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Unformatted text preview: 7. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES 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 = 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 = c n ( x a ) n (  x a < R ), then f [ g ( x )] = summationdisplay n = 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 = ( x 2 ) n = summationdisplay n = ( 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 = ( 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 = parenleftBig x 3 parenrightBig n = summationdisplay n = ( 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 = ( 1) n ( x 1) n 3 n + 1 (  x 1 < 3), and similarly 1 x + 3 = 1 4 + ( x 1) = summationdisplay n = ( 1) n ( x 1) n 4 n + 1 (  x 1 < 4). Combining the last three equations, we deduce that f ( x ) = A summationdisplay n = ( 1) n ( x 1) n 3 n + 1 + B summationdisplay n = ( 1) n ( x 1) n 4 n + 1 = summationdisplay n = 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 = bracketleftbigg ( 1) n + 1 3 n + 1 + 2( 1) n 4 n + 1 bracketrightbigg ( x 1) n (  x 1 < 3). 7.3. Termbyterm differentiation and integration. In this section, our main tool will be the following theorem....
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This note was uploaded on 06/09/2008 for the course M 408d taught by Professor Sadler during the Summer '07 term at University of Texas at Austin.
 Summer '07
 Sadler
 Geometric Series

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