數(shù)值計算方法實驗報告實驗報告題目：newton插值多項式實驗要求用matlab解析Newton插值多項式的程序?qū)嶒灧治觯ò〝?shù)學(xué)原理，小組分析討論后確定實驗方案和實現(xiàn)思路）根據(jù)經(jīng)過n+1個不同的差值點x1,x2,…，x(n+1),構(gòu)造牛頓插值公式實驗步驟（過程）（包括程序及上機的實現(xiàn)的結(jié)果）function[p2,z]=newton(x,y,t)n=length(x);chaS(1)=y(1);fori=2:nx1=x;y1=y;x1(i+1:n)=[];y1(i+1:n)=[];n1=length(x1);s1=0;forj=1:n1t1=1;fork=1:n1ifk==jcontinue;elset1=t1*(x1(j)-x1(k));ehdehds1=s1+y1(j)/t1;endchaS(i)=s1;ehdb(1,:)=[zeros(1,n-1)chaS(1)];c1=cell(1,n-1);fori=2:nu1=1;forj=1:i-1u1=conv(u1,[1-x(j)]);cl{i-1}=ul;endcl{i-1}chaS(i)*cl{i-1};b(i,:)=[zeros(1,n-i),cl{i-1}];endp2=b(1,:);forj=2:np2=p2+b(j,:);endiflengh(t)==1rm=0;fori=1:nrm=rm+p2(i)*t^(n-i);endz=rm;elsekl=length(t);rm=zeros(1,kl)forj=1:k1fori=1:nrm(j)=rm(j)+p2(i)*t(j)^(n-i);endz=rm;endendplot(t,z,’y’,x,y,’*r’)問題：已知數(shù)據(jù)：x00.20.40.60.81.0F(x)1.00001.22141.49181.82212.22552.7183輸入：像x=[0.20.40.60.81.0];y=[exp(0)exp(0.2)exp(0.4)exp(0.6)exp(0.8)exp(1.0)]y=1.00001.22141.49181.82212.22552.7183[c,l]=lagran(x,y)可得出輸出結(jié)果：c=0.01390.03490.17040.49911.00011.0000l=-26.041778.1250-88.541746.8750-11.41671.0000130.2083-364.5833369.7917-160.416725.00000-260.4167677.0833-614.5833222.9167-25.00000260.4167-625.0000510.4167-162.500016.66670-130.2083286.4583-213.541763.5417-6.2500026.0417-52.083336.4583-10.41671.00000有輸出結(jié)果可以得出：P(x)的系數(shù)分別為：a0=0.0139a1=0.0349a2=0.1704a3=0.4991a