Section 1.4. Power Series De…nition. The function de…ned by 1 n f (x) = cn (x a) (1) n=0 X 2 3 = c0 + c1 (x a) + c2 (x a) + c3 (x a) + ::: is called a power series centered at x = a with coe¢ cient sequence cn n1=0 :The domain of this function consists of all real numbers x such thatf theg power series is convergent, i.e., 1 n D (f) = x : cn (x a) is convergent : ( n=0 ) X Example 4.1. Some examples of power series: 1 f (x) = (n!) xn = 1 + x + (2!) x2 + (3!) x3 + (4!) x4 + ::: n=0 X cn = n!; a = 0 n 2 3 1 (x 3) (x 3) (x 3) (x 3) g (x) = = + + + ::: n 1 2 3 n=1 X1 c = ; a = 3 n n n 2 3 1 (x + 1) (x + 1) (x + 1) h (x) = = 1 + (x + 1) + + + ::: n! 2! 3! n=0 X1 cn = ; a = 1: n! Interval of Convergence We now use Ratio Test to study domain of power series (1). To this end, we write, using the standard series notation 1 1 n an = cn (x a) n=0 n=0 X X 1 with n an = cn (x a) and compute n+1 an+1 = cn+1 (x a) and n+1 an+1 cn+1 (x a) cn+1 = n = x a : an cn (x a) cn j j So a c c lim n+1 = lim n+1 x a = lim n+1 x a : n a n c j j n c j j !1 n !1 n !1 n De…ne a number R by 1 c R = = lim n : (2) c n c lim n+1 !1 n+1 n c !1 n (so that c 1 lim n+1 = ): n c R !1 n We call R the Radius of Convergence for the power series. It follows that an+1 cn+1 x a lim = lim x a = j j: n a n c j j R !1 n !1 n By the Ratio Test, the power series is absolutely convergent if an+1 x a lim = j j < 1 or x a < R; n a R j j !1 n and it is divergent if an+1 x a lim = j j > 1 or x a > R: n a R j j !1 n In other words, if we draw a circle centered at a with radius R;then the power series is absolutely convergent inside interval (a R; a + R) and divergent outside the interval. 2 absolutely convergent inside (a•R, a+R) divergent outside a•R a a+R uncertain at endpoints: at each endpoint a•R or a+R, x=a•R and x=a+R it could be convergent or divergent However, at this point we know nothing about what would happen at the endpoints x = a R and x = a + R: Example 4.2. Find radius of convergence for the series in Example 4.1: 1 n (a) f (x) = (n!) x ; cn = n!; a = 0 n=0 X n 1 (x 3) 1 (b) g (x) = ; c = ; a = 3 n n n n=1 X n 1 (x + 1) 1 (c) h (x) = ; c = ; a = 1: n! n n! n=0 X Solution: (a) cn = n!; cn+1 = (n + 1)!;so c n! 1 R = lim n = lim = lim = 0: n c n (n + 1)! n n + 1 !1 n+1 !1 !1 1 1 (b) c = ; c = ; n n n+1 n + 1 c (n + 1) R = lim n = lim = 1: n c n n !1 n+1 !1 1 1 (c) c = ; c = ; n n! n+1 (n + 1)! c (n + 1)! R = lim n = lim = lim (n + 1) = : n c n n! n 1 !1 n+1 !1 !1 3 We summarize the above analysis as follows. Theorem. Consider the power series 1 n f (x) = cn (x a) n=0 X and its Radius of Convergence, if exists, c R = lim n : n c !1 n+1 There are three possibilities: (1) If R = 0; then the power series is convergent only at x = a; (2) If R = ; then the power series is absolutely convergent for all x;and thus f (x) is de…ned1 everywhere; (3) If 0 < R < ; then the power series is absolutely convergent for x a < R;i.e., 1 j j a R < x < a + R; and is divergent for x a > R: It is not clear what happen when x a = R or x = a R: j j j j In short, the domain of the power series (1) is an interval with endpoints x = a R. This interval could be an open interval, a closed interval, or half open half closed interval, and is called Interval of Convergence. Finding Interval of Convergence: Step #1: …nd the radius of convergence R; and then write down an interval of the form a R; a + R Step #2: check convergencef ofg the series f (a R) and f (a + R) at two endpoints x = a R and x a + R Example 4.3. Find interval of convergence for 3nxn (a) f (x) = 1 ; n=0 pn + 1 n P1 n (x + 2) (b) g (x) = n : n=0 5 Solution: (a)P We know that a = 0; and we …rst …nd its radius of conver- gence. To do so, write down 3n 3(n+1) cn = ; cn+1 = ; pn + 1 (n + 1) + 1 p 4 and calculate c 3n pn + 2 1 pn + 2 1 R = lim n = lim = lim = : n c n pn + 1 3(n+1) n 3 pn + 1 3 !1 n+1 !1 !1 We then write down the interval of the form 1 1 a R; a + R = ; f g 3 3 that only indicates the interval of convergence is an interval with endpoints 1 1 and : It could be of any type: open interval, or closed interval, or a half 3 3 open interval. Thus, 1 1 interval of convergence has the form ; ; 3 3 where the type of interval needs to be determined. Next, we check each endpoints. 1 At the left endpoint x = ; 3 n n n 1 n n 1 3 3 ( 1) n 1 1 3 1 3 1 ( 1) f = = = : 3 p p p n=0 n + 1 n=0 n + 1 n=0 n + 1 X X X By Alternating Series Test, we …nd it convergent (but not absolutely conver- gent). 1 At the right endpoint x = ; 3 1 n 3n 1 1 3 1 1 1 1 f = = = : 3 p p pn n=0 n + 1 n=0 n + 1 n=1 X X X This is a divergent p-series. We thus conclude that 1 1 interval of convergence = [ ; ): 3 3 (b) We follow the same process to study n 1 n (x + 2) g (x) = : 5n n=0 X 5 In this case, a = 2; n n + 1 c = ; c = ; n 5n n+1 5n+1 c n 5n+1 5n R = lim n = lim = lim = 5: n c n 5n n + 1 n n + 1 !1 n+1 !1 !1 So the interval of convergence has the form a R; a + R = 7; 3 : f g f g We next check convergence at endpoints. At x = 7; n n n 1 n ( 7 + 2) 1 n ( 5) 1 n ( 1) 5n 1 g ( 7) = = = = ( 1)n n: 5n 5n 5n n=0 n=0 n=0 n=0 X X X X This series is divergent (by Divergence Test). At x = 3; n n 1 n (3 + 2) 1 n (5) 1 g (3) = = = n 5n 5n n=0 n=0 n=0 X X X is obviously divergent. We conclude that interval of convergence = ( 7; 3): Derivatives and Integrals Suppose that a power series 1 n f (x) = cn (x a) n=0 X has the radius of convergence R: Then Theorem. Inside the interval of convergence, i.e., x a < R j j we have 2 df (x) 1 n 1 n 1 d f (x) 1 n 2 = (c (x a) )0 = nc (x a) ; = n (n 1) c (x a) ; ::: dx n n dx2 n n=0 n=1 n=2 X X X 6 and 1 1 c f (x) dx = c (x a)n dx = C+ n (x a)n+1 ;C is an arbitrary constant. n n + 1 n=0 n=0 Z X Z X If c and d are inside the interval of convergence, i.e., c a ; d a < R; then j j j j d d 1 n f (x) dx = cn (x a) dx: c n=0 c Z X Z Example 4.4. For each series, …nd the integral of convergence and derivative and anti-derivative. Find also an explicit formula. 1 1 (a) f (x) = xn; (b) g (x) = 5n+1x3n n=0 n=0 X X Solution: (a) Since cn = 1; the radius of convergence c R = lim n = 1: c n+1 Within the interval of convergence, i.e., ( 1; 1) ; we have 1 1 n n 1 f 0 (x) = (x )0 = nx n=0 n=1 X X 1 1 xn+1 1 xn f (x) dx = xn dx = + C = + C: n + 1 n n=0 n=0 n=1 Z X Z X X Recall the geometric series 1 1 rn = i¤ r < 1: 1 r n=0 j j X Therefore 1 1 f (x) = xn = 1 x n=0 X and interval of convergence = ( 1; 1) 7 (b) We …rst calculate, within the interval of convergence, 1 1 n+1 3n n+1 3n 1 g 0 (x) = 5 x 0 = 3n5 x n=0 n=0 X X 1 1 5n+1 g (x) dx = 5n+1x3ndx = x3n+1 + C 3n + 1 n=0 n=0 Z X Z X We next rewrite the series as 1 1 1 1 g (x) = 5n+1x3n = 5 (5)n x3 n = 5 5x3 n = 5 yn n=0 n=0 n=0 n=0 X X X X where y = 5x3: Thus 1 5 5 g (x) = 5n+1x3n = 5f (y) = = ; 1 y 1 5x3 n=0 X it is convergent for y = 5x3 < 1: So x < 1=p3 5 and j j j j j j 1 1 Interval of convergence = ; : p3 5 p3 5 Example 4.5.