Say I have a collection of points, for example the following:
(1, 167), (2, 11), (3, 255), etc
Is it possible to construct an equation that satisfies all of them? I have a maximum of 32 points.
Say I have a collection of points, for example the following:
(1, 167), (2, 11), (3, 255), etc
Is it possible to construct an equation that satisfies all of them? I have a maximum of 32 points.
Given any $n$ points in the plane, none of which lie on the same vertical line as another, and with at least one lying off the $x$-axis (if they all lie on the $x$-axis, then the $0$ function works), there exists a unique polynomial of degree $n-1$ that passes through all of those points. Also, there are infinitely-many polynomials of any given higher degree passing through those points.
In particular, say the points are $(x_1,y_1),...,(x_n,y_n)$ (where $x_i\neq x_j$ for $i\neq j$). For $1\leq i\leq n$, define the polynomial $P_i(x):=\prod_{j\neq i}\frac{x-x_j}{x_i-x_j}.$ These (and any of their non-0 scalar multiples) are clearly real polynomials of degree $n-1$, and it can be determined rather readily that for any $i,j\in\{1,...,n\}$, we have $P_i(x_j)=\begin{cases}0 & i\neq j\\1 & i=j.\end{cases}$ Consequently, we see that $P(x):=\sum_{j=1}^ny_jP_j(x_j)$ is a polynomial of degree at most $n-1$ such that $P(x_j)=y_j$ for all $1\leq j\leq n$.
As it turns out, the above construction generalizes rather nicely to points in $\Bbb C^2$ (rather than $\Bbb R^2$), though we won't (necessarily) get a real polynomial in this way.
Yes; in fact there are many ways to construct such an equation, depending on the properties you want it to have.
For example, there will be a Lagrange polynomial (of degree at most 31) that passes through all your points; this is likely to be the simplest-looking equation (from an algebraic standpoint) that you can get, but its geometric properties are not so great (it can wiggle up and down a lot, and small changes in your points will potentially lead to large changes in your equation).
If you want something that's geometrically a little better-behaved, you might want to try cubic spline interpolation instead.
I'd prefer that the exponents do not exceed the value of 50 and that the coefficients are integers, but these are not requirements, only preferences.
Here is a (tedious?) way to construct such polynomial. Let $ \{ (x_1, y_1), (x_2, y_2), \ldots, (x_{32}, y_{32}) \} $ denote your data points. Construct the following linear system $Af = y:$ $ \pmatrix{ 1 & x_1 & x_1^2 & x_1^3 & \cdots & x_1^{50} \\ 1 & x_2 & x_2^2 & x_2^3 & \cdots & x_2^{50} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_{32} & x_{32}^2 & x_{32}^3 & \cdots & x_{32}^{50} } \pmatrix{f_0 \\ f_1 \\ \vdots \\ f_{50}} = \pmatrix{y_1 \\ y_2 \\ \vdots \\ y_{32}} $ Both the matrix $A$ and the vector $y$ are built using the input data points. $f$ is the unknown vector. Our goal is to solve for $f,$ and infer the polynomial $f(x) = f_0 + f_1 x + \ldots + f_{50} x^{50}.$ Please note that some $f_i$'s can be zero, including $f_{50}.$ Obviously, you can change that $50$ into something else, or even force some $f_i$'s to be $0$ or $1.$
The system above is an under-determined system. However, it could be solved once we impose more conditions on $f.$
If we want $f$ such that the errors $\{y_i - f(x_i)\}$ are the smallest, in an $\ell_2$ norm fashion, then you can use least-squares. That is, $f = (A^{T}A)^{-1}A^{T}y,$ or using better computational methods cf the Wikipedia page.
Of course, integer (or rational) linear algebra could be used instead of numerical linear algebra, in order to make sure that $f$ is, in fact, an integer vector.