{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Calculus 1 - Midterm 2 Version A Solutions

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

This preview shows pages 1–4. Sign up to view the full content.

University of Ottawa MAT 1330D Midterm Exam March 18, 2009. Duration: 80 minutes. Instructor: Pawel Lorek SOLUTIONS of ver. A 1 ver. A

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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!

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern