使用 Matlab polyfit 拟合抛物线答案

【问题标题】：Fitting a parabola using Matlab polyfit使用 Matlab polyfit 拟合抛物线
【发布时间】：2018-02-20 14:20:52
【问题描述】：

我希望将抛物线拟合到以下数据。

x = [-10:2:16];
y = [0.0334,0.0230,0.0145,0.0079,0.0033,0.0009,0.0006,0.0026,0.0067,0.0130,0.0213,0.0317,0.0440,0.0580];

[p,~,~] = polyfit(x,y,2);

x2 = linspace(-10,16,100);
y2 = polyval(p,x2);

y3 = 0.0003.*x2.^2 -0.0006.*x2 + 0.0011;
figure
plot(x,y,'o',x2,y2,x2,y3)

但是，拟合与数据完全不匹配。将数据放入 excel 并在那里使用二阶多项式进行拟合后，我得到了非常好的拟合。 y = 0.0003x2 - 0.0006x + 0.0011（截断系数的 excel 会稍微偏斜拟合）。使用这些数据的 polyfit 发生了什么？

【问题讨论】：

标签： matlab curve-fitting

【解决方案1】：

解决了。

Matlab 检查用户请求的输出数量。由于即使我没有使用它们，我也请求了三个输出，因此 polyfit 会更改系数以映射到不同的域 xhat。

如果我只是这样做：

p = polyfit(x,y,2);
plot(x2,polyval(p,x2));

然后我会得到适当的结果。使用三个输出恢复相同的答案：

[p2,S,mu] = polyfit(x,y,2);
xhat = (x2-mu(1))./mu(2)
y4 = polyval(p2,xhat)
plot(x2,y4)

【讨论】：

是的...如docs中所述

【解决方案2】：

我会在 matlab 中使用最小二乘法解决这个问题：

x = [-10:2:16]';
Y = [0.0334,0.0230,0.0145,0.0079,0.0033,0.0009,0.0006,0.0026,0.0067,0.0130,0.0213,0.0317,0.0440,0.0580]';

plot(x,Y,'.');

A=[ones(length(x),1) x x.^2];
beta=A\Y;
hold on
plot(x, beta(1)+beta(2)*x+beta(3)*x.^2)
leg_est=sprintf('Estimated (y=%.4f+%.4fx+%.4fx^2',beta(1),beta(2),beta(3))
legend('Data',leg_est)

【讨论】：