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Home , Power series

Arkansas Tech University MATH 2924: II Dr. Marcel B. Finan

9.8 Power ∞ Let {an}n=0 be a of numbers. Then a power series about x = a is a series of the form ∞ 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 of degree m. (b) The 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 ∞ Thus obtaining the sequence {Sn(x)}n=0. If this sequence converges to a number L, i.e. limn→∞ 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

∞ 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 |x − a| < R and diverges for |x − a| > R. The series may or may not converge for |x − a| = 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 = ∞. We call R the .

Finding the radius of convergence The radius of convergence of a power series is found by using the as illustrated in the following examples.

Example 9.8.2 P∞ 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! |x| lim n = lim |x| = lim = 0 n→∞ x n→∞ (n + 1)! n→∞ n + 1 n! for any x. Thus, R = ∞ and the series converges everywhere

2 Example 9.8.3 P∞ 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)|x| = ∞ n→∞ n!xn n→∞ 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 P∞ 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 |x| |x| lim 2 = lim = . n→∞ n x2n n→∞ n 4 4 22n

|x|2 Thus, the series converges if 4 < 1 or |x| < 2. Hence, the radius of con- vergence is R = 2

Endpoint Convergence

Example 9.8.5 P∞ 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

|x−1|n+1 n lim n+1 = lim |x − 1| = |x − 1|. n→∞ |x−1|n n→∞ n + 1 n By the Ratio Test, the given series converges for |x − 1| < 1 so that R = 1. Hence, by Theorem 9.8.1, the series converges for |x−1| < 1 (i.e., 0 < x < 2) and diverges for |x−1| > 1. What about the endpoints x = 0 and x = 2? If we P∞ 1 replace x by 0 we obtain the series − n=1 n which is divergent (Harmonic P∞ n−1 1 series). If we replace x by 2 we obtain the n=1(−1) n

3 which converges by the . 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 , or , of the the original power series. More formally, we have

Theorem 9.8.2 ∞ X n If R > 0 is the radius of convergence of a power series an(x − a) then n=0 the

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

∞ 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 ∞ Z X (x − a)n+1 f(x)dx = C + a n n + 1 n=0 where C is a . 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 ∞ 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 |x| = |x|. n→∞ x n→∞ n n + 1

Thus, the power series f(x) converges for |x| < 1 so that R = 1. Next, we P∞ (−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 P∞ 1 n=1 n which is the Harmonic series (divergent). Hence, the interval of convergence for f is [−1, 1). (b) We have ∞ 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 P∞ (−1)n+1 n=1 n(n+1) which is convergent by the Alternating Series Test. For x = 1, P∞ 1 we obtain the series n=1 n(n+1) which is a convergent . Hence, the interval of convergence of R f(x)dx is [−1, 1]. (c) We have ∞ 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)

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