Algebra: Expressibility of f(x) with 10 Numbers

[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?

 

[matt grime]

There are an infinite number of answers to your homework.

 

[sid_galt]

It's not homework

There are an infinite number of answers to your homework.

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?

 

[fourier jr]

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

 

[matt grime]

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.

 

[HallsofIvy]

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.

 

[Data]

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~