Solution to HW6

# Solution to HW6 - Solution to HW#6 EGM6341 Spring 2009 1....

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

Solution to HW#6 EGM6341 Spring 2009 1. #28a. Find a polynomial p(x) of degree 2 that satisfies p(x 0 )= y 0 , p (x 0 )=y 0 , p (x 1 )=y 1 Given a formula in the form of p(x) = y 0 l 0 (x) + y 0 l 1 (x) + y 1 l 2 (x). Solution: For the suggested p(x), it is seen that p(x 0 )= y 0 => l 0 (x 0 ) = 1 , l 1 (x 0 ) = 0 , and l 2 (x 0 ) =0 ; p (x 0 )=y 0 => l 0 (x 0 ) = 0 , l 1 (x 0 ) = 1 , and l 2 (x 0 ) = 0 ; p (x 1 )=y 1 => l 0 (x 1 ) = 0 , l 1 (x 1 ) = 0 , and l 2 (x 1 ) = 1 . At any interval, l i (x j ) = l i (x j ) = 0 => l i (x)= c i (x-x j ) 2 => l 2 (x)= c 2 (x-x 0 ) 2 Furthermore, l 2 (x 1 ) = 1 => c 2 =1/[2(x 1 -x 0 )] For polynomials of degree 2, there can only one place where p (x)=0. If there are two places where p (x)=0, which is the case for l 1 (x), the polynomial must be a constant. Thus, l 0 (x) =1 . For l 1 (x), l 1 (x 1 ) = 0 => l 1 (x) = a(x - x 1 ) 2 +b (line of symmetry of the parabola at x= x 1 ). l 1 (x 0 ) = 0 => a(x 0 - x 1 ) 2 +b=0 l 1 (x 0 ) = 1 => 2a(x 0 - x 1 )=1 => a= -1/[2(x 1 - x 0 )], b=(x 1 - x 0 )/2 => l 1 (x) = -(x - x 1 ) 2 /[2(x 1 - x 0 )] + (x 1 - x 0 )/2 Finally, p(x) = y 0 + y 0 { -(x - x 1 ) 2 /[2(x 1 - x 0 )] + (x 1 - x 0 )/2 }+ y 1 (x-x 0 ) 2 /[2(x 1 - x 0 )] 2. p185-194 #35 “not-a-knot” condition for cubic spline Solution: a) Given: i x y h i y[1] 0 0 1.4 1 -0.8 1 1 0.6 1 0.4 2 2 1 0.5 -0.7 3 2.5 0.65 0.5 -0.1 4 3 0.6 1 0.4 5 4 1 * General equation:

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 04/23/2009 for the course EGM 6341 taught by Professor Mei during the Spring '09 term at University of Florida.

### Page1 / 7

Solution to HW6 - Solution to HW#6 EGM6341 Spring 2009 1....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online