LN_7_13 - 7. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES...

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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. Term-by-term 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.

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LN_7_13 - 7. REPRESENTATIONS OF FUNCTIONS BY POWER SERIES...

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