Proving linear independence of polynomials asymptotically

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Proving linear independence of polynomials asymptotically

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[QUICK DESCRIPTION]
If you need to prove that a family of polynomials are linearly independent, examine their asymptotic behaviour as the variables tend to $\infty$.

[PREREQUISITES]

High-school algebra

[note article attention]This article needs to be properly classified.[/note]

[GENERAL DISCUSSION]

[note section contributions wanted]This article needs more (and better) examples. Is there a traceable origin to this trick? Is it an example of a more general trick?[/note]

If you need to prove linear independence of a big family of frightening-looking polynomials over $\Q$ or over $\R$, replace your variables $x_1,\ldots,x_n$ with polynomials in $t^{a_1},\ldots,t^{a_n}$ with $a_1>a_2>\cdots>a_n>0$ real numbers, with $a_{i}-a_{i+1}< a_{i+1}-a_{i+2} < 2(a_{i}-a_{i+1})$ for $i=1,\ldots,n-2$ or some permutation thereof, and examine the leading terms of your polynomials as $t$ tends to $\infty$. If each polynomial in your family behaves differently `at $\infty$', you know that they cannot be linearly dependent--- and only the leading terms need to be considered! Experiment with different substitutions in order to uncover the relevant asymptotic properties.

[EXAMPLE]

This example is from Section 4 of [[arxiv:math/0511602|math/0511602]].

Set

$P_2(y_1,y_2,y_3) = (y_1-y_2)^2 + (y_2-y_3)^2 + (y_3-y_1)^2 ,$ 
$P_3(y_1,y_2,y_3) = (y_1-y_2)(y_2-y_3)(y_3-y_1) ,$ 
$P_4(y_1,y_2,y_3,y_4) = (y_1-y_3)(y_2-y_3)(y_1-y_4)(y_2-y_4),$

where $y_1+y_2+y_3+ y_4= 0$.
We would like to show that the following polynomials are linearly independent over $\Q$<comment thread="125" />

$Q^{n,m,k} = \big( P_2(y_1,y_2,y_3)^n P_3(y_1,y_2,y_3)^{2m+3}
+ P_2(y_4,y_3,y_2)^n P_3(y_4,y_3,y_2)^{2m+3}$
         $ \quad + P_2(y_3,y_4,y_1)^n P_3(y_3,y_4,y_1)^{2m+3}
+ P_2(y_2,y_1,y_4)^n P_3(y_2,y_1,y_4)^{2m+3} \big)$
              $ \times \big(
P_4(y_1,y_2,y_3,y_4)^k + P_4(y_1,y_3,y_2,y_4)^k + P_4(y_1,y_4,y_2,y_3)^k \big)$

with $n+2k+3m=d$ and $k>0$ for some fixed $d\in \N$.

We first make the substitution <comment thread="127" />

$y_1 = (3 t^a - t^b - t^c)/4,$ 
$y_2 = (- t^a +3 t^b - t^c)/4,$ 
$y_3 = (- t^a - t^b +3 t^c)/4,$ 
$y_4 = -(t^a + t^b + t^c)/4,$

where $a>b>c>0$ are real numbers and $a-b< b-c < 2(a-b)$. Although other substitutions were also possible, we chose this substitution so as to keep expressions of the form $y_i-y_j$ (with $i,j=1,2,3,4$) as simple as possible, because the polynomials $P_2$, $P_3$, and $P_4$ which make up $Q^{n,m,k}$ are given in terms of such expressions. Thus

$y_1-y_4 = t^a,$ 
$y_2-y_4 = t^b,$ 
$y_3-y_4 = t^c$.

A routine calculation gives the leading term of $Q^{n,m,k}$ as 
$2^n(2m+3)t^{2(n+2m+k+3)a+2(m+k+1)b+c}$. This implies that the only possible linear relations between the $Q^{n,m,k}$'s are between those having the same value of $(n+2m+k,m+k)$.

Repeat the same trick again, this time making the substitution

$y_1 = (2 t^a - t^c)/4 + t^b/2 ,$ 
$y_2 = (2 t^a - t^c)/4 - t^b/2 ,$ 
$y_3 = (-2 t^a + 3 t^c)/4 ,$ 
$y_4 = -(2 t^a + t^c)/4 ,$

where this time $a>b>c>0$ are real numbers satisfying $b-c< a-b < 2(b-c)$. Again, we chose this substitution to keep the differences $y_i-y_j$ as simple as possible, while being different from the first substitution. In this case

$y_1-y_4 = t^a + t^b/2,$ 
$y_2-y_4 = t^a - t^b/2,$ 
$y_3-y_4 = t^c,$ 
$y_1-y_2 = t^b. $

This time the leading term of $Q^{n,m,k}$ in terms of these new variables is 
$2^{n+1}(n+2m+3)t^{(2(n+2m+2k)+5)a+(2m+3)b+c}$. This implies that the only possible linear relations between the $Q^{n,m,k}$'s are between those having the same value of $(n+2m+2k,m)$. Combining both results proves linear independence of the $Q^{n,m,k}$'s.

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Proving linear independence of polynomials asymptotically

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