Given one point $P\equiv (x_0,y_0)$ on a two-dimensional space $\Re^2$, there are infinite functions passing for that point. The question is: given $2$ points $P_0\equiv(x_0,y_0)$ and $P_1\equiv(x_1,y_1)$ is it still correct to say they are infinite? More in general, given $N$ points $P_k\equiv(x_k,y_k)$ with $x_k\neq x_j$, how many functions passing for the $N$ points can be constructed? Thanks.
About the number of functions passing for N points
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0The question seems ill-posed without further restrictions. If your set includes, among $x$-values, all of $\mathbb{R}$ except a single point, then there are uncountably many functions which interpolate your set, each of which map that single unincluded point to a different real number. Must the functions be continuous? Are there any other restrictions to consider? – 2012-07-04
1 Answers
This answer deals with polynomials rather than all functions.
Given any (finite number) of points in the plane, there are always an infinite number of polynomial functions of two variables with zeros at those points. For each point $(a_i,b_i)$, take the polynomial $(x-a_i)(y-b_i)$, then the product of these polynomials will be zero at all the points, as will any higher degree polynomial that it divides, leading to the infinite number.
A more interesting question is to ask how many polynomials of a specified degree have zeros at a collection of points. If the points are in general position, and there are $3d-1$ of them for some $d$, then there are no polynomials of degree less than $d$ zero at all of them, infinitely many of degree more than $d$, and a non-zero finite number of degree $d$.
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0It occurs to me now that you may have meant functions of $1$ variable, with "passing through $(x_0,y_0)$" meaning $y_0=f(x_0)$, in which case the answer would be different (I think this is what Eugene is referring to in the comments). – 2012-07-04