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
Example 4. 11. Compute P3(2.5) in Example 1. 10 using nested multipli cation Using (1.59), we write P3(x)=(-0.1(-4)+0.5)(x-3)-2)(x-1)+5 The values in(1.60)are 0.1 S2=-0.1(2.5-4)+0.5=0.65 S1=0.65(2.5-3)-2=-2.325 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 f(℃ f[k1,]=- f[ck-1.ak-f(ck-2, ck-11 k-2,k-1 T一工 f(k2,221,4=12132 The recursive rule for constructing higher-order divided differences is f{k-+-…,x]-f k k and is used to construct the divided differences in table 4. 8
Table 1. 8 Divided-difference Table fory=f() k 1f{z1]|f Co. cot1. f[r3]f[x2,l fI 31.2, 0,1,2,3
Theorem 1.5 (Newton Polynomial ). Suppose that o, J1,. N are N+1 distinct numbers in a, b]. There exists a unique polynomial PN(ar) of degree at most n with the property that 0.1.N The Newton form of this polynomial is PR(x)=a0+a1(x-m0)+…+aN(x-0)(x-x1)…(x-xN-1),(1.69 where ak=f C0, C1,., w, 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(a), that is, f(a)=PN(ar)+ en(c) If f E CN+a, b, then for each r E a, b] there corresponds a number c= c(a) in (a, 6), so that the error term has the form EN(a) T-70x-21)…(x-xN)代(N+1)(C (N+1)!
