Math 2420 (Spring 2017) Assignment #7 Problem #1: Find the intervals of convergence of the following power series at x = 0: (a) X n3n xn ; n=0 (b) X n=1 xn n10n1 Solution: (a). We apply the Ratio Test: |(n + 1)3n+1 xn+1 | n+1 |an + 1| = lim = 3|x| lim = 3|x| n n n n n n |an | |n3 x | lim So the series converges if 3|x| < 1 i.e. |x| < 13 and is divergent when |x| > 1 . Consequently, the interval of convergence is ( 31 , 13 ) and the radius of 3 convergence is R = 13 (b). Again, we apply the Ratio Test: xn+1 | (n+1)10n | |an + 1| |x| n |x| lim = lim = lim = n x n |an | n | 10 n n + 1 10 | n10n1 Thus, the series converges if |x| < 10, so the interval of convergence is I := (10, 10) and the radius of convergence is R=10. 1 Problem #2: Find the power series expansions of the following functions at x = 0. \u0012 \u0013 1+x (a) f (x) = ln ; 1x (b) f (x) = x ; 1 x 2x2 (c) f (x) = 1 ; 1 + x3 Solution: (a). We have: Z xX Z x Z x X X xn+1 dt n n n n (1) t dt = (1) t dt = (1)n = ln(1+x) = n+1 0 n=0 0 0 1+t n=0 n=0 and Z ln(1 x) = 0 x dt = 1t Z 0 xX Z X t dt = n=0 n n=0 x n t dt = 0 X xn+1 n=0 n+1 thus, we see that \u0012 \u0013 n+1 X X xn+1 1+x n x (1) + f (x) = ln = ln(1 + x) ln(1 x) = 1x n + 1 n=0 n + 1 n=0 X X 2 xn+1 [(1) + 1] = = x2n+1 n + 1 n=0 2n + 1 n=0 n (b). \u0014 \u0015 x x 1 1 1 f (x) = 2 = = + 2x + x 1 (2x 1)(x + 1) 3 (2x 1) x + 1 " # \u0014 \u0015 X 1 1 1 X 1 = (2x)n (x)n = 3 1 2x x + 1 3 n=0 n=0 X1 1X n = (2 (1)n )xn = (2n + (1)n+1 )xn 3 n=0 3 n=0 2 (c). X X 1 3 n f (x) = = (x ) = (1)n x3n 1 + x3 n=0 n=0 3 Problem #3: Use the power series method to find two linearly independent solutions of the following equations near x = 0 ((i) find the recurrence formula for an , (ii) solve the recurrence formula for an , (iii) find a formula for the two power series solutions) (a) y 00 + x2 y = 0; (b) y 00 + 3xy 0 + 3y = 0; (c) (1 x2 )y 00 6xy 0 4y = 0; Solution: We take advantage of the recurrence formula. In this problem we have p(x) = 0 and q(x) = x2 so that pn = 0 for all n, and q2 = 1 and all other qj 's equal zero. Thus we have an+2 = (an2 ) (n + 2)(n + 1) for n > 1 and, we also note that a2 = 0 and a3 = 0 = Solution for y1 : i.e. a0 = 1 and a1 = 0 a0 = 1, a1 = 0, a2 = 0, a3 = 0, a4 = a8 = a1 2 = 1 1 a0 = , a5 = 0, a6 = 0, a7 = 0, 43 43 (1)2 (1)2 1 a4 = = 2 87 (8 4)(7 3) (4 )2!(7 3) 1 (1)3 (1)3 a8 = = 3 12 11 (12 8 4)(11 7 3) (4 )3!(11 7 3) thus we see that for n = 4k we have: a4k = (1)k (1)k = (4k )k!(4k 1)(4k 5) 7 3 (22k )k!(4k 1) 7 3 So that we have y1 defined as follows: y1 = 1 + X (1)k x4k (22k )k!(4k 1) 7 3 k=1 4 Solution for y2 , i.e. a0 = 0 and a1 = 1 a0 = 0, a1 = 1, a2 = 0, a3 = 0, a4 = a6 = 0, a7 = 0, a8 = 0, a9 = a10 = a11 = a12 = 0, a13 = 1 1 (1)1 a0 = 0, a5 = a1 = 43 54 45 (1)2 (1)2 1 a5 = = 2 98 (4 8)(5 9) (4 )2!(5 9) 1 (1)3 (1)3 a9 = = 3 12 13 (4 8 12)(13 9 5) (4 )3!(13 9 5) So that we have: a4k+1 (1)k (1)k = k = 2k (4 )k!(4k + 1)(4k 3) 9 5 (2 )k!(4k + 1)(4k 3) 9 5 Therefore our second solution is: y2 = x + X (1)k x4k+1 (22k )k!(4k + 1)(4k 3) 9 5 k=1 (b). In this case we clearly have: p(x) = 3x p1 = 3 q(x) = 3 q0 = 3 Which gives us the recurrence formula: 1 [3an n + 3an ] (n + 2)(n + 1) 3(n + 1)an = (n + 2)(n + 1) 3an = n+2 an+2 = an+2 an+2 Solution for y1 , i.e. a0 = 1 and a1 = 0 3 3 3 (1)2 32 a0 = 1, a1 = 0, a2 = a0 = , a3 = 0, a4 = a2 = 2 2 4 24 3 (1)3 33 (1)3 33 a5 = 0, a6 = a4 = = 6 246 (23 )3! 5 ... So we have a general formula for an = a2k : a2k (1)k 3k = (2k )k! Therefore the first solution is given by: y1 = 1 + X (3)k x2k (2k )k! k=1 Solution for y2 , i.e. a0 = 0 and a1 = 1 3 (1)1 31 3 a0 = 0, a1 = 1, a2 = 0, a3 = a1 = , a 4 = a1 = 0 3 3 4 3 (3)2 3 (3)3 a5 = a3 = , a6 = 0, a7 = a5 = , a8 = 0 5 53 7 753 So the general case for n = 2k + 1 is: ... (3)k (2k)(2k 2) 4 2 (3)k = (2k + 1) 5 3 (2k + 1)(2k)(2k 1)(2k 2) 5 4 3 2 k k k (1) 3 2 k! (6)k k! = = (2k + 1)! (2k + 1)! a2k+1 = And so we have the second solution: y2 = x + X (6)k k! 2k+1 x (2k + 1)! k=1 (c). To solve this problem we apply directly, the power series method: We assume y = X an x n y = X an n(n 1)x n2 = n=0 0 X an nxn1 and n=0 n=0 00 y0 = X an+2 (n + 2)(n + 1)xn n=0 00 We substitute y, y , y into the original equation to obtain: X X X X n 2 n2 n1 an+2 (n+2)(n+1)x x an n(n1)x 6x an nx 4 an xn = 0 n=0 n=0 n=0 6 n=0 X X X X n n n an+2 (n+2)(n+1)x an n(n1)x 6an nx 4an xn = 0 n=0 n=0 X \u0002 n=0 n=0 \u0003 an+2 (n + 2)(n + 1) an (n2 + 5n + 4) xn = 0 n=0 Therefore each coefficient of the power series must be zero, i.e., we have the following recurrence formula: For n = 0, 1, 2, . . . an+2 = an+2 (n + 2)(n + 1) an (n2 + 5n + 4) = 0 (n + 4)(n + 1) (n + 4) an or an+2 = an (n + 2)(n + 1) (n + 2) Solution for y1 , i.e. a0 = 1 and a1 = 0 4 6 6 4 6 a0 = 1, a1 = 0, a2 = a0 = 2, a3 = 0, a4 = a2 = = 2 4 4 2 2 8 8 a5 = 0, a6 = a4 = 6 6 So, the general recurrence 6 8 10 10 8 10 = , a7 = 0, a8 = a6 = = 2 2 8 8 2 2 formula for n = 2k is: an = a2k = ... 2k + 2 =k+1 2 Therefore, we have y1 : y1 = 1 + X (k + 1)x2k = X (k + 1)x2k k=0 k=1 Solution for y2 , i.e. a0 = 0 and a1 = 1 5 5 7 7 5 7 a0 = 0, a1 = 1, a2 = 0, a3 = a1 = , a4 = 0, a5 = a3 = = 3 3 5 5 3 3 9 7 9 9 a6 = 0, a7 = a5 = = , . . . 7 7 3 3 So, the general recurrence formula for n = 2k + 1 is: an = a2k+1 = 7 2k + 3 3 And we have y2 : y2 = x + X (2k + 1) k=1 3 x 2k+1 = X (2k + 1) k=0 8 3 x2k+1