Lecture note 3 Numerical Analysis ·And so on. ·Example:(xo,0）=(1,8),(c1,h)=(2,1),(r2,2)=(4,5). C0=0=8. c1=(y1-bo)/(x1-x0）=-7 2=2-h--w=3 (x2-x0)(x2-x1) Pn(x)=8-7(x-1)+3(x-1)(x-2) .There exists a closed form for ci,i=0,...,n. 。Define fz=欢 f儿，x+]= f[z+]-f儿] 工i+1-Ti fz,xi+1,+2= f[i+1,i+2]-f[ri,i+1] 工i+2-工i f儿，+1，x+2,+3l= f[z+1,x+2,x+3-f[x,xi+1,x+2] 工i+3-ri The prove will be delayed. co=f[zo] c1=f[ro,1] ci=f[ro,z1,...,xi] cn f[o,1,...,In] divided difference table: 1f] fr1,2] T2 f(2] fx1,x2,x3] f[z2,x3] f[1,x2,T3,T4] fx2,x3,x4] f[x1,x2,x3,x4,x5 f[3,4] fx2,x3,E4,I6] 4 f4] fz3,x4,T6] f[4,x5] 工5f[zsl 9Lecture note 3 Numerical Analysis • And so on. • Example: (x0, y0) = (1, 8), (x1, y1) = (2, 1), (x2, y2) = (4, 5). c0 = y0 = 8. c1 = (y1 − b0)/(x1 − x0) = −7 c2 = y2 − y0 − b1(x2 − x0) (x2 − x0)(x2 − x1) = 3. Pn(x) = 8 − 7(x − 1) + 3(x − 1)(x − 2) • There exists a closed form for ci , i = 0, . . . , n. • Define f[xi ] = yi f[xi , xi+1] = f[xi+1] − f[xi ] xi+1 − xi f[xi , xi+1, xi+2] = f[xi+1, xi+2] − f[xi , xi+1] xi+2 − xi f[xi , xi+1, xi+2, xi+3] = f[xi+1, xi+2, xi+3] − f[xi , xi+1, xi+2] xi+3 − xi • The prove will be delayed. • c0 = f[x0] c1 = f[x0, x1] . . . ci = f[x0, x1, . . . , xi ] . . . cn = f[x0, x1, . . . , xn] • divided difference table: x1 f[x1] f[x1, x2] x2 f[x2] f[x1, x2, x3] f[x2, x3] f[x1, x2, x3, x4] x3 f[x3] f[x2, x3, x4] f[x1, x2, x3, x4, x5] f[x3, x4] f[x2, x3, x4, x5] x4 f[x4] f[x3, x4, x5] f[x4, x5] x5 f[x5] 9