tut5 - Deduce that the zeros of Chebyshev polynomials are...

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cep 2008/09 MT3802 - Numerical Analysis Tutorial Sheet 5 1. Let p 1 ( x ) be the linear interpolant of f ( x ) at the points ( x 0 , f ( x 0 )) and ( x 1 , f ( x 1 )), where x 1 - x 0 = h . Show that for x [ x 0 , x 1 ] k f ( x ) - p 1 ( x ) k Mh 2 8 where k f 00 ( x ) k M . Suppose e x on [ - 1 , 1] is approximated by a piece-wise linear function S 1 ( x ) based on n equal intervals of length h . Estimate the number of sub-intervals required to guarantee an error no larger than 0 . 5 × 10 - 2 . 2. The Chebyshev polynomials satisfy the recurrence relation: T n +1 ( x ) = 2 xT n ( x ) - T n - 1 ( x ) , where T 0 ( x ) = 1 and T 1 ( x ) = x . Show, by induction that T n ( - x ) = ( - 1) n T n ( x ) The polynomials of even degree are Even functions, and those of odd degree are Odd.
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Unformatted text preview: Deduce that the zeros of Chebyshev polynomials are symmetrically placed about the origin. 3. Using the recurrence relation, determine the Chebyshev polynomials { T n ( x ) } n =6 n =0 . Confirm that T 6 ( x ) has 7 extreme points and determine these points directly from the polynomial. Using the properties of the Chebyshev polynomials, sketch T 6 ( x ) ....
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This note was uploaded on 09/28/2009 for the course MATH MATH427 taught by Professor Dr.sharpey during the Spring '09 term at Monmouth IL.

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