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8.6 Power Series Convergence

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8.6 Power Series Convergence Power Series Week 8 8.6 Power Series Convergence Definition A power series about c is a function defined by the infinite sum 1 X n f(x) = an(x − c) n=0 1 where the terms of the sequence fangn=0 are called the coefficients and c is referred to as the center of the power series. Theorem / Definition Every power series converges absolutely for the value x = c. For every power series that converges at more than a single point there is a value R, called the radius of convergence, such that the series converges absolutely whenever jx − cj < R and diverges when jx − cj > R. A series may converge or diverge when jx − cj = R. The interval of x values for which a series converges is called the interval of convergence. By convention we allow R = 1 to indicate that a power series converges for all x values. If the series converges only at its center, we will say that R = 0. Examples f (n)(c) (a) Every Taylor series is also a power series. Their centers coincide and a = . n n! (b) We saw one example of a power series weeks ago. Recall that for jrj < 1, the geometric series converges and 1 X a arn = a + ar + ar2 + ar3 ::: = 1 − r n=0 We can view this as a power series where fang = a and c = 0, 1 X a axn = 1 − x n=0 This converges for jx − 0j = jxj < 1 so the radius of converges is 1. The series does not converge for x = 1 or x = −1. Therefore the interval of convergence is (−1; 1). Testing a Power Series for Convergence Step 1. Use the ratio test or root test to determine the radius of convergence. jx − cj < R Step 2. Plug in x = c + R and x = c − R. Check if the series converges for either of these end points. Step 3. Write out the interval of convergence ([c−R; c+R)] with the appropriate end point signifier chosen. 1 Power Series Week 8 Examples Determine the interval of convergence. 1 X xn (a) p n n3n n=1 By the ratio test the series will converge when p p a xn+1 n n3n n n 1 lim j n+1 j < 1 =) lim p · < 1 =) lim p x < 1 n+1 n n!1 an n!1 (n + 1) n + 13 x n!1 (n + 1) n + 1 3 x =) j j < 1 =) jxj < 3 3 Therefore the radius of convergence is 3 . We now check the end points. When x = 3 we have the series 1 X 1 p n n n=1 which converges as a p series with p > 1. When x = −3 we have the series, 1 X 1 (−1)n p n n n=1 which converges by the alternating series test. This gives that Interval of Convergence: [-3,3] . 1 n X 2 n (b) (4x − 8) n n=1 This strictly isn't expressed as a power series. But we can factor out a 4 from the parenthetical statement to give a series centered at 2, 1 n X 8 n (x − 2) n n=1 p Remember that lim n n = 1. Let's try the root test on this. The series will converge absolutely when n!1 r n pn n 8 n 8 1 lim janj = lim (x − 2) = lim p (x − 2) = j8(x − 2)j < 1 =) jx − 2j < n!1 n!1 n n!1 n n 8 1 =) Radius of Convergence: 8 Now check the endpoints. The left and right endpoints respectively give the series, 1 1 X 1 X 1 (−1)n n n n=1 n=1 The former is the alternating harmonic series and converges. The latter is the harmonic series and diverges. 1 1 15 17 Interval of Convergence: [2 − ; 2 + ) = ; 8 8 8 8 2 Power Series Week 8 1 X 2 (c) 2n (x − 3)n n=0 Apply the ratio test, 2 2n +2n+1(x − 3)n+1 lim = lim 22n + 1(x − 3) = jx − 3j lim 22n+1 n2 n n!1 2 (x − 3) n!1 n!1 This limit goes to infinity of course. Unless, the number we are multiplying by is zero. Therefore this converges only when x = 3. Therefore the radius of converges is 0 and the interval of convergence is [3; 3]. 1 X (x + 1)n (d) n! n=0 Note that the center of this power series is −1. Use the ratio test n+1 (x + 1) n! x + 1 lim · = lim = 0 < 1 n!1 (n + 1)! (x + 1)n n!1 n + 1 This series converges no matter what the value of n is. Therefore, Radius of Convergence: 1 Interval of Convergence: (−∞; 1) 8.7 New from Old: Representing Functions by Power Series Now that we know a bit about power series how can we find the power series for a particular function? So 1 far we've only seen one really, = P xn for jxj < 1. We want to milk this identity for as much as we 1 − x can. One way we can do this is by arithmetic operations and substitution. Examples Determine the power series representation for the given function. Give the interval of convergence for the series. x2 (a) 1 − x We don't have much to work with right now. Let's stare at the power series we know and see if there's a way to transform it into the one we want. Notice that we can do this by multiplication. 1 1 x2 1 X X = x2 · = x2 xn = xn+2 1 − x 1 − x n=0 n=0 There we have it. Our power series representation. Let's think about the interval of convergence. The base series had the interval (−1; 1). We multiplied that series by a value not depending on n. Hence the convergence of the series is not affect. Our interval of convergence is then (−1; 1). 1 (b) 1 + 5y 1 Notice we can get this function by plugging in x = −5y into . Therefore, 1 − x 3 Power Series Week 8 1 1 1 1 X X = = = (−5y)n = (−1)5nyn 1 + 5y 1 − (−5y) n=0 n=0 The original series converges when jxj < 1 so after our substitution our series will converge for j − 5yj < 1 1 =) jyj < 5 . There's nothing special about having chosen y for our variable so we could instead say, 1 1 X 1 = (−1)n5nxn for jxj < 1 + 5x 5 n=0 1 (c) 2 − z Notice that, 1 1 1 1 1 1 X z n X 1 = · = · = zn 2 − z 2 1 − ( z ) 2 2 2n+1 2 n=0 n=0 This will converge when z = jxj < 1 =) jzj < 2 2 4 Power Series Week 8 8.8 Term by Term Differentiation 1 X n Theorem Assume the power series an(x − c) converges for jx − cj < R for some R > 0. For x values n=0 in this interval we can define a function 1 X n f(x) = an(x − c) n=0 f(x) has derivatives of all orders and they are given by differentiating the sum term by term. That is 1 0 X n−1 f (x) = nan(x − c) n=0 Furthermore, the derivative's power series has the same radius of convergence as the original. Note The ability to differentiate term by term is a special property of power series. For example, 1 X sin(n!x) f(x) = n2 n=1 1 The series defining this function converges for all values of x (compare it to n2 ). If we take the derivative of each term we get the series, 1 X cos(n!x) n! n2 n=1 which diverges for all x by the divergence test. Examples Determine a power series for the given function. 1 (a) (1 − x)2 Let's use our shiny new toy in the form of the theorem of this section. We want to think if we know a function whose derivative will give us the above function. We do. 1 1 1 1 d 1 d X X X = = xn = nxn−1 = (n + 1)nxn (for − 1 < x < 1) (1 − x)2 dx 1 − x dx n=0 n=0 n=0 k! (b) (1 − x)k+1 Notice that 1 1 1 k! dk 1 dk X X n! X (n + k)! = = xn = xn−k = xn (for − 1 < x < 1) (1 − x)k+1 dxk 1 − x dxk (n − k)! n! n=0 n=k n=0 Note The theorem says that the radius of convergence is unaffected by differentiation. However, the end points of convergence may be affect. 1 X xn f(x) = n2 n=1 5 Power Series Week 8 converges for −1 ≤ x ≤ 1. It's radius of convergence is 1 and it converges at both endpoints. On the other hand, 1 X xn−1 f 0(x) = n n=1 has radius of converges 1 but converges for −1 ≤ x < 1. Since setting x = 1 gives us the harmonic series which diverges. 8.9 Term by Term Integration 1 X n Theorem Assume the power series an(x − c) converges for jx − cj < R for some R > 0. For x values n=0 in this interval we can define a function 1 X n f(x) = an(x − c) n=0 f(x) is integrable and Z 1 X an f(x) dx = C + (x − c)n+1 n + 1 n=0 Furthermore, the integral's power series has the same radius of convergence as the original.
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