11.11 Taylor Series Applications-solutions

11.11 Taylor Series Applications-solutions - im(gi768 11.11...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
im (gi768) – 11.11 Taylor Series Applications – sadun – (53088) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is a longer-than-usual homework, but you have until Thursday to do it, since Wednesday is a review day. 001 10.0points Determine the degree 3 Taylor polynomial T 3 ( x ) centered at the origin for the function f ( x ) = sin 3 x. 1. T 3 ( x ) = 3 parenleftBig x - 3 x 3 2 parenrightBig correct 2. T 3 ( x ) = 1 - 9 x 2 2 3. T 3 ( x ) = 1 - 2 x 2 4. T 3 ( x ) = 2 parenleftBig x - 2 x 3 3 parenrightBig 5. T 3 ( x ) = x - x 3 6 6. T 3 ( x ) = 1 - x 2 2 Explanation: For a function f the degree 3 Taylor poly- nomial centered at the origin is given by T 3 ( x ) = f (0) + f (0) x + 1 2! f ′′ (0) x 2 + 1 3! f ′′′ (0) x 3 . Now, when f ( x ) = sin 3 x , f ( x ) = 3 cos 3 x, f ′′ ( x ) = - 9 sin 3 x, f ′′′ ( x ) = - 27 cos 3 x, so f (0) = 0 , f (0) = 3 , f ′′ (0) = 0 , f ′′′ (0) = - 27 . Consequently, T 3 ( x ) = 3 parenleftBig x - 3 x 3 2 parenrightBig . 002 10.0points Compute the degree 2 Taylor polynomial for f centered at x = 1 when f ( x ) = x. 1. T 2 ( x ) = 1 + 1 4 ( x - 1) + 1 4 ( x - 1) 2 2. T 2 ( x ) = 1 - 1 2 ( x - 1) + 1 8 ( x - 1) 2 3. T 2 ( x ) = 1 - 1 4 ( x - 1) + 1 8 ( x - 1) 2 4. T 2 ( x ) = 1 + 1 2 ( x - 1) - 1 4 ( x - 1) 2 5. T 2 ( x ) = 1 - 1 4 ( x - 1) - 1 4 ( x - 1) 2 6. T 2 ( x ) = 1 + 1 2 ( x - 1) - 1 8 ( x - 1) 2 correct Explanation: The degree 2 Taylor polynomial centered at x = 1 for a general f is given by T 2 ( x ) = f (1)+ f (1) ( x - 1)+ f ′′ (1) 2! ( x - 1) 2 . Now when f ( x ) = x , f ( x ) = 1 2 x , f ′′ ( x ) = - 1 4 x x , in which case, f (1) = 1 , f (1) = 1 2 , while f ′′ (1) 2! = - 1 8 .
Image of page 1

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

View Full Document Right Arrow Icon
im (gi768) – 11.11 Taylor Series Applications – sadun – (53088) 2 Consequently, T 2 ( x ) = 1 + 1 2 ( x - 1) - 1 8 ( x - 1) 2 . 003 10.0points Determine the degree 2 Taylor polynomial T 2 ( x ) centered at x = 1 for the function f when f ( x ) = radicalbig 3 + x 2 . 1. T 2 ( x ) = 2 + 1 2 ( x - 1) + 3 8 ( x - 1) 2 2. T 2 ( x ) = 2 + 1 2 ( x + 1) + 3 8 ( x + 1) 2 3. T 2 ( x ) = 2 - 1 2 ( x - 1) + 3 16 ( x - 1) 2 4. T 2 ( x ) = 2 + 1 2 ( x - 1) + 3 16 ( x - 1) 2 correct 5. T 2 ( x ) = 2 - 1 2 ( x + 1) + 3 8 ( x + 1) 2 6. T 2 ( x ) = 2 - 1 2 ( x + 1) + 3 16 ( x + 1) 2 Explanation: For a function f the degree 2 Taylor poly- nomial centered at x = 1 is given by T 2 ( x ) = f (1) + f (1)( x - 1) + 1 2 f ′′ (1)( x - 1) 2 . Now when f ( x ) = radicalbig 3 + x 2 , f ( x ) = x 3 + x 2 , while f ′′ ( x ) = 3 + x 2 - x 2 3 + x 2 3 + x 2 = 3 (3 + x 2 ) 3 / 2 . But then f (1) = 2 , f (1) = 1 2 , f ′′ (1) = 3 8 . Consequently, T 2 ( x ) = 2 + 1 2 ( x - 1) + 3 16 ( x - 1) 2 . 004(part1of2)10.0points (i) Compute the degree 2 Taylor polynomial for f centered at x = 1 when f ( x ) = x. 1. T 2 ( x ) = 1 - 1 2 ( x - 1) + 1 8 ( x - 1) 2 2. T 2 ( x ) = 1 - 1 4 ( x - 1) - 1 4 ( x - 1) 2 3. T 2 ( x ) = 1 - 1 4 ( x - 1) + 1 8 ( x - 1) 2 4. T 2 ( x ) = 1 + 1 2 ( x - 1) - 1 4 ( x - 1) 2 5. T 2 ( x ) = 1 + 1 4 ( x - 1) + 1 4 ( x - 1) 2 6. T 2 ( x ) = 1 + 1 2 ( x - 1) - 1 8 ( x - 1) 2 correct Explanation: The degree 2 Taylor polynomial centered at x = 1 for a general f is given by T 2 ( x ) = f (1)+ f (1) ( x - 1)+ f ′′ (1) 2!
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern