數(shù)值計(jì)算方法第5講牛頓插值總結(jié)課件

數(shù)值計(jì)算方法第5講牛頓插值總結(jié)課件NumericalComputationalMethod數(shù)值計(jì)算方法NumericalComputationalMethod插值法第二

章數(shù)值計(jì)算方法國(guó)家精品課程主講教師：楊愛(ài)民8/eol/jpk/course/welcome.jsp?courseId=1220插值法第二章數(shù)值計(jì)算方法國(guó)家精品課程主講教師：楊愛(ài)民問(wèn)題的引入思考1問(wèn)題的由來(lái)

問(wèn)題的實(shí)質(zhì)思考2提法的抽象

新概念的誕生

新概念的初識(shí)學(xué)習(xí)計(jì)算方法的建議思考3思考4準(zhǔn)確理解概念特性（獨(dú)有的性質(zhì)）問(wèn)題的引入思考1問(wèn)題的由來(lái)

新算法研究

算法的警示思考5思考6

算法的應(yīng)用

算法的進(jìn)一步研究思考7思考8算法原理警示：A！B!C!能解決的專業(yè)問(wèn)題聯(lián)想與展望學(xué)習(xí)建議新算法研究算法的警示思考5思考6

第二章插值法

插值法的一般理論Newton插值Lagrange插值

分段低次插值Hermite插值、樣條插值13425第二章插值法插值法的一般理論Newton插值插值法Lagrange插值Newton插值樣條插值誤差估計(jì)分段插值兩點(diǎn)式點(diǎn)斜式等距節(jié)點(diǎn)算法比較推廣方法均差差分知識(shí)結(jié)構(gòu)圖Chapter2Interpolation一般理論插值多項(xiàng)式

Newton前插后插公式插LagrangeNewton樣條插值誤差估計(jì)分段插值兩點(diǎn)式差商及其性質(zhì)Newdon插值多項(xiàng)式的構(gòu)造Newdon插值多項(xiàng)式余項(xiàng)差分及其應(yīng)用Newdon插值法的基本思路三、牛頓插值法差商及其性質(zhì)Newdon插值多項(xiàng)式的構(gòu)造Newdon插值多項(xiàng).,0的一階差商（亦稱均差）.關(guān)于點(diǎn)kxx)()()(],[000為函數(shù)稱kkkxfxxxfxfxxf--=牛頓插值法

差商及其性質(zhì)

差商定義二階差商:K階差商:.,0的一階差商（亦稱均差）.關(guān)于點(diǎn)kxx)()()(],[特別地

差商記號(hào)牛頓插值法

差商及其性質(zhì)特別地差商記號(hào)牛頓插值法差商及其性質(zhì)差商具有線性牛頓插值法

差商及其性質(zhì)差商可表示為函數(shù)值的線性組合差商與函數(shù)值的關(guān)系差商具有線性牛頓插值法差商及其性質(zhì)差商可表示為函數(shù)值的線性牛頓插值法

差商及其性質(zhì)差商與函數(shù)值的關(guān)系觀察與思考牛頓插值法差商及其性質(zhì)差商與函數(shù)值的關(guān)系觀察與思考牛頓插值法

差商及其性質(zhì)差商與所含節(jié)點(diǎn)的順序無(wú)關(guān)建議記憶差商與節(jié)點(diǎn)的關(guān)系牛頓插值法差商及其性質(zhì)差商與所含節(jié)點(diǎn)的順序無(wú)關(guān)建議記憶差商牛頓插值法

差商及其性質(zhì)差商與導(dǎo)數(shù)的關(guān)系則N階差商和N階導(dǎo)數(shù)密切相關(guān)！證明見(jiàn)后牛頓插值法差商及其性質(zhì)差商與導(dǎo)數(shù)的關(guān)系則N階差商和N階導(dǎo)數(shù)牛頓插值法

差商性質(zhì)總結(jié)牛頓插值法差商性質(zhì)總結(jié)牛頓插值法Newdon插值法的基本思路建立Newdon插值公式的理由

具有規(guī)律性又有承襲性不具有承襲性每當(dāng)增加一個(gè)節(jié)點(diǎn)時(shí),不僅要增加求和的項(xiàng)數(shù),而且以前的各項(xiàng)也必須重新計(jì)算.具有嚴(yán)格的規(guī)律性便于記憶優(yōu)點(diǎn)缺點(diǎn)牛頓插值法Newdon插值法的基本思路建立Newdon插值公使其滿足：牛頓插值法Newdon插值法的基本思路通過(guò)節(jié)點(diǎn)思考題需要引入新的概念使其滿足：牛頓插值法Newdon插值法的基本思路通過(guò)節(jié)點(diǎn)牛頓插值法Newdon插值多項(xiàng)式的構(gòu)造牛頓插值法Newdon插值多項(xiàng)式的構(gòu)造Newdon插值多項(xiàng)式的構(gòu)造歸納構(gòu)造多項(xiàng)式：Newdon插值多項(xiàng)式展開(kāi)即為Newdon插值多項(xiàng)式的構(gòu)造歸納構(gòu)造多項(xiàng)式：Newdon由插值多項(xiàng)式存在唯一性可知：Newdon插值多項(xiàng)式的余項(xiàng)從而：牛頓插值余項(xiàng)由此可得差商的另一性質(zhì)差商與導(dǎo)數(shù)的關(guān)系由插值多項(xiàng)式存在唯一性可知：Newdon插值多項(xiàng)式的余項(xiàng)從而說(shuō)明每增加一個(gè)結(jié)點(diǎn)，Newton插值多項(xiàng)式只增加一項(xiàng)，具有承襲性！觀察與思考說(shuō)明每增加一個(gè)結(jié)點(diǎn)，Newton插值多項(xiàng)式只增加一項(xiàng)，具有插

值法N_L插值多項(xiàng)式的比較說(shuō)明每增加一個(gè)結(jié)點(diǎn)，Newton插值多項(xiàng)式只增加一項(xiàng)，具有承襲性！NewdonLagrange基函數(shù)簡(jiǎn)便承襲性計(jì)算簡(jiǎn)便基函數(shù)規(guī)律可分析性計(jì)算量較大誤差可估誤差可估比較插值法N_L插值多項(xiàng)式的比較說(shuō)明每增加一個(gè)結(jié)點(diǎn)，New牛頓插值法一階均差二階均差三階均差四階均差差商計(jì)算表Newdon插值的計(jì)算牛頓插值法一階均差二階均差三階均差四階均差差商計(jì)算表Newd例2.4解題步驟(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出插值；(4)估計(jì)誤差；典型例題分析依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.0.400.410750.550.578150.650.696750.800.888110.901.026521.051.25382(5)給出直觀解釋。例2.4解題步驟(1)完成差商表；(2)求出插值多項(xiàng)式；解析(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出插值；(4)估計(jì)誤差；Newdon插值的例題一階均差二階均差三階均差四階均差五階均差0.400.410750.550.578151.116000.650.696751.186000.280000.800.888111.275730.358930.1973330.901.026521.384100.433480.2129520.03123811.051.253821.515330.524930.2286670.03142860.000380952(5)給出直觀解釋。依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4解析(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出Newdon插值的例題依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.第1步完成差商表一階均差二階均差三階均差四階均差五階均差0.400.410750.550.578151.116000.650.696751.186000.280000.800.888111.275730.358930.1973330.901.026521.384100.433480.2129520.03123811.051.253821.515330.524930.2286670.03142860.000380952例2.4Newdon插值的例題依照的函數(shù)表，求5Newdon插值的例題依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.一階均差二階均差三階均差四階均差五階均差0.400.410750.550.578151.116000.650.696751.186000.280000.800.888111.275730.358930.1973330.901.026521.384100.433480.2129520.03123811.051.253821.515330.524930.2286670.03142860.000380952第2步求出插值多項(xiàng)式例2.41

基函數(shù)Newdon插值的例題依照的函數(shù)表，求5Newdon插值的例題依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.第3步求出插值N(0.596)=0.6319174965773844例2.4Newdon插值的例題依照的函數(shù)表，求5例2.4Newdon插值的例題依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.第4步估計(jì)誤差N(0.596)=0.6319174965773844例2.4Newdon插值的例題依照的函例2.4解題步驟(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出插值；(4)估計(jì)誤差；典型例題的求解實(shí)驗(yàn)依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.0.400.410750.550.578150.650.696750.800.888110.901.026521.051.25382(5)幾何直觀驗(yàn)證。例2.4解題步驟(1)完成差商表；(2)求出插值多項(xiàng)式；第1步完成差商表Clear[x,k,y]{x[0],x[1],x[2],x[3],x[4],x[5]}={0.4,0.55,0.65,0.80,0.90,1.05};{y[0],y[1],y[2],y[3],y[4],y[5]}={0.41075,0.57815,0.69675,0.88811,1.02652,1.25382};Table[y[k],{k,0,5}]//N;f[i_,j_]:=(y[j]-y[i])/(x[j]-x[i])Table[f[i,i+1],{i,0,4}]//N;f[i_,j_,k_]:=(f[j,k]-f[i,j])/(x[k]-x[i])Table[f[i,i+1,i+2],{i,0,3}]//N;f[i_,j_,k_,l_]:=(f[j,k,l]-f[i,j,k])/(x[l]-x[i])Table[f[i,i+1,i+2,i+3],{i,0,2}]//N;f[i_,j_,k_,l_,m_]:=(f[j,k,l,m]-f[i,j,k,l])/(x[m]-x[i])Table[f[i,i+1,i+2,i+3,i+4],{i,0,1}]//N;f[i_,j_,k_,l_,m_,p_]:=(f[j,k,l,m,p]-f[i,j,k,l,m])/(x[m]-x[i])Table[f[i,i+1,i+2,i+3,i+4,i+5],{i,0,0}]//N;程序設(shè)計(jì)依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)第1步完成差商表Clear[x,k,y]程序設(shè)計(jì)依照程序設(shè)計(jì)A={{y[0],y[1],y[2],y[3],y[4],y[5]},{0,f[0,1],f[1,2],f[2,3],f[3,4],f[4,5]},{0,0,f[0,1,2],f[1,2,3],f[2,3,4],f[3,4,5]},{0,0,0,f[0,1,2,3],f[1,2,3,4],f[2,3,4,5]},{0,0,0,0,f[0,1,2,3,4],f[1,2,3,4,5]},{0,0,0,0,0,f[0,1,2,3,4,5]}};Transpose[A]//N;MatrixForm[%]0.41075000000.578151.11600000.696751.1860.280000.888111.275730.3589330.197333001.026521.38410.4334670.2129520.031238101.253821.515330.5249330.2286670.03142860.000380952第1步完成差商表依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)程序設(shè)計(jì)A={{y[0],y[1],y[2],y[3],y[程序設(shè)計(jì)0.41075000000.578151.11600000.696751.1860.280000.888111.275730.3589330.197333001.026521.38410.4334670.2129520.031238101.253821.515330.5249330.2286670.03142860.000380952第2步求出插值多項(xiàng)式a[0]=y[0];a[1]=f[0,1];a[2]=f[0,1,2];a[3]=f[0,1,2,3];a[4]=f[0,1,2,3,4];a[5]=f[0,1,2,3,4,5];NU[x]=a[0]+Sum[a[k]*Product[(x-x[m]),{m,0,k-1}],{k,1,5}]//NExpand[%]依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)程序設(shè)計(jì)0.410750第3步求出插值N[%145/.x->0.596,20]依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)第3步求出插值N[%145/.x->0.596,20估計(jì)誤差第4步程序設(shè)計(jì)Clear[A,g1,g2]xx={0.40,0.55,0.65,0.80,0.90,1.05};yy={0.41075,0.57815,0.69675,0.88811,1.02652,1.25382};A=Table[{xx[[k]],yy[[k]]},{k,1,6}];g1=ListPlot[A,Prolog->AbsolutePointSize[15]];Interpolation[A,InterpolationOrder->5]g2=Plot[%[x],{x,0.40,1.05}]Show[g1,g2]N[%%%[0.596],20]插值原理計(jì)算的數(shù)值牛頓插值數(shù)值依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)估計(jì)誤差第4步程序設(shè)計(jì)Clear[A,g1,g2]插值原理Clear[A,g1,g2]xx={0.40,0.55,0.65,0.80,0.90,1.05};yy={0.41075,0.57815,0.69675,0.88811,1.02652,1.25382};A=Table[{xx[[k]],yy[[k]]},{k,1,6}];g1=ListPlot[A,Prolog->AbsolutePointSize[15]];Interpolation[A,InterpolationOrder->5]g2=Plot[%[x],{x,0.40,1.05}]Show[g1,g2]程序設(shè)計(jì)幾何直觀驗(yàn)證第5步依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)Clear[A,g1,g2]程序設(shè)計(jì)幾何直觀驗(yàn)證第5步依照程序設(shè)計(jì)課后實(shí)驗(yàn)課題(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出插值；(4)估計(jì)誤差；(5)幾何直觀驗(yàn)證。已知求4次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.實(shí)驗(yàn)課題借助mathematica要求解題參考{x[0],x[1],x[2],x[3],x[4]}={0.001,11,12,13,14};y[k_]:=Log[x[k]]Table[y[k],{k,0,4}]//N;MatrixForm[%]f[i_,j_]:=(y[j]-y[i])/(x[j]-x[i])Table[f[i,i+1],{i,0,3}]//N;MatrixForm[%]f[i_,j_,k_]:=(f[j,k]-f[i,j])/(x[k]-x[i])Table[f[i,i+1,i+2],{i,0,2}]//N;MatrixForm[%]f[i_,j_,k_,l_]:=(f[j,k,l]-f[i,j,k])/(x[l]-x[i])Table[f[i,i+1,i+2,i+3],{i,0,1}]//N;MatrixForm[%]f[i_,j_,k_,l_,m_]:=(f[j,k,l,m]-f[i,j,k,l])/(x[m]-x[i])Table[f[i,i+1,i+2,i+3,i+4],{i,0,0}]//N;MatrixForm[%]程序設(shè)計(jì)課后實(shí)驗(yàn)課題(1)完成差商表；(2)求出插值多項(xiàng)程序設(shè)計(jì)課后實(shí)驗(yàn)課題第1步完成差商表Clear[A]A={{y[0],y[1],y[2],y[3],y[4]},{0,f[0,1],f[1,2],f[2,3],f[3,4]},{0,0,f[0,1,2],f[1,2,3],f[2,3,4]},{0,0,0,f[0,1,2,3],f[1,2,3,4]},{0,0,0,0,f[0,1,2,3,4]}};Transpose[A]//N;MatrixForm[%]-6.9077600002.39790.8460450002.484910.0870114-0.0632581002.564950.0800427-0.003484330.0045983302.639060.074108-0.002967370.000172322-0.000316166程序設(shè)計(jì)課后實(shí)驗(yàn)課題第1步完成差商表Clear[A]-6.程序設(shè)計(jì)課后實(shí)驗(yàn)課題第2步求出插值多項(xiàng)式a[0]=y[0];a[1]=f[0,1];a[2]=f[0,1,2];a[3]=f[0,1,2,3];a[4]=f[0,1,2,3,4];N[x]=a[0]+Sum[a[k]*Product[(x-x[m]),{m,0,k-1}],{k,1,4}]//NExpand[%]程序設(shè)計(jì)課后實(shí)驗(yàn)課題第2步求出插值多項(xiàng)式a[0]=y[0];程序設(shè)計(jì)課后實(shí)驗(yàn)課題第3步求出插值估計(jì)誤差第4步Clear[x,xx,yy,y,g1,g2]x={0.001,11,12,13,14};y[x_]:=Log[x]yy={y[0.001],y[11],y[12],y[13],y[14]};A=Table[{x[[k]],yy[[k]]},{k,1,5}];g1=ListPlot[A,Prolog->AbsolutePointSize[15]];Interpolation[A,InterpolationOrder->4]g2=Plot[%[x],{x,0,14}]Show[g1,g2]N[%%%[0.00125],20]插值原理計(jì)算的數(shù)值牛頓插值數(shù)值誤差程序設(shè)計(jì)課后實(shí)驗(yàn)課題第3步求出插值估計(jì)誤差第4步C程序設(shè)計(jì)課后實(shí)驗(yàn)課題幾何直觀驗(yàn)證第5步程序設(shè)計(jì)課后實(shí)驗(yàn)課題幾何直觀驗(yàn)證第5步程序設(shè)計(jì)Newdon插值的通用程序設(shè)計(jì)基于Mathematica9.0插值節(jié)點(diǎn)坐標(biāo)xx={*,*,*}插值函數(shù)值yy={*,*,*}程序設(shè)計(jì)Newdon插值的通用程序設(shè)計(jì)基于Mathemati數(shù)值計(jì)算方法第5講牛頓插值總結(jié)課件數(shù)值計(jì)算方法第5講牛頓插值總結(jié)課件NumericalComputationalMethod數(shù)值計(jì)算方法NumericalComputationalMethod插值法第二

章數(shù)值計(jì)算方法國(guó)家精品課程主講教師：楊愛(ài)民8/eol/jpk/course/welcome.jsp?courseId=1220插值法第二章數(shù)值計(jì)算方法國(guó)家精品課程主講教師：楊愛(ài)民問(wèn)題的引入思考1問(wèn)題的由來(lái)

問(wèn)題的實(shí)質(zhì)思考2提法的抽象

新概念的誕生

新概念的初識(shí)學(xué)習(xí)計(jì)算方法的建議思考3思考4準(zhǔn)確理解概念特性（獨(dú)有的性質(zhì)）問(wèn)題的引入思考1問(wèn)題的由來(lái)

新算法研究

算法的警示思考5思考6

算法的應(yīng)用

算法的進(jìn)一步研究思考7思考8算法原理警示：A！B!C!能解決的專業(yè)問(wèn)題聯(lián)想與展望學(xué)習(xí)建議新算法研究算法的警示思考5思考6

第二章插值法

插值法的一般理論Newton插值Lagrange插值

分段低次插值Hermite插值、樣條插值13425第二章插值法插值法的一般理論Newton插值插值法Lagrange插值Newton插值樣條插值誤差估計(jì)分段插值兩點(diǎn)式點(diǎn)斜式等距節(jié)點(diǎn)算法比較推廣方法均差差分知識(shí)結(jié)構(gòu)圖Chapter2Interpolation一般理論插值多項(xiàng)式

Newton前插后插公式插LagrangeNewton樣條插值誤差估計(jì)分段插值兩點(diǎn)式差商及其性質(zhì)Newdon插值多項(xiàng)式的構(gòu)造Newdon插值多項(xiàng)式余項(xiàng)差分及其應(yīng)用Newdon插值法的基本思路三、牛頓插值法差商及其性質(zhì)Newdon插值多項(xiàng)式的構(gòu)造Newdon插值多項(xiàng).,0的一階差商（亦稱均差）.關(guān)于點(diǎn)kxx)()()(],[000為函數(shù)稱kkkxfxxxfxfxxf--=牛頓插值法

差商及其性質(zhì)

差商定義二階差商:K階差商:.,0的一階差商（亦稱均差）.關(guān)于點(diǎn)kxx)()()(],[特別地

差商記號(hào)牛頓插值法

差商及其性質(zhì)特別地差商記號(hào)牛頓插值法差商及其性質(zhì)差商具有線性牛頓插值法

差商及其性質(zhì)差商可表示為函數(shù)值的線性組合差商與函數(shù)值的關(guān)系差商具有線性牛頓插值法差商及其性質(zhì)差商可表示為函數(shù)值的線性牛頓插值法

差商及其性質(zhì)差商與函數(shù)值的關(guān)系觀察與思考牛頓插值法差商及其性質(zhì)差商與函數(shù)值的關(guān)系觀察與思考牛頓插值法

差商及其性質(zhì)差商與所含節(jié)點(diǎn)的順序無(wú)關(guān)建議記憶差商與節(jié)點(diǎn)的關(guān)系牛頓插值法差商及其性質(zhì)差商與所含節(jié)點(diǎn)的順序無(wú)關(guān)建議記憶差商牛頓插值法

差商及其性質(zhì)差商與導(dǎo)數(shù)的關(guān)系則N階差商和N階導(dǎo)數(shù)密切相關(guān)！證明見(jiàn)后牛頓插值法差商及其性質(zhì)差商與導(dǎo)數(shù)的關(guān)系則N階差商和N階導(dǎo)數(shù)牛頓插值法

差商性質(zhì)總結(jié)牛頓插值法差商性質(zhì)總結(jié)牛頓插值法Newdon插值法的基本思路建立Newdon插值公式的理由

具有規(guī)律性又有承襲性不具有承襲性每當(dāng)增加一個(gè)節(jié)點(diǎn)時(shí),不僅要增加求和的項(xiàng)數(shù),而且以前的各項(xiàng)也必須重新計(jì)算.具有嚴(yán)格的規(guī)律性便于記憶優(yōu)點(diǎn)缺點(diǎn)牛頓插值法Newdon插值法的基本思路建立Newdon插值公使其滿足：牛頓插值法Newdon插值法的基本思路通過(guò)節(jié)點(diǎn)思考題需要引入新的概念使其滿足：牛頓插值法Newdon插值法的基本思路通過(guò)節(jié)點(diǎn)牛頓插值法Newdon插值多項(xiàng)式的構(gòu)造牛頓插值法Newdon插值多項(xiàng)式的構(gòu)造Newdon插值多項(xiàng)式的構(gòu)造歸納構(gòu)造多項(xiàng)式：Newdon插值多項(xiàng)式展開(kāi)即為Newdon插值多項(xiàng)式的構(gòu)造歸納構(gòu)造多項(xiàng)式：Newdon由插值多項(xiàng)式存在唯一性可知：Newdon插值多項(xiàng)式的余項(xiàng)從而：牛頓插值余項(xiàng)由此可得差商的另一性質(zhì)差商與導(dǎo)數(shù)的關(guān)系由插值多項(xiàng)式存在唯一性可知：Newdon插值多項(xiàng)式的余項(xiàng)從而說(shuō)明每增加一個(gè)結(jié)點(diǎn)，Newton插值多項(xiàng)式只增加一項(xiàng)，具有承襲性！觀察與思考說(shuō)明每增加一個(gè)結(jié)點(diǎn)，Newton插值多項(xiàng)式只增加一項(xiàng)，具有插

值法N_L插值多項(xiàng)式的比較說(shuō)明每增加一個(gè)結(jié)點(diǎn)，Newton插值多項(xiàng)式只增加一項(xiàng)，具有承襲性！NewdonLagrange基函數(shù)簡(jiǎn)便承襲性計(jì)算簡(jiǎn)便基函數(shù)規(guī)律可分析性計(jì)算量較大誤差可估誤差可估比較插值法N_L插值多項(xiàng)式的比較說(shuō)明每增加一個(gè)結(jié)點(diǎn)，New牛頓插值法一階均差二階均差三階均差四階均差差商計(jì)算表Newdon插值的計(jì)算牛頓插值法一階均差二階均差三階均差四階均差差商計(jì)算表Newd例2.4解題步驟(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出插值；(4)估計(jì)誤差；典型例題分析依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.0.400.410750.550.578150.650.696750.800.888110.901.026521.051.25382(5)給出直觀解釋。例2.4解題步驟(1)完成差商表；(2)求出插值多項(xiàng)式；解析(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出插值；(4)估計(jì)誤差；Newdon插值的例題一階均差二階均差三階均差四階均差五階均差0.400.410750.550.578151.116000.650.696751.186000.280000.800.888111.275730.358930.1973330.901.026521.384100.433480.2129520.03123811.051.253821.515330.524930.2286670.03142860.000380952(5)給出直觀解釋。依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4解析(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出Newdon插值的例題依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.第1步完成差商表一階均差二階均差三階均差四階均差五階均差0.400.410750.550.578151.116000.650.696751.186000.280000.800.888111.275730.358930.1973330.901.026521.384100.433480.2129520.03123811.051.253821.515330.524930.2286670.03142860.000380952例2.4Newdon插值的例題依照的函數(shù)表，求5Newdon插值的例題依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.一階均差二階均差三階均差四階均差五階均差0.400.410750.550.578151.116000.650.696751.186000.280000.800.888111.275730.358930.1973330.901.026521.384100.433480.2129520.03123811.051.253821.515330.524930.2286670.03142860.000380952第2步求出插值多項(xiàng)式例2.41

基函數(shù)Newdon插值的例題依照的函數(shù)表，求5Newdon插值的例題依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.第3步求出插值N(0.596)=0.6319174965773844例2.4Newdon插值的例題依照的函數(shù)表，求5例2.4Newdon插值的例題依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.第4步估計(jì)誤差N(0.596)=0.6319174965773844例2.4Newdon插值的例題依照的函例2.4解題步驟(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出插值；(4)估計(jì)誤差；典型例題的求解實(shí)驗(yàn)依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.0.400.410750.550.578150.650.696750.800.888110.901.026521.051.25382(5)幾何直觀驗(yàn)證。例2.4解題步驟(1)完成差商表；(2)求出插值多項(xiàng)式；第1步完成差商表Clear[x,k,y]{x[0],x[1],x[2],x[3],x[4],x[5]}={0.4,0.55,0.65,0.80,0.90,1.05};{y[0],y[1],y[2],y[3],y[4],y[5]}={0.41075,0.57815,0.69675,0.88811,1.02652,1.25382};Table[y[k],{k,0,5}]//N;f[i_,j_]:=(y[j]-y[i])/(x[j]-x[i])Table[f[i,i+1],{i,0,4}]//N;f[i_,j_,k_]:=(f[j,k]-f[i,j])/(x[k]-x[i])Table[f[i,i+1,i+2],{i,0,3}]//N;f[i_,j_,k_,l_]:=(f[j,k,l]-f[i,j,k])/(x[l]-x[i])Table[f[i,i+1,i+2,i+3],{i,0,2}]//N;f[i_,j_,k_,l_,m_]:=(f[j,k,l,m]-f[i,j,k,l])/(x[m]-x[i])Table[f[i,i+1,i+2,i+3,i+4],{i,0,1}]//N;f[i_,j_,k_,l_,m_,p_]:=(f[j,k,l,m,p]-f[i,j,k,l,m])/(x[m]-x[i])Table[f[i,i+1,i+2,i+3,i+4,i+5],{i,0,0}]//N;程序設(shè)計(jì)依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)第1步完成差商表Clear[x,k,y]程序設(shè)計(jì)依照程序設(shè)計(jì)A={{y[0],y[1],y[2],y[3],y[4],y[5]},{0,f[0,1],f[1,2],f[2,3],f[3,4],f[4,5]},{0,0,f[0,1,2],f[1,2,3],f[2,3,4],f[3,4,5]},{0,0,0,f[0,1,2,3],f[1,2,3,4],f[2,3,4,5]},{0,0,0,0,f[0,1,2,3,4],f[1,2,3,4,5]},{0,0,0,0,0,f[0,1,2,3,4,5]}};Transpose[A]//N;MatrixForm[%]0.41075000000.578151.11600000.696751.1860.280000.888111.275730.3589330.197333001.026521.38410.4334670.2129520.031238101.253821.515330.5249330.2286670.03142860.000380952第1步完成差商表依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)程序設(shè)計(jì)A={{y[0],y[1],y[2],y[3],y[程序設(shè)計(jì)0.41075000000.578151.11600000.696751.1860.280000.888111.275730.3589330.197333001.026521.38410.4334670.2129520.031238101.253821.515330.5249330.2286670.03142860.000380952第2步求出插值多項(xiàng)式a[0]=y[0];a[1]=f[0,1];a[2]=f[0,1,2];a[3]=f[0,1,2,3];a[4]=f[0,1,2,3,4];a[5]=f[0,1,2,3,4,5];NU[x]=a[0]+Sum[a[k]*Product[(x-x[m]),{m,0,k-1}],{k,1,5}]//NExpand[%]依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)程序設(shè)計(jì)0.410750第3步求出插值N[%145/.x->0.596,20]依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)第3步求出插值N[%145/.x->0.596,20估計(jì)誤差第4步程序設(shè)計(jì)Clear[A,g1,g2]xx={0.40,0.55,0.65,0.80,0.90,1.05};yy={0.41075,0.57815,0.69675,0.88811,1.02652,1.25382};A=Table[{xx[[k]],yy[[k]]},{k,1,6}];g1=ListPlot[A,Prolog->AbsolutePointSize[15]];Interpolation[A,InterpolationOrder->5]g2=Plot[%[x],{x,0.40,1.05}]Show[g1,g2]N[%%%[0.596],20]插值原理計(jì)算的數(shù)值牛頓插值數(shù)值依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)估計(jì)誤差第4步程序設(shè)計(jì)Clear[A,g1,g2]插值原理Clear[A,g1,g2]xx={0.40,0.55,0.65,0.80,0.90,1.05};yy={0.41075,0.57815,0.69675,0.88811,1.02652,1.25382};A=Table[{xx[[k]],yy[[k]]},{k,1,6}];g1=ListPlot[A,Prolog->AbsolutePointSize[15]];Interpolation[A,InterpolationOrder->5]g2=Plot[%[x],{x,0.40,1.05}]Show[g1,g2]程序設(shè)計(jì)幾何直觀驗(yàn)證第5步依照的函數(shù)表，求5次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.例2.4典型例題的求解實(shí)驗(yàn)Clear[A,g1,g2]程序設(shè)計(jì)幾何直觀驗(yàn)證第5步依照程序設(shè)計(jì)課后實(shí)驗(yàn)課題(1)完成差商表；(2)求出插值多項(xiàng)式；(3)求出插值；(4)估計(jì)誤差；(5)幾何直觀驗(yàn)證。已知求4次牛頓插值多項(xiàng)式，并由此計(jì)算的近似值.實(shí)驗(yàn)課題借助