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Unformatted text preview: ide by xr , and separate out the ﬁrst term from the ﬁrst summation,
obtaining
∞ (2r − 1)ra0 x−1 + (2n + 2r − 1)(n + r)an xn−1 + n=1 ∞ an xn = 0.
n=0 If we let n = m + 1 in the ﬁrst inﬁnite sum, this becomes
∞ (2r − 1)ra0 x−1 + ∞ (2m + 2r + 1)(m + r + 1)am+1 xm +
m=0 an xn = 0.
n=0 Finally, we replace m by n, obtaining
(2r − 1)ra0 x−1 + ∞ ∞ (2n + 2r + 1)(n + r + 1)an+1 xn +
n=0 an xn = 0.
n=0 The coeﬃcient of each power of x must be zero. In particular, we must have
(2r − 1)ra0 = 0, (2n + 2r + 1)(n + r + 1)an+1 + an = 0. (1.16) If a0 = 0, then all the coeﬃcients must be zero from the second of these equations, and we don’t get a nonzero solution. So we must have a0 = 0 and hence
(2r − 1)r = 0.
19 This is called the indicial equation . In this case, it has two roots
r1 = 0, r2 = 1
.
2 The second half of (1.16) yields the recursion formula
an+1 = − 1
an ,
(2n + 2r + 1)(n + r + 1) for n ≥ 0. We can try to ﬁnd a generalized power series solution for either root of the
indic...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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