0,soa 1 is formally infinite and the method fails. We discuss how tofind a second linearly independent solution in the next section. ◭One particular case is worth mentioning. If the point about which the solution543