1.4 Newton polynomial
1.4 Newton Polynomial
(x)=00+a1(x-x0 P(x)=0+a1(x-x0)+a2(x-x0)(x-x1) P3(x)=a0+a1(x-x0)+a2(x-0)(x-x1) +a2(x-x0)(x-2x1)(x-x2) 56 Px(x)=a0+a1(x-0)+a2(x-0)(x-x1) +a2(x-x0)(x-x1)(x-2) +a4(x-x0)(x-x1)(x-m2)+ +aNlc- -N-1 Here the polynomial PN(a)is obtained from PN-1(a)using the recursive rela- tianshi FRx(x)=PN-1(x)+aN(x-0)(x-x2)…(x-xN-1).(1.58) The polynomial (1.57) is said to be a Newton polynomial with N centers
Example 4.10. Given the centers 3 =1.1=3. c2-4 and c?= 4.5 and the coefficients a0=5, a1=-2, a2=0.5, 03=-0. 1, and aA=0.003, find P1(c), P2(a), P3(a), and P4()and evaluate P (2.5) for k=1, 2, 3, 4
sing formulas(1.54)through(1.57), we have P2(x)=5-2(x-1)+0.5(x-1)(x P(x)=P(x)-0.1(x-1)(x-3)(x-4), P4(x)=P3(x)+0.003(x-1)(x-3)(x-4)(x-4.5) Evaluating the polynomials at .c = 2.5 results in P1(25)=5-2(1.5) P2(25)=B1(25)+0.5(1.5)(-0.5)=1.625, P3(25)=P(2.5)-0.1(1.5)(-0.5)(-15)=1.5125 P4(25)=P2(25)+0.003(1.5)(-0.5)(-1.5)(-2.0)=150575
1. 4.1 Nested Multiplication 3(x)=(2(x-2)+02)(x-x1)+01)(x-0)+a To evaluate P3() for a given value of c, start with the innermost grouping and form successively the quantities xxx ))) +++ The quantity So is now P3 (a)
1.4.1 Nested Multiplication