# Algebra: Expressibility of f(x) with 10 Numbers

• sid_galt
In summary, the conversation discusses the possibility of expressing a function of a series of numbers as a polynomial equation, either in one or two variables, and the use of Lagrange's interpolation formula or Newton's divided difference scheme to find it. It is also mentioned that while a finite sequence of points can be represented as a polynomial in two variables, it may not necessarily be expressible as a product of two polynomials.
sid_galt
Let's say you have ten numbers

$$f(1) = 1$$
$$f(2) = 100$$
$$f(3) = 45$$
$$f(4) = 9000$$
$$f(5) = 999$$
$$f(6) = 46$$
$$f(7) = 47$$
$$f(8) = 48$$
$$f(9) = 59$$
$$f(10) = 60$$

Is f(x) expressible in the form

$$f(x)=a_nx^n+a_{n-1}x^{n-1}...a_1x+a_0$$
or perhaps
$$f(x)=(a_nx^n+a_{n-1}x^{n-1}...a_1x+a_0)(b_ny^n+b_{n-1}y^{n-1}...b_1y+b_0)$$
Why? Why not?

If it is, is there any way to find it?

It's not homework

matt grime said:

So can any function of a random series of numbers be expressed as a product of two or more polynomial equations if f is a function of two variables or one polynomial equation if f is a function of one variable?

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i don't really know the answer but i think "lagrange interpolation" might have something to do with this.

Firstly, you shouldn't have an input x into f(x) and an output in two variables.

And of course given a finite number of points x, f(x) there are an infinite number of polynomials through those points.

Given any finite number, n, of points (x, y) there exist an infinite number of functions (and polynomials) whose graphs pass through those points (i.e. y= f(x)).

However, there exist a unique polynomial of degree n+1 (or lower if the points are not "independent") whose graph passes through those points.

As Fourier jr. said, Lagrange's interpolation formula will give that polynomial. Newton's divided difference scheme will also work.

A finite sequence of points (x, y, z) CAN be represented as a polynomial in the two variables (x,y). However, I do not believe that it can necessarily be represented as a polynomial in x TIMES a polynomial in y.

Not quite true. The x's corresponding to distinct y's must be distinct (ie. if $(x_n, y_n), \ (x_m, y_m)$ are some of the points and $x_n = x_m$ then in order to have a set $(x, \ f(x))$ for a polynomial $f(x)$ containing both points you need $y_n=y_m$), then it's fine~