Pre-Calc Homework Solutions 376

Pre-Calc Homework Solutions 376 - 376 Section 9.3 3. Since...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online