1. 4 Newton Polynomial
1.4 Newton Polynomial
P(x)=00+a1(x-20) P2(x)=a0+a1(x-x0)+a2(x-x0)( f3(x)=a0+a1(x-x0)+a2(x-x0)(x-x1) +a2(x-x0)(x-x1)(x-x2 Px(x)=0+a1(x-xo)+a2(x-x0)(x-x1) +a2(x-0)(x-x1)(x-x2) +a4(x-0)(x-x1)(x-x2)+… +aN(x-x0)…(x-xN-1) Here the polynomial PN(a)is obtained from PN-1(a) using the recursive rela. tionship (x)+aN(x-x0)(x-2)…(x-N-1) The polynomial (1.57)is said to be a Newton polynomial with N centers
Example 4.10. Given the centers =1. 1=3, C2-4, and C3=4.5 and the coefficients a0=5, 01=-2, a2=0.5, a3 =-0 1, and a4=0.003, find P1(), P2(a), P3(a), and PA(a) and evaluate P: (2.5) for k= 1, 2, 3, 4
Using formulas(1.54)through(1.57) we have P1(x)=5-2(x-1 P2(x) (x-1)+0.5(x-1)(x-3) P3(x)=P(x)-0.1(x-1)(x-3)(x-4), P4(x)=B3(x)+0.003(x-1)(x-3)(x-4)(x-45) Evaluating the polynomials at a=2.5 results in P1(25)=5-2(1.5), P2(25)=B(25)+0.5(15)(-0.5)=1.625 P3(25)=P2(25)-0.1(1.5)(-0.5)(-1.5)=1.5125 P4(25)=P3(25)+0.003(1.5)(-0.5)(-1.5)(-2.0)=1.50575
1. 4. 1 Nested Multiplication P3(x)=(a3(x-x2)+02(x-x1)+a1)(x-x0)+a(1.59 To evaluate P3(a) for a given value of c, start with the innermost grouping and form successively the quantities S3(x-x2) S1=S2(x-x1) +++ -t The quantity So is now P3(ar)
1.4.1 Nested Multiplication