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《matlab基础及其应用（陈姿羽）第3章习题答案》由会员上传分享，免费在线阅读，更多相关内容在工程资料-天天文库。
1、第三章习题答案1.代码：a=[1-1-1];roots(a)结果：ans=-0.61801.61802・代码:x=0:10;y=sin(x);xi=0:0.15:10;%选取了67个插值点，要增加n,只需减小步长即可y0二sin(xi);%算精确值yl=interpl(x,y,xi);%分段线性插值y2二interpl(x,y,xi,'spline，);%三次样条插值plot(xi,yO,，0，,xi,yl,xi,y2,J，)legend('精确值','分段线性插值',、三次样条插值')结果:10.80.60.40.20-0.2-0.4-0.6-0.823456789精确值
2、分段线性插值三次样条插值■1o3.理论公式为：p=1.0332*exp(-(x+500)/7756),所以拟合模型可写为：p=a*exp(-k*x+b)式中，a,k,b为常数，两边同时取自然对数，得:log(p)=-k*x+b+log(a)问题转化为线性模型。注意：自然对数是log(x),以10为底的对数是loglO(x)代码：clear;x=[0300600100015002000];p二[0.96890.93220.89690.85190.79890.7491];lnp=log(p);%转化为p的自然对数值，模型转化为线性模型pk=polyfit(x,lnp,1);%线
3、性拟合，得到模型的斜率pk(l)和常数pk(2)模型为:p=exp(pk(1)*x)*exp(pk(2))xi二0:50:2000;pO二1.0332*exp(-(xi+500)/7756);%理论值pl=exp(pk(1)*xi+pk⑵)；％拟合模型值p2=interpl(x,p,xi,'spline，);%三次样条插值plot(x,p,'p',xi,pO,xi,pl,，一',xi,p2,，_.，);1egendC测量值','理论值','拟合值','三次样条值')；formatlong%数据显示格式为15位有效数字x2二0:200:2000%取10个点，比较差异ppi二1
4、.0332*exp(-(x2+500)/7756)%理论值pp2=exp(pk(1)*x2+pk(2))%拟合值pp3=interpl(x,p,x2,'spline，)%样条插值errl=sum(abs(pp2-ppl)."2)%拟合值的误差绝对值总和err2=sum(abs(pp3-ppl).^2)%样条值的误差绝对值总和结果:0.950.90.850.80.75测最值理论值拟合值三次样条值q7•0200400600800100012001400160018002000从图像上，都符合得很好，但很难看出差异。取11个点，看其差异X020040060080010001200
5、14001600180020000.96860.94400.92000.89650.87370.85150.82980.80870.78810.76800.7485950355351056029400825578583849152437383429132678259707627615102985109224798164357552963963103336067154853280517砧907757760.96880.94420.92020.89690.87410.85190.83020.80920.78860.76860.74917796557051468804091467
6、133493583375749644200866875648022627380479227726757362402512695861645926299345037377189271611187310样0.96890.94420.92020.89690.87410.85190.83020.80920.78870.76860.7491条0000007335008220540000000564970000008278872739430060566921120000004白0000073237823190000017514000000056550283497189943500000
7、从数据表可看出，拟合法和样条法，均有某些点与理论值更接近，为比较误差的差异，取它们与理论值差值的绝对值之和:拟合法：errl=0.004472633610315样条法：err2=0.004501295874100可看出，拟合法更好。4.梯形法：clear;x=-3:0.1:3;y=exp(-x.*x/2)/2/pi;t=trapz(x,y)%梯形法结果：t=0.397856386182920辛普森法：clear;f=exp(-x.*x/2)/2/pi,;q二quad(f,-3,3)%辛普森法结果：q=0.39786
