LN_7_13 - 7 R EPRESENTATIONS OF F UNCTIONS BY P OWER S ERIES 7.1 Introduction We start by recalling Theorem 4.4 Replacing r by x we can restate(4.4 in

# LN_7_13 - 7 R EPRESENTATIONS OF F UNCTIONS BY P OWER S...

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7. R EPRESENTATIONS OF F UNCTIONS BY P OWER S ERIES 7.1. Introduction. We start by recalling Theorem 4.4. Replacing r by x , we can restate (4.4) in the form 1 1 - x = summationdisplay n = 0 x n ( | x |< 1), (7.1) that is, the geometric series represents the function f ( x ) = (1 - x ) - 1 when | x |< 1. In this lecture, we will learn how to derive new power series representations from old ones. 7.2. Substitution. We start with a very simple idea: if f ( x ) = summationdisplay n = 0 c n ( x - a ) n ( | x - a |< R ), then f [ g ( x )] = summationdisplay n = 0 c n ( g ( x ) - a ) n ( | g ( x ) - a |< R ). If g ( x ) is chosen wisely, this simple idea can be used to a great advantage. Examples. (A) Substituting ( - x 2 ) mapstochar→ x in (7.1), we find that 1 1 + x 2 = 1 1 - ( - x 2 ) = summationdisplay n = 0 ( - x 2 ) n = summationdisplay n = 0 ( - 1) n x 2 n , provided that vextendsingle vextendsingle - x 2 vextendsingle vextendsingle < 1 | x | 2 < 1 | x |< 1. That is, we obtain the power series representation 1 1 + x 2 = summationdisplay n = 0 ( - 1) n x 2 n ( | x |< 1). (B) By (7.1) with ( - x /3) mapstochar→ x , 1 3 + x = 1/3 1 - ( - x /3) = 1 3 summationdisplay n = 0 parenleftBig - x 3 parenrightBig n = summationdisplay n = 0 ( - 1) n x n 3 n + 1 , provided that |- x /3 |< 1 | x |< 3. (C) We now want to represent the function f ( x ) = x + 1 x 2 + 5 x + 6 37
as a power series in ( x - 1). We know that f has a partial fraction decomposition of the form x + 1 x 2 + 5 x + 6 = A x + 2 + B x + 3 (7.2) for some constants A and B . Replacing x by ( x - 1) in the previous example, we obtain 1 x + 2 = 1 3 + ( x - 1) = summationdisplay n = 0 ( - 1) n ( x - 1) n 3 n + 1 ( | x - 1 |< 3), and similarly 1 x + 3 = 1 4 + ( x - 1) = summationdisplay n = 0 ( - 1) n ( x - 1) n 4 n + 1 ( | x - 1 |< 4). Combining the last three equations, we deduce that f ( x ) = A summationdisplay n = 0 ( - 1) n ( x - 1) n 3 n + 1 + B summationdisplay n = 0 ( - 1) n ( x - 1) n 4 n + 1 = summationdisplay n = 0 bracketleftbigg A ( - 1) n 3 n + 1 + B ( - 1) n 4 n + 1 bracketrightbigg ( x - 1) n ( | x - 1 |< 3). (Note that x must satisfy the more restrictive of the conditions | x - 1 |< 3 and | x - 1 |< 4: namely, the former.) Thus, it remains to find the coefficients A and B in (7.2), or equiva- lently, in the identity x + 1 = A ( x + 3) + B ( x + 2). Plugging in x =- 2 and x =- 3, we obtain the equations - 1 = A and - 2 =- B , respectively. Therefore, A =- 1, B = 2, and f ( x ) = summationdisplay n = 0 bracketleftbigg ( - 1) n + 1 3 n + 1 + 2( - 1) n 4 n + 1 bracketrightbigg ( x - 1) n ( | x - 1 |< 3). 7.3. Term-by-term differentiation and integration. In this section, our main tool will be the following theorem. Theorem. Suppose that f ( x ) = summationdisplay n = 0 c n ( x - a ) n ( | x - a |< R ). (7.3) Then f is differentiable (and therefore continuous) in the interval ( a - R , a + R ) and satisfies (i) d dx [ f ( x )] = summationdisplay n = 0 c n d dx [( x - a ) n ] = summationdisplay n = 1 nc n ( x - a ) n - 1 ( | x - a |< R ) ; (ii) integraldisplay x a f ( t ) dt = summationdisplay n = 0 c n integraldisplay x a ( t - a ) n dt = summationdisplay n = 0 c n n + 1 ( x - a ) n + 1 ( | x - a |< R ) . 38
In particular, the series appearing on the right sides of (i) and (ii) have radii of convergence R. Remarks. 1. Conceptually, the theorem says that we can differentiate and integrate a power series term-by-term: this is the meaning of the first equalities in (i) and (ii). The second equali- ties in (i) and (ii) simply record the almost trivial observations that d dx [( x - a ) n ] = n ( x - a ) n - 1 and integraldisplay x a ( t - a ) n dt = ( x - a ) n + 1 n + 1 .