- #1
ihggin
- 14
- 0
Here is a potentially neat problem. Let [tex]x(t),y(t)[/tex] (for all [tex]t\in \mathbb{R}[/tex]) be polynomials in [tex]t[/tex]. Prove that for any [tex]x(t),y(t)[/tex] there exists a non-zero polynomial [tex]f(x,y)[/tex] in 2 variables such that [tex]f(x(t),y(t))=0[/tex] for all [tex]t[/tex]. The strategy is to show that for [tex]n[/tex] sufficiently large, the polynomials [tex]x(t)^{i}y(t)^{j}[/tex] with [tex]0\leq i,j \leq n[/tex] are linearly dependent.
For example, suppose we are given [tex]x(t)=t[/tex] and [tex]y(t)=t^2 + 1[/tex]. Then the polynomial [tex]f(x,y)=1-y+x^2[/tex] would be a non-zero polynomial such that [tex]f(x(t),y(t)) = 1 - (t^2 +1) + t^2 = 0[/tex] for all [tex]t \in \mathbb{R}[/tex].
My attempt: say [tex]x(t)[/tex] is of degree [tex]a[/tex] and [tex]y(t)[/tex] is of degree [tex]b[/tex]. Then we can take [tex]n=ab[/tex], as we will then have two terms: [tex]c_{0a}y^{a}[/tex] and [tex]c_{b0}x^{b}[/tex] with the same highest degree of [tex]t[/tex]: [tex]ab[/tex]. I then tried to prove that the number of ordered pairs [tex](i,j)[/tex], such that [tex]ia+jb \leq ab[/tex], is greater than [tex]ab[/tex], so that we would have at least as many variables [tex]c_{ij}[/tex] as we have equations ([tex]ab[/tex]) to solve, so that we can always find a solution to [tex]f(x(t),y(t))=\sum_{i,j} c_{ij} x(t)^i y(t)^j[/tex] being zero. However, I played around with trying to prove this inequality and I don't think it's true.
Does anyone have any ideas on how to solve this problem?
For example, suppose we are given [tex]x(t)=t[/tex] and [tex]y(t)=t^2 + 1[/tex]. Then the polynomial [tex]f(x,y)=1-y+x^2[/tex] would be a non-zero polynomial such that [tex]f(x(t),y(t)) = 1 - (t^2 +1) + t^2 = 0[/tex] for all [tex]t \in \mathbb{R}[/tex].
My attempt: say [tex]x(t)[/tex] is of degree [tex]a[/tex] and [tex]y(t)[/tex] is of degree [tex]b[/tex]. Then we can take [tex]n=ab[/tex], as we will then have two terms: [tex]c_{0a}y^{a}[/tex] and [tex]c_{b0}x^{b}[/tex] with the same highest degree of [tex]t[/tex]: [tex]ab[/tex]. I then tried to prove that the number of ordered pairs [tex](i,j)[/tex], such that [tex]ia+jb \leq ab[/tex], is greater than [tex]ab[/tex], so that we would have at least as many variables [tex]c_{ij}[/tex] as we have equations ([tex]ab[/tex]) to solve, so that we can always find a solution to [tex]f(x(t),y(t))=\sum_{i,j} c_{ij} x(t)^i y(t)^j[/tex] being zero. However, I played around with trying to prove this inequality and I don't think it's true.
Does anyone have any ideas on how to solve this problem?