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
Example 4. 11. Compute P3(2.5)in Example 1.10 using nested multipli- cation Using(1.59), we write P3(x)=(-0.1(x-4)+0.5)(x-3)-2)(x-1)+5 The values in(1.60)are 0.1 S2=-0.1(25-4)+05=0.65 S1=0.65(25-3)-2=-2325 S0=-2325(25-1)+5=1.5125 Therefore, P3(2.5)=1.5125
Definition 4.1(Divided Differences). The divided differences for a func tion f(a)are defined as follows k f{xk-1,m」 f[=k-flack-1 Ck-Ik_1 7 f k-2, k-1, CkI flrk-1 Ik]-f ak-2, Tk-1 (1.66) k一k-2 k-3k=2、k-1Ok≈k=2k-1-和=3x=2b-1 The recursive rule for constructing higher-order divided differenc k-+1,…,÷2 k-j+1 ·· x小-f{xk-j k-1 and is used to construct the divided differences in table 4.8
Table 1. 8 Divided-difference Table for y=f() k ICo. C1. 0,T1,T f[c3] f[ 2, c3] f[ 1, I2, c3] f[co, T1 4 f[ca] f[23, 24] f[ 2, I3, C4] f[c1, 2, 3, 4] f[co, 1, 2, 3, 4]
Theorem 1.5(Newton Polynomial). Suppose that to, I1,..., IN are N+ distinct numbers in a, b]. There exists a unique polynomial PN()of degree at most N with the property that f(i)= PN(ifor j=0 The Newton form of this polynomial is Px(x)=a0+a1(x-x0)+…+aN(x-x0)(x-x1)…(x-xN-1),(1.69) where ak= f co, 21, ...,k], for k=0, 1,..., N
Corollary 1. 2(Newton Approximation). Assume that PN(a)is the New- ton polynomial given in Theorem 1.5 and is used to approximate the function f(r), that is, f(a)=PN()+en(a) If fE CN+a, bl, then for each a E a, b there corresponds a number=c( in(a, b), so that the error term has the form EN(a) (x-x0)(x-2x1)…(x-xrx)f(N+(c) (N+1)
