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So I am working on this problem with fitting a second degree polynomial of the form $y=a_1x^2+a_2x+a_3$ to four points using least squares. One of the parts of the problem is to write out the matrix equation that describes the least squares problem. Basically we have the equation

$$\begin{bmatrix} x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1 \\ x_3^2 & x_3 & 1 \\ x_4^2 & x_4 & 1 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{bmatrix}$$

Let's call the matrix $A$ so that the problem is $A\mathbf{a}=\mathbf{y}$. Then the least squares equation is $A^TA\mathbf{a}=A^T\mathbf{y}$. If we write that out explicitly we get

$$\begin{bmatrix} \sum_{i=1}^4x_i^4 & \sum_{i=1}^4x_i^3 & \sum_{i=1}^4x_i^2 \\ \sum_{i=1}^4x_i^3 & \sum_{i=1}^4x_i^2 & \sum_{i=1}^4x_i \\ \sum_{i=1}^4x_i^2 & \sum_{i=1}^4x_i & 4 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^4x_i^2y_i \\ \sum_{i=1}^4x_iy_i \\ \sum_{i=1}^4y_i \end{bmatrix} $$

This new matrix makes no intuitive sense to me whatsoever but is striking. Why do we care about, for instance, the sum of the 4th powers of $x_i$? Is there a way to interpret why this is the way it is, maybe in terms of calculus or something else?

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There's an error in the definition of your $A$ matrix. It should be: $$A=\begin{bmatrix} x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1 \\ x_3^2 & x_3 & 1 \\ x_4^2 & x_4 & 1 \end{bmatrix}$$ – Dolma Apr 22 '13 at 0:11
I'd be surprised if there was a simple, intuitive explanation of the form, though I may well be wrong. It's sort of tied up with the moments (see for example) of the coordinates, I suppose. – Sharkos Apr 22 '13 at 0:30
@crf, Once you reduce the problem to $A\mathbf{x}=\mathbf{y}$, you are breaking with the knowledge that the elements of $A$ come from powers of $x$. The expression of the normal equations basically projects the entire problem onto the column space of $A$. As long as the columns of $A$ are linearly independent, it does not matter where they came from. You are translating the problem from one domain (i.e. curve fitting) to another (linear algebra) to simplify the solution. You should not insist that notions from one domain maintain their meaning in the other. – Tpofofn Apr 22 '13 at 0:47
@Dolma thanks I fixed that – crf Apr 22 '13 at 0:48
I don't know what does the matrix means. However, if $x_1,\ldots,x_n$ are all the roots of a real polynomial, then this symmetric matrix defines a quadratic form called the Hermite form. The signature of this matrix is the number of real roots and its rank is the number of distinct complex roots. – achille hui Apr 22 '13 at 0:51

The normal equations for the least squares problem $Ax = b$ is given by $A^T Ax = A^T b$, as you clearly already know. It's not really the fourth powers of $x_i$ that are significant, but rather the matrix $A^T A$ that is significant, from a purely linear algebra point of view. One way to look at the matrix $A^T A$ is that it reduces the overdetermined problem $Ax = b$ (notice that you have three variables but four equations) down to an invertible problem, by taking each output $Ax$ and then dotting it by the columns of $A$, giving you precisely $A^T Ax$. Ultimately the least squares problem can be phrased as:

Given $Ax = b$, find $\hat{b}$ in the column space of $A$ closest to $b$, and then find $\hat{x}$ such that $A\hat{x} = \hat{b}$. We call $\hat{x}$ a least squares solution.

There is, however, a calculus way of looking at it. From an optimization point of view, we can say that $\hat{x}$ is a least squares solution if $\|Ax - b\|_2^2 = (Ax - b)^T (Ax - b)$ is minimized. We can calculate when this is minimized by taking the derivative in $x$. Let $f(x) = \|Ax - b\|_2^2$. Then \begin{align} Df_x(y) & = \lim_{h \rightarrow 0} \frac{1}{h} (f(x+hy) - f(x)) \\ & = \lim_{h \rightarrow 0} \frac{1}{h} \left( \|Ax - b\|_2^2 + h(Ay)^T (Ax - b) + h(Ax - b)^T Ay - \|Ax - b\|_2^2 \right) \\ & = (Ay)^T (Ax - b) + (Ax - b)^T Ay \\ & = y^T (A^T Ax - A^T b) + (A^T Ax - A^T b)^T y \end{align} We set this derivative to zero to solve for the minimum; this can only be zero for every choice of $y$ if $A^T Ax - A^T b = 0$. Thus we have derived the normal equations.

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I agree that the geometrical minimization gives an understanding of the form of the normal equations, but I interpret the question as Why does $A^T A$ look the way it does? and in particular Why do we get moments of the $x_i$s? – Sharkos Apr 22 '13 at 0:37

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