Arkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan 9.8 Power Series 1 Let fangn=0 be a sequence of numbers. Then a power series about x = a is a series of the form 1 X n 2 an(x − a) = a0 + a1(x − a) + a2(x − a) + ··· n=0 Note that a power series about x = a always converges at x = a with sum equals to a0. Example 9.8.1 Which of the following is a power series? (a) A polynomial of degree m: (b) The geometric series 1 + x + x2 + ··· : 1 1 1 (c) The series x + x2 + x3 + ··· : (d) The series 1 + x + (x − 1)2 + (x − 2)3 + (x − 3)4 + ··· : Solution. (a) A polynomial of degree m is a power series about x = 0 since 2 m p(x) = a0 + a1x + a2x + ··· + amx : Note that an = 0 for n ≥ m + 1: (b) The geometric series 1 + x + x2 + ··· is a power series about x = 0 with an = 1 for all n. 1 1 1 (c) The series x + x2 + x3 + ··· is not a power series since it has negative powers of x: (d) The series 1 + x + (x − 1)2 + (x − 2)3 + (x − 3)4 + ··· is not a power series since each term is a power of a different quantity Convergence of Power Series To study the convergence of a power series about x = a one starts by fixing x and then constructing the partial sums S0(x) =a0 S1(x) =a0 + a1(x − a) 2 S2(x) =a0 + a1(x − a) + a2(x − a) . 2 n Sn(x) =a0 + a1(x − a) + a2(x − a) + ··· + an(x − a) : 1 1 Thus obtaining the sequence fSn(x)gn=0. If this sequence converges to a number L; i.e. limn!1 Sn(x) = L, then we say that the power series con- verges to L for the specific value of x: Otherwise, we say that the power series diverges. Power series may converge for some values of x and diverge for other values. The following theorems provide a tool for determining the values of x for which a power series converges and those for which it diverges. Theorem 9.8.1 Given a power series 1 X n 2 an(x − a) = a0 + a1(x − a) + a2(x − a) + ··· n=0 Then one of following is true: (i) The series converges only at x = a; (ii) the series converges for all x; (iii) there is some positive number R such that the series converges absolutely for jx − aj < R and diverges for jx − aj > R: The series may or may not converge for jx − aj = R: That is for the values x = a − R and x = a + R: The largest interval for which the power series converges is called the inter- val of convergence. If a power series converges at only the point x = a then we define R = 0: If a power series converges for all values of x then we define R = 1: We call R the radius of convergence. Finding the radius of convergence The radius of convergence of a power series is found by using the Ratio Test as illustrated in the following examples. Example 9.8.2 P1 xn Find the radius of convergence of the series n=0 n! : Solution. Using the Ratio Test, we find n+1 x (n+1)! n! jxj lim n = lim jxj = lim = 0 n!1 x n!1 (n + 1)! n!1 n + 1 n! for any x: Thus, R = 1 and the series converges everywhere 2 Example 9.8.3 P1 n Find the radius of convergence of the series n=0 n!x : Solution. Using the Ratio Test, we find n+1 (n + 1)!x lim = lim (n + 1)jxj = 1 n!1 n!xn n!1 for x 6= 0: That is, the series diverges for all x 6= 0: If x = 0: the series converges to 0. Hence, R = 0 Example 9.8.4 P1 n2 2n What is the radius of convergence of the series n=0 22n x ? Solution. Using the Ratio Test we find 2 (n+1) x2n+2 2 2 2 22(n+1) n + 1 jxj jxj lim 2 = lim = : n!1 n x2n n!1 n 4 4 22n jxj2 Thus, the series converges if 4 < 1 or jxj < 2: Hence, the radius of con- vergence is R = 2 Endpoint Convergence Example 9.8.5 P1 n−1 (x−1)n Find the interval of convergence of the series n=1(−1) n : Solution. First, we find the radius of convergence. We have jx−1jn+1 n lim n+1 = lim jx − 1j = jx − 1j: n!1 jx−1jn n!1 n + 1 n By the Ratio Test, the given series converges for jx − 1j < 1 so that R = 1: Hence, by Theorem 9.8.1, the series converges for jx−1j < 1 (i.e., 0 < x < 2) and diverges for jx−1j > 1: What about the endpoints x = 0 and x = 2? If we P1 1 replace x by 0 we obtain the series − n=1 n which is divergent (Harmonic P1 n−1 1 series). If we replace x by 2 we obtain the alternating series n=1(−1) n 3 which converges by the alternating series test. Thus, the interval of conver- gence is 0 < x ≤ 2 Differentiation and Integration of Power Series Convergent power series can be differentiated, or integrated, term-by-term to obtain a new power series that has the same radius of convergence as the original power series. The new power series is a representation of the derivative, or antiderivative, of the the original power series. More formally, we have Theorem 9.8.2 1 X n If R > 0 is the radius of convergence of a power series an(x − a) then n=0 the function 2 X n f(x) = a0 + a1(x − a) + a2(x − a) + ··· = an(x − a) n=0 is differentiable on the interval (a − R; a + R) with derivative 1 0 X n−1 f (x) = nan(x − a) : n=1 Since f(x) is differentiable, it is therefore continuous and hence integrable with antiderivative 1 Z X (x − a)n+1 f(x)dx = C + a n n + 1 n=0 where C is a constant of integration. Both new series have radii of conver- gence R (but not necessarily the same interval of convergence). Example 9.8.6 Consider the power series 1 X xn f(x) = : n n=1 Find the interval of convergence of each of the following: (a) f(x) (b) R f(x)dx (c) f 0(x): 4 Solution. (a) By The Ratio Test, we find xn+1 n+1 n lim n = lim jxj = jxj: n!1 x n!1 n n + 1 Thus, the power series f(x) converges for jxj < 1 so that R = 1: Next, we P1 (−1)n check the endpoints. For x = −1; we obtain the series n=1 n which is convergent by the Alternating Series Test. For x = 1; we obtain the series P1 1 n=1 n which is the Harmonic series (divergent). Hence, the interval of convergence for f is [−1; 1): (b) We have 1 Z X xn+1 f(x)dx = n(n + 1) n=1 with radius of convergence equals to 1. For x = −1; we obtain the series P1 (−1)n+1 n=1 n(n+1) which is convergent by the Alternating Series Test. For x = 1; P1 1 we obtain the series n=1 n(n+1) which is a convergent Telescoping Series. Hence, the interval of convergence of R f(x)dx is [−1; 1]: (c) We have 1 X f 0(x) = xn−1: n=1 The radius of convergnce is 1. The series diverges at x = ±1: The interval of convergence is (−1; 1) 5.