{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MATH2051_05 - University of Durham EXAMINATION PAPER date...

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

View Full Document Right Arrow Icon
University of Durham EXAMINATION PAPER date May/June 2005 exam code MATH2051/01 description NUMERICAL ANALYSIS II Time allowed: 3 hours Examination material provided: None Instructions: Credit will be given for the best FOUR answers from Section A and the best THREE answers from Section B. Questions in Section B carry TWICE as many marks as those in Section A. Approved electronic calculators may be used. ED01/2005 University of Durham Copyright continued
Background image of page 1

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

View Full Document Right Arrow Icon
2 MATH2051/01 SECTION A 1. (a) Let f ( x ) = e x - 4 x . Using the Intermediate Value Theorem prove that the equation f ( x ) = 0 has a unique solution, p , in the interval (0 , 1). (b) One rearrangement of this equation is x = g ( x ) := 0 . 25 e x . Show that the iterative method p n +1 = g ( p n ) will converge to p for suitable starting values. (c) Starting with p 0 = 0 . 36, estimate p correct to three decimal places. 2. Let f ( x ) be a smooth function defined on the interval [ - 1 , 1]. The interpolating Hermite polynomial, p 3 ( x ), of degree 3 is defined by p 3 ( - 1) = f ( - 1) , p 3 (1) = f (1) , p 0 3 ( - 1) = f 0 ( - 1) , p 0 3 (1) = f 0 (1) and satisfies the following error estimate f ( x ) - p 3 ( x ) = f (4) ( c ) 4! ( x + 1) 2 ( x - 1) 2 , c ( - 1 , 1) .
Background 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 ]}