{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sample_short - Final Exam Math 151A/2 Winter 2001 UCLA...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Final Exam, Math 151A/2, Winter 2001, UCLA, 03/21/2001, 8am-11am I. (a) Let x 0 , x 1 , ..., x n be n +1 distinct points in [ a, b ], with x 0 = a and x n = b , and f C n +1 [ a, b ]. Let P ( x ) = P 0 , 1 ,...,n ( x ) be the Lagrange polynomial interpolating the points x 0 , x 1 , ..., x n , such that P ( x i ) = f ( x i ), for all i = 0 , 1 , ..., n . Express f ( x ) in terms of P ( x ) and a remainder term (the error formula). (b) Consider the case n = 2 and the data x x 0 = 0 x 1 = 1 x 2 = 2 f ( x ) = ln( x + 1) 0 ln 2 ln 3 (i) Find polynomials L i ( x ), i = 0 , 1 , 2, such that L i ( x i ) = 1 and L i ( x j ) = 0 if i negationslash = j . (ii) Deduce the Lagrange polynomial P ( x ) = P 0 , 1 , 2 ( x ) interpolating this data points. (iii) Write the error formula and find a bound for the error | f (0 . 5) P (0 . 5) | . II. (a) Let i, j be two distinct integers in { 0 , 1 , ..., n } . Express P 0 , 1 ,...,n ( x ) in terms of P 0 , 1 ,...,i 1 ,i +1 ,...,n ( x ) and of P 0 , 1 ,...,j 1 ,j +1 ,...,n ( x ) (Neville’s method). (b) Suppose x j = j for j = 0, 1, 2, 3 and it is known that P 0 , 1 ( x ) = x + 1 , P 1 , 2 ( x ) = 3 x 1 , and P 1 , 2 , 3 (1 . 5) = 4 . Find P 0 , 1 , 2 , 3 (1 . 5). III. For a function f the forward divided-differences are given by x 0 = 0 f [ x 0 ] = f [ x 0 , x 1 ] = x 1 = 1 f [ x 1 ] = f [ x 0 , x 1 , x 2 ] = 4 f [ x
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
Image of page 2
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