作业帮我在MATLAB中输入了实验数据和一条拟合曲线,如何对比拟合曲线相对于真实数据的精确度?程序如clear;clc;%%%实验数据点x=[12,26,38,120,382,436,1266,11870,76120];y=[0.04419,0.03343,0.02762,0.01936,0.01484,0.01031,0.00
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 12:09:56
![我在MATLAB中输入了实验数据和一条拟合曲线,如何对比拟合曲线相对于真实数据的精确度?程序如clear;clc;%%%实验数据点x=[12,26,38,120,382,436,1266,11870,76120];y=[0.04419,0.03343,0.02762,0.01936,0.01484,0.01031,0.00](/uploads/image/z/11830097-65-7.jpg?t=%E6%88%91%E5%9C%A8MATLAB%E4%B8%AD%E8%BE%93%E5%85%A5%E4%BA%86%E5%AE%9E%E9%AA%8C%E6%95%B0%E6%8D%AE%E5%92%8C%E4%B8%80%E6%9D%A1%E6%8B%9F%E5%90%88%E6%9B%B2%E7%BA%BF%2C%E5%A6%82%E4%BD%95%E5%AF%B9%E6%AF%94%E6%8B%9F%E5%90%88%E6%9B%B2%E7%BA%BF%E7%9B%B8%E5%AF%B9%E4%BA%8E%E7%9C%9F%E5%AE%9E%E6%95%B0%E6%8D%AE%E7%9A%84%E7%B2%BE%E7%A1%AE%E5%BA%A6%3F%E7%A8%8B%E5%BA%8F%E5%A6%82clear%3Bclc%3B%25%25%25%E5%AE%9E%E9%AA%8C%E6%95%B0%E6%8D%AE%E7%82%B9x%3D%5B12%2C26%2C38%2C120%2C382%2C436%2C1266%2C11870%2C76120%5D%3By%3D%5B0.04419%2C0.03343%2C0.02762%2C0.01936%2C0.01484%2C0.01031%2C0.00)
我在MATLAB中输入了实验数据和一条拟合曲线,如何对比拟合曲线相对于真实数据的精确度?程序如clear;clc;%%%实验数据点x=[12,26,38,120,382,436,1266,11870,76120];y=[0.04419,0.03343,0.02762,0.01936,0.01484,0.01031,0.00
我在MATLAB中输入了实验数据和一条拟合曲线,如何对比拟合曲线相对于真实数据的精确度?程序如
clear;
clc;
%%%实验数据点
x=[12,26,38,120,382,436,1266,11870,76120];
y=[0.04419,0.03343,0.02762,0.01936,0.01484,0.01031,0.00822,0.00626,0.00459];
hold on;
plot(x,y,'ro');
set(gca,'xscale','log','yscale','log');
grid on
for n=1:1:80000;
e(n)=0.18*(2*n)^(-0.6)+0.012*(2*n)^(-0.08567);
n=n+1;
end
n=1:80000;
plot(n,e,'linewidth',2);
set(gca,'xscale','log','yscale','log');
grid on;
我在MATLAB中输入了实验数据和一条拟合曲线,如何对比拟合曲线相对于真实数据的精确度?程序如clear;clc;%%%实验数据点x=[12,26,38,120,382,436,1266,11870,76120];y=[0.04419,0.03343,0.02762,0.01936,0.01484,0.01031,0.00
求个相对误差不就行了吗
ex=0.18*(2*x).^(-0.6)+0.012*(2*x).^(-0.08567);
delta=(y-ex)./ex;
plot(x,delta);