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Assignment 1 - solution

# Assignment 1 - solution - University of Texas at Dallas...

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Unformatted text preview: University of Texas at Dallas Assignment #1 Last Name: First Name and Initial: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Course Name: Number: Mathematical Analysis 2 MATH 4302 Instructor: Due Date: Wieslaw Krawcewicz January 31, 2012 E-mail Address: Student’s Signature: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Question Weight Your Score Comments 1. 10 2. 10 3. 10 4. 10 5. 10 6. 10 7. 10 8. 10 9. 10 10. 10 Total: 100 2 Problem 1. Consider the function f : [ − a, a ] → R , a > 0, defined by f ( x ) = cosh (3 x ) , x ∈ [ − a, a ] . (a): Compute the n-th derivative of the function f . Solution: Notice that we have (by simple induction) cosh(3 x ) = e 3 x + e − 3 x 2 , d dx cosh(3 x ) = 3 1 e 3 x + ( − 1) 1 e − 3 x 2 d 2 dx 2 cosh(3 x ) = 3 2 e 3 x + ( − 1) 2 e − 3 x 2 , d n dx n cosh(3 x ) = 3 n e 3 x + ( − 1) n e − 3 x 2 . (b): For a given n ∈ N , write the Taylor polynomial T n ( x ) of f (centered at x o = 0). Solution: Then we have T n ( x ) = ⌊ n 2 ⌋ ∑ k =0 3 k (2 k )! x 2 k . (c): For a given ε > 0, find n ∈ N such that sup {| f ( x ) − T n ( x ) | : x ∈ [ − a, a ] } < ε. Solution: Notice that by the Lagrange formula for the remainder of the Taylor approximation, we have for some c ∈ ( − a, a ) r n ( x ) = f ( x ) − T n ( x ) = 3 n +1 e 3 c + ( − 1) n +1 e − 3 c 2( n + 1)! x n +1 Put t o = 3 ae − 2. Put M := e 3 a +1 2 . Then the inequality ( n + 1)! ≥ ( n +2) n +1 e n +1 implies | r n ( x ) | ≤ (3 a ) n +1 e 3 a + 1 2( n + 1)! ≤ M (3 a ) n +1 ( n + 1)! ≤ M (3 ae ) n +1 ( n + 2) n +1 ≤ M 3 ae n + 2 < ε, whenever n ≥ max { t o , 3 aeM ε − 2 } . 3 Problem 2. Consider the function f : [ − 1 , 1] → R , defined by f ( x ) = 1 √ 1 + 1 2 x , x ∈ [ − 1 , 1] . (a): Compute the n-th derivative of the function f . Solution: Consider first the function g ( x ) = (1 + x ) − 1 2 . Clearly, by applying principle of mathematical induction, we obtain g ′ ( x ) = ( − 1 2 ) (1 + x ) − 3 2 , g ′′ ( x ) = ( − 1 2 )( − 3 2 ) (1 + x ) − 5 2 , g ′′′ ( a ) = ( − 1 2 )( − 3 2 )( − 5 2 ) (1 + x ) − 7 2 g ( n ) ( x ) = ( − 1 2 )( − 3 2 ) . . . ( − 2 n − 1 2 ) (1 + x ) − 2 n +1 2 = ( − 1) n (2 n )!...
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Assignment 1 - solution - University of Texas at Dallas...

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