sole13 - -x 2 8 x 3 16(1 tx 5 2 t ∈(0 1 5 By...

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ANALYSIS EXERCISE 13–SOLUTIONS 1. For the function f ( t ) = t p write the Mean value theorem f ( x ) - f ( y ) = f 0 ( ξ )( x - y ), where x > ξ > y > 0, i.e. x p - y p = p - 1 ( x - y ). It is enough to notice that y p - 1 < ξ p - 1 < x p - 1 . 2. (a) lim x 1 x n - 1 x 2 - x = lim x 1 nx n - 1 2 x - 1 = n. (b) lim x 1 2 (1 - x ) m - x m (1 - x ) n - x n = lim x 1 2 m [ - (1 - x ) m - 1 - x m - 1 ] n [ - (1 - x ) n - 1 - x n - 1 ] = m n 2 n - m . (c) lim x 1 nx n +2 - ( n + 1) x n +1 + x ( x - 1) 2 = lim x 1 n ( n + 2) x n +1 - ( n + 1) 2 x n + 1 2( x - 1) = lim x 1 n ( n + 2)( n + 1) x n - n ( n + 1) 2 x n - 1 2 = n ( n + 1) 2 . 3. (a) f ( x ) = x 2 - 4 x - 9 ,f 0 ( x ) = 2 x - 4 ,f 00 ( x ) = 2, f (3) = - 12 ,f 0 (3) = 2 ,f 00 (3) = 2. Hence f ( x ) = - 12 + 2( x - 3) + ( x - 3) 2 . (b) f ( x ) = x 5 ,f 0 ( x ) = 5 x 4 ,f 00 ( x ) = 20 x 3 ,f (3) ( x ) = 60 x 2 , f (4) ( x ) = 120 x, f (5) ( x ) = 120, f (3) = 243 ,f 0 (3) = 405 ,f 00 (3) = 540 ,f (3) (3) = 540, f (4) (3) = 360 ,f (5) (3) = 120. Hence f ( x ) = 243+405( x - 3)+270( x - 3) 2 + 90( x - 3) 3 + 15( x - 3) 4 + ( x - 3) 5 . 4. (a) 1 - x + x 2 - x 3 (1+ tx ) 4 , t (0 , 1). (b) 1 + x 2
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Unformatted text preview: -x 2 8 + x 3 16(1+ tx ) 5 / 2 , t ∈ (0 , 1). 5. By l’Hopital’s rule lim h → f ( a + h ) + f ( a-h )-2 f ( a ) h 2 = lim h → f ( a + h )-f ( a-h ) 2 h = f 00 ( a ) . 6. (a) No, since lim n →∞ a n 6 = 0. (b) Yes. Comparison test: √ n +1-√ n n = 1 n ( √ n +1+ √ n ) < 1 n 3 / 2 . (c) Yes. Use d’Alembert’s test. 7. (a) Comparison test and definition of the absolute convergence. (b) Note that | a n x n | ≤ a n . Then use comparison test. 1...
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Winter '10 term at Bristol Community College.

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