# mixedmath

Explorations in math and programming
David Lowry-Duda

FtYou writes

Hello everyone ! There is a concept I have a hard time getting my head wrap around. If you have a Vector Space V and a subspace W, I understand that you can find the least square vector approximation from any vector in V to a vector in W. And this correspond to the projection of V to the subspace W. Now , for data fitting ... Let's suppose you have a bunch of points (xi, yi) where you want to fit a set a regressors so you can approximate yi by a linear combination of the regressors lets say ( 1, x, x2 ... ). What Vector space are we talking about ? If we consider the Vector space of function R -> R, in what subspace are we trying to map these vectors ? I have a hard time merging these two concepts of projecting to a vector space and fitting the data. In the latter case what vector are we using ? The functions ? If so I understand the choice of regressors ( which constitute a basis for the vector space ) But what's the role of the (xi,yi) ? I want to point out that I understand completely how to build the matrices to get Y = AX and solving using least square approx. What I miss is the big picture. The linear algebra picture. Thanks for any help !

We'll go over this by closely examining and understanding an example. Suppose we have the data points ${(x_i, y_i)}$

$\displaystyle \begin{cases} (x_1, y_1) = (-1,8) \\ (x_2, y_2) = (0,8) \\ (x_3, y_3) = (1,4) \\ (x_4, y_4) = (2,16) \end{cases},$

and we have decided to try to find the best fitting quadratic function. What do we mean by best-fitting? We mean that we want the one that approximates these data points the best. What exactly does that mean? We'll see that before the end of this note - but in linear algebra terms, we are projecting on to some sort of vector space - we claim that projection is the ''best-fit'' possible.

So what do we do? A generic quadratic function is ${f(t) = a + bt + ct^2}$. Intuitively, we apply what we know. Then the points above become

$\displaystyle \begin{cases} f(-1) = a - b + c = 8 \\ f(0) = a = 8 \\ f(1) = a + b + c = 4 \\ f(2) = a + 2b + 4c = 16 \end{cases},$

and we want to find the best ${[a b c]}$ we can that ''solves'' this. Of course, this is a matrix equation:

$\displaystyle \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 8 \\ 8 \\ 4 \\ 16 \end{pmatrix}.$

And so you see how the algorithm would complete this. But now let's get down the ''linear algebra picture,'' as you say.

We know that quadratic polynomials ${f(t) = a + bt + ct^2}$ are a three dimensional vector space (which I denote by ${P_2)}$ spanned by ${1, t, t^2}$. We know we have four data points, so we will define a linear transformation ${A}$ to be the transformation taking a quadratic polynomial ${f}$ to ${\mathbb{R}^4}$ by evaluating ${f}$ at ${-1, 0, 1, 2}$ (i.e. the ${x_i}$). In other words,

$\displaystyle A : P_2 \longrightarrow \mathbb{R}^4$

where

$\displaystyle A(f) = \begin{pmatrix} f(-1) \\ f(0) \\ f(1) \\ f(2) \end{pmatrix}.$

We interpret ${f}$ as being given by three coordinates, ${a, b, c \in \mathbb{R}^3}$, so we can think of ${A}$ as a linear transformation from ${\mathbb{R}^3 \longrightarrow \mathbb{R}^4}$. In fact, ${A}$ is nothing more than the matrix we wrote above.

Then a solution to

$\displaystyle A^T A \begin{pmatrix} a \\ b \\ c \end{pmatrix} = A^T \begin{pmatrix} 8 \\ 8 \\ 4 \\ 16 \end{pmatrix}$

is the projection of the space of quadratic polynomials on ${\mathbb{R}^4}$ (which in this case is the space of evaluations of quadratic polynomials at four different points). If ${f^* }$ is the found projection, and I denote the ${y_i}$ coordinate vector as ${y^ *}$, then this projection minimizes

$\displaystyle || y^* - Af^*||^2 = (y_1 - f^*(x_1))^2 + \ldots + (y_4 - f^*(x_4))^2,$

and it is in this sense that we mean we have the ''best-fit.'' (This is roughly interpreted as the distances between the ${y_i}$ and ${f^*(x_i)}$ are minimized; really, it's the sum of the squares of the distances - hence ''Least-Squares'').

So in short: ${A}$ is a matrix evaluating quadratic polynomials at different points. The columns vectors correspond to a basis for the space of quadratic polynomials, ${1, t, t^2}$. The codomain is ${\mathbb{R}^4}$, coming from the evaluation of the input polynomial at the four different ${x_i}$. The projection of the set of quadratic polynomials onto their evaluation space minimizes the sum of the squares of the distances between ${f(x_i)}$ and ${y_i}$.

Does that make sense?

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