01spex3 - Math 222 Exam III Answers I(30 points formulas...

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

View Full Document Right Arrow Icon
Math 222, Exam III, March 30, 2001 Answers I. (30 points.) Complete the following paragraph by writing in the correct formulas. The Taylor series of a function f ( x ) about the point a is the infinite series X k =0 f ( k ) ( a )( x - a ) k k ! It may be used to compute f ( x ) when x is sufficiently close to a . The n th degree Taylor polynomial of f ( x ) about the point a is the polynomial f n ( x ) of degree n which best approximates f ( x ) for x near a . It is f n ( x ) = n X k =0 f ( k ) ( a )( x - a ) k k ! The n th remainder (error) is defined by f ( x ) = f n ( x ) + R n ( x, a ) . Three formulas for the remainder are R n ( x, a ) = X k = n +1 f ( k ) ( a )( x - a ) k k ! R n ( x, a ) = Z x a ( x - t ) n n ! f ( n +1) ( t ) dt and R n ( x, a ) = f ( n +1) ( c )( x - a ) n +1 ( n + 1)! for some number c with a < c < x or x < c < a . The last is called Lagrange’s formula for the remainder. 1
Image of page 1

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

View Full Document Right Arrow Icon
II. (30 points.) (1) Find all solutions of dy dx + 2 y = 0 . Answer: By separation of variables dy y = - 2 dx so ln y = - 2 x + C so y = y 0 e - 2 x where y 0 = e C . (2) Find all solutions of dy dx + 2 y = e - x .
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