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33. continued (c) The coefficient is } f ( k k ) ! (0) } 5 (d) f (0) 5 1, f 9 (0) 5 m , and we’re done by part (c). 34. Because f ( x ) 5 (1 1 x ) m is a polynomial of degree m. Alternately, observe that f ( k ) (0) 5 0 for k \$ m 1 1. Section 9.3 Taylor’s Theorem (pp. 480–487) Exploration 1 Your Turn 1. We need to consider what happens to R n ( x ) as n . By Taylor’s Theorem, R n ( x ) 5 } f ( n ( n 1 1 1) 1 ( c )! ) } ( x 2 0) n , where f ( n 1 1) ( c ) is the ( n 1 1)st derivative of cos x evaluated at some c between x and 0. As with sin x , we can say that f ( n 1 1) ( c ) lies between 2 1 and 1 inclusive. Therefore, no matter what x is, we have ) R n ( x ) ) 5 ) } f ( n ( n 1 1 1) 1 ( c )! ) } ( x 2 0) n ) # ) } ( n 1 1 1)! } x n ) 5 } ( n 1 ) x ) n 1)! } . The factorial growth in the denominator, as noted in Example 3, eventually outstrips the power growth in the numerator, and we have } ( n 1 ) x ) n 1)! } 0 for all x . This means that R n ( x ) 0 for all x , which completes the proof. Exploration 2 Euler’s Formula 1. e ix 5 1 1 ix 1 } ( i 2 x ! ) 2 } 1 } ( i 3 x ! ) 3 } 1 1 } ( i n x ! ) n } 1 5 1 1 ix 2 } x 2 2 ! } 2 i } x 3 3 ! } 1 } x 4 4 ! } 1 i } x 5 5 ! } 2 1 ( i ) n } x n n ! } 1 2. If we isolate the terms in the series that have i as a factor, we get: e ix 5 1 1 ix 2 } x 2 2 ! } 2 i } x 3 3 ! } 1 } x 4 4 ! } 1 i } x 5 5 ! } 2 1 ( i ) n } x n n ! } 1 5 1 2 } x 2 2 ! } 1 } x 4 4 ! } 2 } x 6 6 ! } 1 1 ( 2 1) n } (2 x n 2 n )! } 1 1 i 1 x 2 } x 3 3 ! } 1 } x 5 5 ! } 2 } x 7 7 ! } 1 1 ( 2 1) n } (2 x n 2 1 n 1 1 1)! } 1 2 5 cos x 1 i sin x . (We are assuming here that we can rearrange the terms of a convergent series without affecting the sum. It happens to be true in this case, but we will see in Section 9.5 that it is not always true.) 3. e i p 5 cos p 1 i sin p 5 2 1 1 0 5 2 1 Thus, e i p 1 1 5 0 Quick Review 9.3 1. Since ) f ( x ) ) 5 ) 2 cos (3 x ) ) # 2 on [ 2 2 p , 2 p ] and f (0) 5 2, M 5 2. 2. Since f ( x ) is increasing and positive on [1, 2], M 5 f (2) 5 7. 3. Since f ( x ) is increasing and positive on [ 2 3, 0], M 5 f (0) 5 1. 4. Since the minimum value of f ( x ) is f ( 2 1) 5 2 } 1 2 } and the maximum value of f ( x ) is f (1) 5 } 1 2 } , M 5 } 1 2 } . 5. On [ 2 3, 1], the minimum value of f ( x ) is f ( 2 3) 5 2 7 and the maximum value of f ( x ) is f (0) 5 2. On (1, 3], f is increasing and positive, so the maximum value of f is f (3) 5 5. Thus ) f ( x ) ) # 7 on [ 2 3, 3] and M 5 7. 6. Yes, since each expression for an n th derivative given by the Quotient Rule will be a rational function whose denominator is a power of x 1 1. 7. No, since the function f ( x ) 5 ) x 2 2 4 ) has a corner at x 5 2. 8. Yes, since the derivatives of all orders for sin x and cos x are defined for all values of x . 9. Yes, since the function f ( x ) 5 e 2 x has derivatives of the form f ( n ) ( x ) 5 2 e 2 x for odd values of n and f ( n ) ( x ) 5 e 2 x for even values of n , and both of these expressions are defined for all values of x . 10. No, since f ( x ) 5 x 3/2 , we have f 9 ( x ) 5 } 3 2 } x 1/2 and f 0 ( x ) 5 } 3 4 } x 2 1/2 , so f 0 (0) is undefined. Section 9.3 Exercises 1. f (0) 5 e 2 2 x ) x 5 0 5 1 f 9 (0) 5 2 2 e 2 2 x ) x 5 0 5 2 2 f 0 (0) 5 4 e 2 2 x ) x 5 0 5 4, so } f 0 2 ( !

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