# Dx2 n0 substitution into equation13 yields n 2n

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Unformatted text preview: y nan xn−1 . = a1 + 2a2 x + 3a3 x2 + · · · = dx n=1 (Note that the last summation only goes from 1 to ∞, since the term with n = 0 drops out of the sum.) Diﬀerentiating again yields ∞ d2 y = 2a2 + 3 · 2a3 x + 4 · 3a4 x2 + · · · = n(n − 1)an xn−2 . dx2 n=2 We can replace n by m + 2 in the last summation so that ∞ ∞ d2 y = (m + 2)(m + 2 − 1)am+2 xm+2−2 = (m + 2)(m + 1)am+2 xm . dx2 m+2=2 m=0 7 The index m is a “dummy variable” in the summation and can be replaced by any other letter. Thus we are free to replace m by n and obtain the formula ∞ d2 y = (n + 2)(n + 1)an+2 xn . dx2 n=0 Substitution into equation(1.3) yields ∞ ∞ (n + 2)(n + 1)an+2 xn + n=0 or an xn = 0, n=0 ∞ [(n + 2)(n + 1)an+2 + an ]xn = 0. n=0 Recall that a polynomial is zero only if all its coeﬃcients are zero. Similarly, a power series can be zero only if all of its coeﬃcients are zero. It follows that (n + 2)(n + 1)an+2 + an = 0, or an+2 = − an . (n + 2)(n + 1) (1.5) This is called a recursion formula for the coeﬃcients an . Th...
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