數(shù)據(jù)擬合插值matlab

1、Lagrange 插值對(duì)給定的n個(gè)插值節(jié)點(diǎn)(x , j ),(x , j ),.,(x , j ),利用n次Lagrange插值多項(xiàng)式 1122n n公式，則對(duì)插值區(qū)間內(nèi)任意X的函數(shù)值y可通過下式求得：j( x)=&E 二)k=1k j=i xk - xjj豐k編寫Lagrange插值函數(shù)的m文件%拉格朗日插值(lagrange insert)function y=lagrange(x0,y0,x)n=length(x0);m=length(x);for i=1:mz=x(i);s=0.0;for k=1:np=1.0;for j=1:nif j=kp=p*(z-x0(j)/(x0(k)-x0

2、(j);endends=p*y0(k)+s;endy(i)=s;end給出f (x) = Inx的數(shù)值表，如下表所示，求Lagrange插值多項(xiàng)式，并用Lagrange插值計(jì)算ln(0.54)的近似值。X0.40.50.60.70.8ln(x)-0.916291-0.693147-0.510826-0.357765-0.223144 syms t x=0.4:0.1:0.8; y=-0.916291 -0.693147 -0.510826 -0.357765 -0.223144; L=lagrange(x,y,t)%求 Lagrange 插值多項(xiàng)式L =-62809452203122625/1

3、688849860263936*(5/2*t-1)*(t-1/2)*(t-3/5)*(t-7/10)+402807580171551375/2251799813685248*(10/3*t-4/3)*(t-1/2)*(t-3/5)*(t-4/5)-287569472906395125/1125899906842624*(5*t-2)*(t-1/2)*(t-7/10)*(t-4/5)+390207071364122125/3377699720527872*(10*t-4)*(t-3/5)*(t-7/10)*(t-4/5)+515825975770367375/13510798882111488*

4、(-10*t+5)*(t-3/5)*(t-7/10)*(t-4/5) expand(L)%將上式展開ans =153809433971751925/27021597764222976*t-148241949004410125/27021597764222976*tA2+8809885296067375/3377699720527872*tA3-2091359076960625/6755399441055744*tA4-1392941972647469/562949953421312 vpa(ans)ans =5.6920925000000575459206212750966*t-5.48605

5、41666668140766323820874580*tA2+2.6082500000001636782561339108118*tA3-.30958333333340008290216853007 829*tA4-2.4743620000000081660118667059578lagrange(x,y,0.54)%用 Lagrange 插值計(jì)算 ln(0.54)的近似值。ans =-0.6160 log(0.54)%計(jì)算 ln(0.54)ans =-0.6162并非插值多項(xiàng)式的次數(shù)越高越逼近逼近函數(shù)f (x)，這種現(xiàn)象叫做Runge現(xiàn)象。下面例子反映了這一現(xiàn)象：x e - 5, 5 x=-

6、5:1:5; y=1./(1+x.A2); x0=-5:0.1:5; y0=lagrange(x,y,x0); y1=1./(1+x0.A2); plot(x0,y0,-r) hold on plot(x0,y1,-b)分段線性插值所謂分段線性插值就是通過插值點(diǎn)用折線段連接起來逼近原曲線，這也是計(jì) 算機(jī)繪圖的基本原理。一維插值：（1）yi=interp1（x,y,xi） 對(duì)一組節(jié)點(diǎn)（x,y）進(jìn)行插值，計(jì)算插值點(diǎn)xi的函數(shù)值。（2）yi=interp1（y,xi） 此格式默認(rèn) x=1:n。（3） yi=interp1（x,y,xi,method）method用來指定插值算法，默認(rèn)為線性算法。插值

7、算法：1.nearest線性最近項(xiàng)插值2.linear線性插值spline三次樣條插值cubic三次插值 x=0:0.1:10; y=sin(x); xi=0:0.25:10; yi=interp1(x,y,xi); plot(x,y,xi,yi,ro)下面用分段插值擬合上面Runge現(xiàn)象例子的函數(shù)，并和Lagrange插值擬合效果進(jìn)行 較。/ (尤)=土尤 - 5, 5 x=-5:1:5;y=1./(1+x.A2);x0=-5:0.1:5;y0=lagrange(x,y,x0);y1=1./(1+x0.A2); y2=interp1(x,y,x0); plot(x0,y1,-b)%繪制原函數(shù)

8、曲線 hold on plot(x0,y0,-r)%制繪 Lagrange 插值擬合曲線 plot(x0,y2,-.m) %繪制分段線性插值擬合曲線可見，分段線性插值精度要差些，但不會(huì)出現(xiàn)Runge現(xiàn)象,這在實(shí)際計(jì)算中很重 要。Hermite插值(實(shí)踐表明，高次Hermite插值也會(huì)產(chǎn)生Runge現(xiàn)象)在節(jié)點(diǎn)上函數(shù)值相等，一階(甚至多階)導(dǎo)數(shù)值也相等，滿足這一要求的插值多 項(xiàng)式叫做Hermite插值多項(xiàng)式。下面只討論函數(shù)值與一階導(dǎo)數(shù)個(gè)數(shù)相等的情況。對(duì)給定的n個(gè)插值節(jié)點(diǎn)(x , j ),(x , j ),.,(x , j )，及其對(duì)應(yīng)的一階導(dǎo)數(shù)值1122n nj, j，,j，則計(jì)算插值在區(qū)域內(nèi)任

9、意x的函數(shù)值y的Hermite插值公式:12 n,、 nj(X) = h (x - x)(2aj j) + j i ii i i i= 1i=1h 上(二)2;a 其中，i x - x 1 J編寫Hermite插值函數(shù)的m文件(注意：函數(shù)名務(wù)必為hermite) % 哈密頓插值(Hermite insert) function y=hermite(x0,y0,y1,x) n=length(x0);m=length(x);for k=1:myy=0.0;for i=1:n h=1.0;a=0.0;for j=1:nif j=ih=h*(x(k)-x0(j)/(x0(i)-x0(j)八2;a=1/

10、(x0(i)-x0(j)+a;endendyy=yy+h*(x0(i)-x(k)*(2*a*y0(i)-y1(i)+y0(i);endy(k)=yy;end對(duì)給定的數(shù)據(jù)表如下，試構(gòu)造Hermite多項(xiàng)式，并給出sin0.34的近似值。x0.300.320.35sinx0.295520.314570.34290(sinx)=cosx0.955340.949240.93937 x0=0.3 0.32 0.35; y0=0.29552 0.31457 0.34290; y1=0.95534 0.94924 0.93937; x=0.3:0.005:0.35; y=hermite(x0,y0,y1,x

11、); plot(x,y) y=hermite(x0,y0,y1,0.34)y =0.3335 sin(0.34)ans =0.3335 y2=sin(x); hold on plot(x,y2,-r) syms x y=hermite(x0,y0,y1,x)y =(-50*x+16)A2*(-20*x+7)A2*(-5454862782959501467/439804651110400000+2978578055176345/70368744177664*x)+(50*x-15)A2*(-100/3*x+35/3)A2*(1850602607564 03941/5497558138880000

12、0-1342132736951433/140737488355328*x)+(20*x-6)A2*( 100/3*x-32/3)A2*(563641348323090491/43980465111040000-626926224956017/17 592186044416*x) p=expand(y)P =28281198279123245/39582418599936*xA4-8634831314984375/19791209299968*x A5-740425907428585007/1583296743997440*xA3+3023214825494586299/1979120 9299

13、968000*xA2-6314108835312459499/263882790666240000*x+8934539859570 6789/54975581388800000 vpa(p,7)ans =714.4889*xA4-436.2963*xA5-467.6482*xA3+152.7554*xA2-23.92770*x+1.625183 subs(p,0.34)ans =0.3335最小二乘擬合最小二乘擬合通常有如下兩種途徑：(1)利用ployfit函數(shù)進(jìn)行多項(xiàng)式擬合格式：polyfit(x,y,n)注：n次多項(xiàng)式a xn + a xn-1 +. + a擬合，輸出系數(shù)aa . ann-

14、10nn-10（2）利用矩陣除法解決復(fù)雜型函數(shù)擬合設(shè)j = span1, x, x 2,用最小二乘法擬合下表所示的數(shù)據(jù)。x0.51.01.52.02.53.0y1.752.453.814.808.008.60 x=0.5 1.0 1.5 2.0 2.5 3.0; y=1.75 2.45 3.81 4.80 8.00 8.60; a=polyfit(x,y,2)a =0.49001.25010.8560 x1=0.5:0.05:3.0; y1=a(3)+a(2)*x1+a(1)*x1.A2;%或 y1=polyval(a,x1)(更簡(jiǎn)潔) plot(x,y,*) hold on plot(x1,

15、y1,-r)用最小二乘法求一個(gè)形如j = a + bx2的經(jīng)驗(yàn)公式，使它與下表所示的數(shù)據(jù)相擬合。x1925313844y19.032.349.073.398.8 x=19 25 31 38 44; y=19.0 32.3 49.0 73.3 98.8; x1=x.A2x1 =36162596114441936 x2=ones(5,1),x1x2 = TOC o 1-5 h z 1361162519611144411936 ab=x2yab =0.59370.0506 x0=19:0.2:44; y0=ab(1)+ab(2)*x0.A2; plot(x,y,o) hold on plot(x0,y0,-r)這之前的最小擬合問題也可用這種方法，擬合結(jié)果完全一致: x=0.5 1.0 1.5 2.0 2.5 3.0;y=1.75 2.45 3.81 4.80 8.00 8.60; x1=(x.A2),x,ones(6,1)x1 =0.25000.50001.00001.00001