# Question in algebra

1. Apr 22, 2005

### 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?

2. Apr 22, 2005

### matt grime

There are an infinite number of answers to your homework.

3. Apr 22, 2005

### sid_galt

It's not 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?

Last edited: Apr 22, 2005
4. Apr 22, 2005

### fourier jr

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

5. Apr 23, 2005

### 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.

6. Apr 23, 2005

### 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.

7. Apr 23, 2005

### 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~