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Calculus 1 - Midterm 2 Version A Solutions

Calculus 1 - Midterm 2 Version A Solutions - University of...

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University of Ottawa MAT 1330D Midterm Exam March 18, 2009. Duration: 80 minutes. Instructor: Pawel Lorek SOLUTIONS of ver. A 1 ver. A
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Question 1. [4p] Use Newton’s method to find an approximation of the equilibrium of the following DTDS: x t +1 = h ( x t ) = x 2 t + 4 x t 8 x t Use x 0 = 2 and make at least two iterations (i.e. calculate x 2 ) Solution: We have to solve the following equation: h ( x ) = x ( x ) 2 + 4 x 8 x = x ( x ) 2 + 3 x 8 x = 0 It means we have to solve the following equation: f ( x ) = x 2 + 3 x 8 x = 0 . We have f ( x ) = 2 x + 3 8 · 1 x 2 = 2 x + 3 + 8 x 2 . Therefore, the Newton’s method discrete-time dynamical system is: x t +1 = x t f ( x t ) f ( x t ) = x t x 2 t + 3 x t 8 x t 2 x + 3 + 8 x 2 t We start with x 0 = 2. x 1 = x 0 x 2 0 + 3 x 0 8 x 0 2 x 0 + 3 + 8 x 2 0 = 2 2 2 + 3 · 2 8 2 2 · 2 + 3 + 8 2 2 = 2 4 + 8 4 4 + 3 + 2 = 2 6 9 = 2 2 3 = 4 3 x 2 = x 1 x 2 1 + 3 x 1 8 x 1 2 x 1 + 3 + 8 x 2 1 = 4 3 ( 4 3 ) 2 + 3 · 4 3 8 4 3 2 · 4 3 + 3 + 8 ( 4 3 ) 2 = 4 3 16 9 + 4 8 · 3 4 8 3 + 3 + 8 · 9 16 = 4 3 16 9 + 4 6 8 3 + 3 + 9 2 = 4 3 16+36 6 · 9 9 16+18+27 6 = 4 3 2 9 · 6 61 = 4 3 + 4 183 = 248 183 = 1 . 355191257 (calculator could be used to calculate above numbers faster) 2 ver. A
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Question 2. [5 points] Consider the function f ( x ) = 1 x (a) [2 points] Find the tangent line approximation with base point a = 1 . (b) [2 points] Find the Taylor polynomial of degree three with base point a = 1 . (c) [1 point] Calculate approximation of 1 1 . 1 with the approximation from the tangent line and from the Taylor polynomial of degree 3 (found in (a) and (b)) Solution: Let’s recall the taylor polynomial matching n first derivatives at x = a : P n ( x ) = f ( a ) + f ( a )( x a ) + f ′′ ( a ) 2 ( x a ) 2 + f (3) ( a ) 3! ( x a ) 3 + ... + f ( n ) ( a ) n ! ( x a ) n , where f ( k ) denotes k th derivative of f . For simplicity let’s write: f ( x ) = x 1 2 . We will also need the following: f ( x ) = 1 2 x - 3 2 f ′′ ( x ) = 1 2 · ( 3 2 ) x - 5 2 = 3 4 x 5 2 f ′′′ ( x ) = 3 4 · ( 5 2 ) x 7 2 = 15 8 and f (1) = 1 , f (1) = 1 2 , f ′′ (1) = 3 4 , f ′′′ (1) = 15 8 a) Tangent line approximation with base point a = 1 is just P 1 ( x ): P 1 ( x ) = f (1) + f (1)( x 1) = 1 1 2 ( x 1) b) P 3 ( x ) = f (1)+ f (1)( x 1)+ f ′′ (1) 2 ( x 1) 2 + f ′′′ (1) 3!
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