西安建筑科技大学：《高等数学计算方法》课程教学资源（PPT课件讲稿）Chapter 3.3 Lagrange Approximation

Chapter 3 Interpolation ana Polynomial Approximation 4.3 Lagrange Approximation

Chapter 3 Interpolation and Polynomial Approximation 4.3 Lagrange Approximation

Example 1.6. Consider the graph y=f(a)=cos(=)over[0.0, 1.2) (a) Use the nodes 20=0.0 and r1=1. 2 to construct a linear interpolating polynomia ial P1(r) (b) Use the nodes 0=0. 2 and = 1.0 to construct a linear approximating polynomial Q1(a) USing(1.22) with the abscissas zo=0.0 and 21=1.2 and the ordinates yo co8(0.0)=10000d01=co8(.2)=0.362358 produces P1(x)=100 x-0.0 00-15+0.362358 1.2-0.0 =-0.8333-1.2)+0.301965(x-0.0) When the nodes 0=0.2 and 21=1.0 with yo = cos(0.2)=0.980067 and =co8(1 (1.0 )=0.540302 are used, the result is x-1.0 x-0.2 Q(x)=09067 +0.540302 0.2-1.0 1.0-0.2 =-1.225083(x-1.0)+0675378(x-0.2)

y=f(x) 3 Figure 1.11(a) Figure1.11(b) Figure 1.11(a) The linear approximation of y= P1(r)where th he nodes 0=0.0 and 1.2 are the end points of the interval [a, b].( b)The linear approximation of y=Q1(a) where the nodes xo=0.2 and 11= 1.0 lie inside the interval [a, b

Table 1.6 Comparison of f(r)=cos(a )and the Linear Approximations Pi(a)and Q1() Tk f(ak)=cos(ak) Pl(ak) f(ak)-Pi(ck) Q1(k) f(ak)-Q1(ak) 0.0 1.000000 10000000.00000 1.090008 0.090008 0.1 0.995004 0.946863 0.048141 1.035037 0.040033 0.2 0.980067 0.893726 0.086340 0.980067 0.000000 0.3 0.955336 0.840589 0.114747 0.925096 0.030240 0.4 0.921061 0.787453 0.133608 0.870126 0.050935 0.5 0.877583 0.734316 0.143267 0.815155 0.062428 0.60.825336 0.681179 0.144157 0.760184 0.065151 0.7 0.764842 0.628042 0.136800 0.705214 0.059628 00 0.696707 0.544905 0.121802 0.650243 0.046463 0.621610 0.521768 0.099842 0.595273 0.026337 100.5403020.4686310.0716710.5403020.0000 110.45359604154950038102048322031736 1.2 0.362358 0.362358 0.0000004303610.068003

The generalization of(1. 25) is the construction of a polynomial of a polynomial PN(a)of degree at most N that passes through the N+1 points(co, yo), (=1, y1) (aN, JN)and has the form P()=∑LNk() where ln h is the lagrange coefficient polynomial based on these nodes x-0)…(-k-1)(-k+1)…(T-TN) IN (xk-x0)…(xk-xk-1)(xk-xk+1)…(xk-xN (1.27) It is understood that the terms (r-Tk) and(ak-ak do not appear on the right side of equation(1.27). It is approximate to introduce the product notation for(1.27) and we write N =0,≠k(-x LN k=0, #(=k-2 (1.28

ne Lagrange quadratic interpolating polynomial through the three points(=o, go) (1, 91), and (r2, y2)is -0)(x- C-T)(- P2 (x-x1)(x-x2)( =90 +9 +y2 T0-1(0-2 1-0)(x1-2 T2-0)(2-T1 1.3 The Lagrange cubic interpolating polynomial through the four points(zo, 3o), (21, 31), I 2, 32), and(=3, 33)is P3(x)=90 r-1)x-x2)(x-03)+yy(-r0/(x1-2)(21-T3) x-T0)(-T2)(-T3 (x0-x1)(x0-x2)(x0-3) x-x0)(x-x1)(x-x3) x-r0)(x-1)(-2 +y2 (x2-x0)(x2-x1)x2-x3) t 93 (x3-x0)(x3-x1)(x3-2 132)

Example 1.7. Consider y=f(a)=cos(r)over[0.0, 1.2 (a)Use the three nodes T0=0.0, 1=0.6, and r2=1. 2 to construct a quadratic interpolation polynomial P2(a) (b)Use the four nodes 0=0.0, 1=0.4, 2=0.8, and r=1. 2 to construct a cubic interpolation polynomial P3(a)

Using o=0.0,x1=0.6,x2=1.2 and yo=cos(0.0)=1,n=cos(0.6)=0.825336 and y2=cos(1. 2)=0.362358 in equation(1. 31) produces P2(x)=1.0 (x-0.6)(x-1.2) +0.825336 (x-0.0)(x-1.2) (0.0-0.6)(0.0-1.2 (0.6-0.0)(0.6-1.2 +0.362358 (x-0.0)(x-0.6) (1.2-0.0)(1.2-0.6 =138889(x-0.6)(x-1.2)-2.292599(x-00(x-1.2) +0.503275(x-0.00)(x-0.6)

Using To=0.0,x1=0.4,x2=0.8,x3=12 and yo=co(0.0)=1.0,y=cos(0.4) 0.921061,2=c0s(0.8)=0.696707, and y3=cos(1.2)=0.362358 in equation(1.32) produces P3(x)=1.00000 x-0.4)(x-0.8)(x-1.2 0.0-0.4)(0.0-0.8)(0.0-1.2) +0.921061 0.0)(x-0.8)(x-12) (0.4-0.0)(0.4-0.8)(0.4-1.2) 0.696707 x-0.0)(x-0.4)(x-1.2) (0.8-0.0)(0.8-0.4)(0.8-1.2 +0.362358 (x-0.0)(x-0.4)(x-0.8) 12-0.0)(1.2-0.4(1.2-0.8) 2.604167(x-0.4)(x-0.8)(x-1.2) +7195789(x-0.0)(x-0.8)(x-1.2) 5443021(x-0.0)(x-0.4)(x-1.2) +0.93641(x-0.0)(x-0.4)(x-0.8)

y=f(x) y=f(x) y=P,(x) Figure 1. 12(a) Figure 1.12(b) Figure 1.12(a) The quadratic approximation polynomial y= P2(a) based on the nodes 20=0.0, 1=0.6, and 2=1.2 (b) The cubic approximation polynomial y= P1()based on the nodes o=0.0, 11=0.4, 12=0.8, and r3=12