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Unable to find the nonlinear least squares |
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| Apr4-09, 12:53 AM | #1 |
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Unable to find the nonlinear least squares
1. The problem statement, all variables and given/known data
We have the following x, y values x ||| y 1.0 -0.15 1.5 0.24 2.0 0.68 2.5 1.04 3.0 1.21 3.5 1.15 4.0 0.86 4.5 0.41 5.0 -0.08 How can you find the equation [tex]y(x) = ax^2 + bx + c[/tex] by least squares? 3. The attempt at a solution I know how to calculate the equation for a line by solving Ax = b taking transposes of A at the both sides [tex]A^TAx = A^Tb[/tex] and then solving for x. My second attempt I made a 9 x 3 matrix for A where the first two columns are ones, 3 x 1 for x and 9 x 1 for b. However, I get a singular matrix for [tex]A^TA.[/tex] Apparently, my method is not right. I could make 3 equations such as y(0), y(1) and y(2) and solve for a, b and c. However, I see that the method is not least squares and also rather inaccurate, since not all points are considered. |
| Apr4-09, 05:54 AM | #2 |
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To find [itex]y= ax^2+ bx+ c[/itex] that gives the best fit, the equation you are trying to solve is AX= Y:
[tex]\begin{bmatrix}x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1\\\cdot & \cdot & \cdot \\\cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ x_n^2 & x_n & 1\end{bmatrix}\begin{bmatrix} a \\ b \\ c\end{bmatrix}\begin{bmatrix}y_1 \\ y_2 \\\cdot \\\cdot\\\cdot \\ y_n\end{bmatrix}[/tex] Multiplying by the transpose of A on both sides gives an equation with a 3 by 3 matrix you can solve: [tex]\begin{bmatrix} \sum x_i^4 & \sum x_i^3 & \sum x_i^2 \\ \sum x_i^3 & \sum x_i^2 & \sum x_i \\ \sum x_i^2 & \sum x_i & n\end{bmatrix}\begin{bmatrix}a \\ b \\ c\end{bmatrix}= \begin{bmatrix} \sum x_i^2y_i \\ \sum x_iy_i \\ \sum y_i \end{bmatrix}[/tex] |
| Apr7-09, 08:58 AM | #3 |
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Is it wrong to write the columns as A3, A2, A1? I have always set the column with the lowest degree to be the first column, and so on. |
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