后小霞2012214023

1、數(shù)值分析上機(jī)實(shí)驗(yàn)報(bào)告 函數(shù)插值方法 現(xiàn)代農(nóng)業(yè)工程學(xué)院 農(nóng)業(yè)水土工程 2012214023 后小霞任課教師：羅志強(qiáng)課題名稱課題五 函數(shù)插值法一、 問題提出 對于給定的一元函數(shù) 的n+1個(gè)節(jié)點(diǎn)值 。試用Lagrange公式求其插值多項(xiàng)式或分段二次Lagrange插值多項(xiàng)式。數(shù)據(jù)如下：（1）0.40.550.650.800.951.050.410750.578150.696750.901.001.25382求五次Lagrange多項(xiàng)式L，和分段三次插值多項(xiàng)式，計(jì)算 的值。 （2）12345670.3680.1350.0500.0180.0070.0020.001 試構(gòu)造Lagrange多項(xiàng)式L，計(jì)算

2、的值。 結(jié)果0.165299 0.00213348 一、 要求1、利用Lagrange插值公式 編寫出插值多項(xiàng)式程序；2、給出插值多項(xiàng)式或分段三次插值多項(xiàng)式的表達(dá)式；3、根據(jù)節(jié)點(diǎn)選取原則，對問題（2）用三點(diǎn)插值或二點(diǎn)插值，其結(jié)果如何；4、對此插值問題用Newton插值多項(xiàng)式其結(jié)果如何。三、班級(jí)、姓名、學(xué)號(hào)現(xiàn)代農(nóng)業(yè)工程學(xué)院 農(nóng)業(yè)水土工程 后小霞 2012214023四、目的與意義1、學(xué)會(huì)常用的插值方法，求函數(shù)的近似表達(dá)式，以解決其它實(shí)際問題； 2、明確插值多項(xiàng)式和分段插值多項(xiàng)式各自的優(yōu)缺點(diǎn)； 3、熟悉插值方法的程序編制； 五、計(jì)算公式與方法 1.Lagrange 插值法 n次的拉格朗日插值多項(xiàng)式

3、為，其中是插值基函數(shù)，且 插值基函數(shù)計(jì)算公式為： 2. Newton插值法 Newton均差插值多項(xiàng)式為：六、結(jié)構(gòu)程序設(shè)計(jì)對于題目要求做了一下內(nèi)容(1) Lagrange插值多項(xiàng)式function lk,L=lagrange(X,Y)a=length(X); L=ones(a,a);for n=1: a; V=1; for i=1:a if n=i V=conv(V,poly(X(i)/(X(k)-X(i); endendL1(k,:)=V; l(k,:)=poly2sym (V);endlk=Y*L1;L=Y*l;其中L為拉格朗日插值多項(xiàng)式, lk為基函數(shù)。.(2)求表1中五次Lagrang

4、e多項(xiàng)式L 在窗口中輸入：>> X=0.4 0.55 0.65 0.80 0.95 1.05;>> Y=0.41075 0.57815 0.69675 0.90 1.00 1.25382;>> lk,L=lagrange(X,Y)運(yùn)行結(jié)果C和L：lk = 121.6264 -422.7503 572.5667 -377.2549 121.9718 -15.0845L =(3913328*x5)/32175-取L121.6264-422.7503+572.5667-377.2549+121.9718-15.0845計(jì)算表1 lagranger插值多項(xiàng)式對應(yīng)插值

5、點(diǎn)的值>> x=0.596; M=1; X=0.4,0.55,0.65,0.80,0.95,1.05;>> Y=0.41075,0.57815,0.69675,0.90,1.00,1.25382;>> y,R=lagranger(X,Y,x,M)運(yùn)行結(jié)果：y =0.6257R =2.2170e-008在命令窗口輸入程序：>> x=0.99; M=1; X=0.4,0.55,0.65,0.80,0.95,1.05;>> Y=0.41075,0.57815,0.69675,0.90,1.00,1.25382;>> y,R=la

6、granger(X,Y,x,M)運(yùn)行結(jié)果：y =1.0542R =5.5901e-008（3）求表二的六次lagrange多項(xiàng)式在窗口中輸入：>> M=1:1:7;>> Y=0.368 0.135 0.050 0.018 0.007 0.002 0.001;>> lk, L=lagrange(X,Y)運(yùn)行結(jié)果C和L：lk = 0.0001 -0.0016 0.0186 -0.1175 0.4419 -0.9683 0.9950L =(7*x6)/120000 - (193*x5)/120000 + (223*x4)/12000 - (2821*x3)/240

7、00 + (53023*x2)/120000 - (19367*x)/20000 + 199/200取計(jì)算表2 lagranger插值多項(xiàng)式對應(yīng)插值點(diǎn)的值在命令窗口輸入:x=1.8; M=1; X=1,2,3,4,5,6,7;Y=0.368,0.135,0.050,0.018,0.007,0.002,0.001;y,R=lagranzi(X,Y,x,M) 得到運(yùn)算結(jié)果為y =0.1648R =0.0059在命令窗口輸入:x=6.15; M=1; X=1,2,3,4,5,6,7;Y=0.368,0.135,0.050,0.018,0.007,0.002,0.001;y,R=lagranzi(X,

8、Y,x,M)得到運(yùn)算結(jié)果為：y =0.0013R =0.0042（4）利用牛頓插值方法求function A,C,L,wcgs,Cw= newploy(X,Y)n=length(X); A=zeros(n,n); A(:,1)=Y's=0.0; p=1.0; q=1.0; c1=1.0;for j=2:nfor i=j:nA(i,j)=(A(i,j-1)- A(i-1,j-1)/(X(i)-X(i-j+1);endb=poly(X(j-1);q1=conv(q,b); c1=c1*j; q=q1;endC=A(n,n); b=poly(X(n); q1=conv(q1,b);for k

9、=(n-1):-1:1C=conv(C,poly(X(k); d=length(C); C(d)=C(d)+A(k,k);endL(k,:)=poly2sym(C); Q=poly2sym(q1);syms Mwcgs=M*Q/c1; Cw=q1/c1;輸入：x=0.4 0.55 0.65 0.80 0.95 1.05;y=0.41075 0.57815 0.69675 0.90 1.00 1.25382;A,C,L,wcgs,Cw= newploy(x,y)syms x;ezplot(L,0 1.1);運(yùn)行結(jié)果如下，A =0.4108 0 0 0 0 00.5782 1.1160 0 0 0

10、 00.6967 1.1860 0.2800 0 0 00.9000 1.3550 0.6760 0.9900 0 01.0000 0.6667 -2.2944 -7.4261 -15.3020 01.2538 2.5382 7.4861 24.4514 63.7551 121.6264C =121.6264 -422.7503 572.5667 -377.2549 121.9718 -15.0845L =wcgs =(M*(x6 - (22*x5)/5 + (1583*x4)/200 - (3721*x3)/500 +Cw =0.0014 -0.0061 0.0110 -0.0103 0.0

11、054 -0.0014 0.0002在命令窗口中輸入x=1 2 3 4 5 6 7;y=0.368 0.135 0.050 0.018 0.007 0.002 0.001;A,C,L,wcgs,Cw= newploy(x,y)syms x;ezplot(L,0 8);9運(yùn)行結(jié)果如下， A =0.3680 0 0 0 0 0 00.1350 -0.2330 0 0 0 0 00.0500 -0.0850 0.0740 0 0 0 00.0180 -0.0320 0.0265 -0.0158 0 0 00.0070 -0.0110 0.0105 -0.0053 0.0026 0 00.0020 -

12、0.0050 0.0030 -0.0025 0.0007 -0.0004 00.0010 -0.0010 0.0020 -0.0003 0.0005 -0.0000 0.0001C =0.0001 -0.0016 0.0186 -0.1175 0.4419 -0.9683 0.9950L =wcgs =(M*(x7 - 28*x6 + 322*x5 - 1960*x4 + 6769*x3 - 13132*x2 + 13068*x - 5040)/5040Cw =0.0002 -0.0056 0.0639 -0.3889 1.3431 -2.6056 2.5929 -1.0000（5）對于問題2

13、求分段三次Hermite插值多項(xiàng)式Hermite插值程序function f,f0 = Hermite1(x,y,y_1)syms t;f = 0.0;if(length(x) = length(y) if(length(y) = length(y_1) n = length(x); else disp('y和y的導(dǎo)數(shù)的維數(shù)不相等'); return; endelse disp('x和y的維數(shù)不相等！ '); return;end for i=1:n h = 1.0; a = 0.0; for j=1:n if( j = i) h = h*(t-x(j)2/(x

14、(i)-x(j)2); a = a + 1/(x(i)-x(j); end end f = f + h*(x(i)-t)*(2*a*y(i)-y_1(i)+y(i);end f0 = subs(f,'t');其中x為給定點(diǎn)橫坐標(biāo)數(shù)組，y為給定點(diǎn)縱坐標(biāo)數(shù)組，y_1為原函數(shù)在給定點(diǎn)處的導(dǎo)數(shù)數(shù)組(該導(dǎo)數(shù)數(shù)組通過拉格朗日多項(xiàng)式求得)。>> y=0.41075 0.57815 0.69675 0.90 1.00 1.25382;>> x=0.4 0.55 0.65 0.80 0.95 1.05;>> y=0.41075 0.57815 0.69675 0.90 1.00 1.25382;>> y_1=2.3440 0.9032 1.4329 0.9903 0.9170 5.1439;>> f,f0 = Hermite1(x,y,y_1);>> f0 f0 =七、結(jié)果討論和分析 插值法是一種古老的數(shù)學(xué)方法，它