数值计算方法O课程组庆学学与信息学学O|(肇院)O1 / 41:?多: 过给定节点的多;唯一5: 过n + 1个节点次不过n的?多唯一;Lagrange?多:pn(x) =Pn i=0yiLi(x) =Pn i=0yinQj=0 j,ix xj xi xj:fx0,x1, ,xk =fx1,x2, ,xk fx0,x1, ,xk1 xk x0,其fxi = f(xi);表: 略Newton?多:pn(x) =nPi=0fx0,x1, ,xi (x x0)(x x1)(x xi1).O|(肇院)O2 / 411八讲Hermite?值式O|(肇院)O3 / 4181Hermite?Hermite?多 n次Hermite?2n次样条?背景 n次样条?的概g n弯矩3本?结O|(肇院)O4 / 411Hermite?Hermite?多 n次Hermite?2n次样条?3本?结O|(肇院)O5 / 41插例子例 以-2和1为?节点(=给定点(2,4)和(1,1),构E函x2的?多 ,d此0.82和0.92的近q.)N易构E?多p1(x) = 2 xuk0.82 p1(0.8) = 2 0.8 = 1.2,0.92 p1(0.9) = 2 0.9 = 1.1姑 不论们的误 大?,就其近q0.82 0.92S不 的.O|(肇院)O5 / 41插例子例 以-2和1为?节点(=给定点(2,4)和(1,1),构E函x2的?多 ,d此0.82和0.92的近q.)N易构E?多p1(x) = 2 xuk0.82 p1(0.8) = 2 0.8 = 1.2,0.92 p1(0.9) = 2 0.9 = 1.1姑 不论们的误 大?,就其近q0.82 0.92S不 的.O|(肇院)O5 / 41插例子例 以-2和1为?节点(=给定点(2,4)和(1,1),构E函x2的?多 ,d此0.82和0.92的近q.)N易构E?多p1(x) = 2 xuk0.82 p1(0.8) = 2 0.8 = 1.2,0.92 p1(0.9) = 2 0.9 = 1.1姑 不论们的误 大?,就其近q0.82 0.92S不 的.O|(肇院)O5 / 41?因3u: x23x 0单调递O的, ?多p1(x) = 2 x%单调递的.一地,对u?多,不论其表y为Lagrange.还Newton.,?条都保y了被?函3?节点处得 同的函,并不U保y们3 点k同的O变化.以,k必要讨论3?节点处也给定导的?多.O|(肇院)O6 / 41?因3u: x23x 0单调递O的, ?多p1(x) = 2 x%单调递的.一地,对u?多,不论其表y为Lagrange.还Newton.,?条都保y了被?函3?节点处得 同的函,并不U保y们3 点k同的O变化.以,k必要讨论3?节点处也给定导的?多.O|(肇院)O6 / 41?因3u: x23x 0单调递O的, ?多p1(x) = 2 x%单调递的.一地,对u?多,不论其表y为Lagrange.还Newton.,?条都保y了被?函3?节点处得 同的函,并不U保y们3 点k同的O变化.以,k必要讨论3?节点处也给定导的?多.O|(肇院)O6 / 41Hermite插5-4 已函f(x)3k + 1个互异节点xi(i = 0,1, ,k)处的函f(xi) 和到mi阶的导f(j)(xi) (j = 1, ,mi).e存3次不过n的多 pn(x),满vp(j)n(xi) = f(j)(xi), j = 0,1, ,mi; i = 0,1, ,kKpn(x)为f(x)的Hermite?(Hermite Interpolation Poly -nomial).待定X也一解,但对具体问题还不同的 解.O|(肇院)O7 / 41Hermite插5-4 已函f(x)3k + 1个互异节点xi(i = 0,1, ,k)处的函f(xi) 和到mi阶的导f(j)(xi) (j = 1, ,mi).e存3次不过n的多 pn(x),满vp(j)n(xi) = f(j)(xi), j = 0,1, ,mi; i = 0,1, ,kKpn(x)为f(x)的Hermite?(Hermite Interpolation Poly -nomial).待定X也一解,但对具体问题还不同的 解.O|(肇院)O7 / 41例例5-3 已f(1) = 3, f0(1) = 5, f00(1) = 2, f(2) = 4,3次Hermite?多.)d?条p3(1) = 3, p03(1) = 5和p00 3(1) = 2,f(x)3x = 1 处的2阶Taylor为T2(x) = f(1) + f0(1)(x 1) +f00(1) 2(x 1)2= 3 + 5(x 1) (x 1)2已满v3x = 1点的k条.3次?多为p3(x) = T2(x) + c(x 1)3= 3 + 5(x 1) (x 1)2+ c(x 1)3其Xc待定.w,p3(x)也满v3x = 1点的k?条.2 dp3(2) = 4解得c = 3.故Hermite?多为p3(x) = 3 + 5(x 1) (x 1)2 3(x 1)3O|(肇院)O8 / 41例例5-3 已f(1) = 3, f0(1) = 5, f00(1) = 2, f(2) = 4,3次Hermite?多.)d?条p3(1) = 3, p03(1) = 5和p00 3(1) = 2,f(x)3x = 1 处的2阶Taylor为T2(x) = f(1) + f0(1)(x 1) +f00(1) 2(x 1)2= 3 + 5(x 1) (x 1)2已满v3x = 1点的k条.3次?多为p3(x) = T2(x) + c(x 1)3= 3 + 5(x 1) (x 1)2+ c(x 1)3其Xc待定.w,p3(x)也满v3x = 1点的k?条.2 dp3(2) = 4解得c = 3.故Hermite?多为p3(x) = 3 + 5(x 1) (x 1)2 3(x 1)3O|(肇院)O8 / 41例例5-3 已f(1) = 3, f0(1) = 5, f00(1) = 2, f(2) = 4,3次Hermite?多.)d?条p3(1) = 3, p03(1) = 5和p00 3(1) = 2,f(x)3x = 1 处的2阶Taylor为T2(x) = f(1) + f0(1)(x 1) +f00(1) 2(x 1)2= 3 + 5(x 1) (x 1)2已满v3x = 1点的k条.3次?多为p3(x) = T2(x) + c(x 1)3= 3 + 5(x 1) (x 1)2+ c(x 1)3其Xc待定.w,p3(x)也满v3x = 1点的k?条.2 dp3(2) = 4解得c = 3.故Hermite?多为p3(x) = 3 + 5(x 1) (x 1)2 3(x 1)3O|(肇院)O8 / 41例例5-3 已f(1) = 3, f0(1) = 5, f00(1) = 2, f(2) = 4,3次Hermite?多.)d?条p3(1) = 3, p03(1) = 5和p00 3(1) = 2,f(x)3x = 1 处的2阶Taylor为T2(x) = f(1) + f0(1)(x 1) +f00(1) 2(x 1)2= 3 + 5(x 1) (x 1)2已满v3x = 1点的k条.3次?多为p3(x) = T2(x) + c(x 1)3= 3 + 5(x 1) (x 1)2+ c(x 1)3其Xc待定.w,p3(x)也满v3x = 1点的k?条.2 dp3(2) = 4解得c = 3.故Hermite?多为p3(x) = 3 + 5(x 1) (x 1)2 3(x 1)3O|(肇院)O8 / 41例例5-3 已f(1) = 3, f0(1) = 5, f00(1) = 2, f(2) = 4,3次Hermite?多.)d?条p3(1) = 3, p03(1) = 5和p00 3(1) = 2,f(x)3x = 1 处的2阶Taylor为T2(x) = f(1) + f0(1)(x 1) +f00(1) 2(x 1)2= 3 + 5(x 1) (x 1)2已满v3x = 1点的k条.3次?多为p3(x) = T2(x) + c(x 1)3= 3 + 5(x 1) (x 1)2+ c(x 1)3其Xc待定.w,p3(x)也满v3x = 1点的k?条.2 dp3(2) = 4解得c = 3.故Hermite?多为p3(x) = 3 + 5(x 1) (x 1)2 3(x 1)3O|(肇院)O8 / 41例例5-3 已f(1) = 3, f0(1) = 5, f00(1) = 2, f(2) = 4,3次Hermite?多.)d?条p3(1) = 3, p03(1) = 5和p00 3(1) = 2,f(x)3x = 1 处的2阶Taylor为T2(x) = f(1) + f0(1)(x 1) +f00(1) 2(x 1)2= 3 + 5(x 1) (x 1)2已满v3x = 1点的k条.3次?多为p3(x) = T2(x) + c(x 1)3= 3 + 5(x 1) (x 1)2+ c(x 1)3其Xc待定.w,p3(x)也满v3x = 1点的k?条.2 dp3(2) = 4解得c = 3.故Hermite?多为p3(x) = 3 + 5(x 1) (x 1)2 3(x 1)3O|(肇院)O8 / 41例例5-3 已f(1) = 3, f0(1) = 5, f00(1) = 2, f(2) = 4,3次Hermit