作业帮那位高手能用matlab帮我求一下下面两列数字绘出的曲线的拟合方程啊?不胜感激!数字按行左右对应.0.97720.0323622.24110.1019013.69880.2288694.66520.3393025.62590.4659397.05370.6784687.52560.7573589.38861.08039511.6521.5
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那位高手能用matlab帮我求一下下面两列数字绘出的曲线的拟合方程啊?不胜感激!数字按行左右对应.0.97720.0323622.24110.1019013.69880.2288694.66520.3393025.62590.4659397.05370.6784687.52560.7573589.38861.08039511.6521.5
那位高手能用matlab帮我求一下下面两列数字绘出的曲线的拟合方程啊?不胜感激!数字按行左右对应.
0.97720.0323622.24110.1019013.69880.2288694.66520.3393025.62590.4659397.05370.6784687.52560.7573589.38861.08039511.6521.544076
0.9772，0.032362\2.2411，0.101901\3.6988，0.228869\4.6652，0.339302\5.6259，0.465939\7.0537，0.678468\7.5256，0.757358\9.3886，1.080395\11.652，1.544076
每两个数字一组，谢谢！你好，是不是把下面的数据公式输入METLAB的命令行里？
那位高手能用matlab帮我求一下下面两列数字绘出的曲线的拟合方程啊?不胜感激!数字按行左右对应.0.97720.0323622.24110.1019013.69880.2288694.66520.3393025.62590.4659397.05370.6784687.52560.7573589.38861.08039511.6521.5
x = [0.9772 2.2411 3.6988 4.6652 5.6259 7.0537 7.5256 9.3886 11.652];
y = [0.032362 0.101901 0.228869 0.339302 0.465939 0.678468 0.757358 1.080395 1.544076];
p = polyfit(x,y,3);
xx = min(x):.1:max(x);
yy = polyval(p,xx);
plot(x,y,'bo'); hold on;
plot(xx,yy,'r-'); hold off;
grid on;
legend('原始数据点','拟合曲线','Location','NorthWest');
>>X = [0.9772 2.2411 3.6988 4.6652 5.6259 7.0537 7.5256 9.3886 11.652];
>>Y = [0.032362 0.101901 0.228869 0.339302 0.465939 0.678468 0.757358 1.080395 1.544076];
>>cftool
Linear model Poly...
全部展开
>>X = [0.9772 2.2411 3.6988 4.6652 5.6259 7.0537 7.5256 9.3886 11.652];
>>Y = [0.032362 0.101901 0.228869 0.339302 0.465939 0.678468 0.757358 1.080395 1.544076];
>>cftool
Linear model Poly2:
f(x) = p1*x^2 + p2*x + p3
where x is normalized by mean 5.87 and std 3.415
Coefficients (with 95% confidence bounds):
p1 = 0.09164 (0.08291, 0.1004)
p2 = 0.4656 (0.4569, 0.4743)
p3 = 0.4995 (0.4883, 0.5107)
Goodness of fit:
SSE: 0.0005821
R-square: 0.9997
Adjusted R-square: 0.9996
RMSE: 0.00985
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