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edited Oct 11, 2012 at 9:30

Markus Schweighofer

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As Vicki already pointed out above, Polya-Szegö (Problems and Theorems in Analysis II. Theory of Functions. Zeros. Polynomials. Determinants. Number Theory. Geometry.) is a good reference for this particular result.

See §6, Problem 49 in that book: By the Fundamental Theorem of Algebra you can restrict to quadratic polynomials. For lines it is trivial. For parabolas opening down it is easy. The only nontrivial case happens for parabolas opening up. Parabolas opening up whose minimum lies above the interval are also just a little exercise. The only case remaining is parabolas opening up without zero on the real line.

Now determine a polynomial $h \in \mathbb Z[K,L,A,B,C]_ 4$ such that for all $a,b,c\in\mathbb R$ and all $\ell\in\mathbb N_{\ge2}$ one has $$aX^2+bX+c=\sum_{k=0}^\ell\frac{(\ell-2)!}{k!(\ell-k)!}h(k,\ell,a,b,c)X^k(1-X)^{\ell-k}.$$

Show that each $f\in\mathbb R[X]_ 2$ with $f>0$ on $\mathbb R$ has the desired representation by comparing the discriminants of $f=aX^2+bX+c\in\mathbb R[X]_ 2$ and $h(K,\ell,a,b,c)\in\mathbb Z[K]_ 2$ for $a,b,c\in\mathbb R$ and large $\ell\in\mathbb N_{\ge2}$.

This approach unfortunately does not carry over for the numerous generalizations of the theorem.

As Vicki already pointed out above, Polya-Szegö (Problems and Theorems in Analysis II. Theory of Functions. Zeros. Polynomials. Determinants. Number Theory. Geometry.) is a good reference for this particular result.

See §6, Problem 49 in that book: By the Fundamental Theorem of Algebra you can restrict to quadratic polynomials. For lines it is trivial. For parabolas opening down it is easy. The only nontrivial case happens for parabolas opening up.

Now determine a polynomial $h \in \mathbb Z[K,L,A,B,C]_ 4$ such that for all $a,b,c\in\mathbb R$ and all $\ell\in\mathbb N_{\ge2}$ one has $$aX^2+bX+c=\sum_{k=0}^\ell\frac{(\ell-2)!}{k!(\ell-k)!}h(k,\ell,a,b,c)X^k(1-X)^{\ell-k}.$$

Show that each $f\in\mathbb R[X]_ 2$ with $f>0$ on $\mathbb R$ has the desired representation by comparing the discriminants of $f=aX^2+bX+c\in\mathbb R[X]_ 2$ and $h(K,\ell,a,b,c)\in\mathbb Z[K]_ 2$ for $a,b,c\in\mathbb R$ and large $\ell\in\mathbb N_{\ge2}$.

This approach unfortunately does not carry over for the numerous generalizations of the theorem.

As Vicki already pointed out above, Polya-Szegö (Problems and Theorems in Analysis II. Theory of Functions. Zeros. Polynomials. Determinants. Number Theory. Geometry.) is a good reference for this particular result.

See §6, Problem 49 in that book: By the Fundamental Theorem of Algebra you can restrict to quadratic polynomials. For lines it is trivial. For parabolas opening down it is easy. The only nontrivial case happens for parabolas opening up. Parabolas opening up whose minimum lies above the interval are also just a little exercise. The only case remaining is parabolas opening up without zero on the real line.

Now determine a polynomial $h \in \mathbb Z[K,L,A,B,C]_ 4$ such that for all $a,b,c\in\mathbb R$ and all $\ell\in\mathbb N_{\ge2}$ one has $$aX^2+bX+c=\sum_{k=0}^\ell\frac{(\ell-2)!}{k!(\ell-k)!}h(k,\ell,a,b,c)X^k(1-X)^{\ell-k}.$$

Show that each $f\in\mathbb R[X]_ 2$ with $f>0$ on $\mathbb R$ has the desired representation by comparing the discriminants of $f=aX^2+bX+c\in\mathbb R[X]_ 2$ and $h(K,\ell,a,b,c)\in\mathbb Z[K]_ 2$ for $a,b,c\in\mathbb R$ and large $\ell\in\mathbb N_{\ge2}$.

This approach unfortunately does not carry over for the numerous generalizations of the theorem.

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created Oct 11, 2012 at 0:39

Markus Schweighofer

806
8
11

As Vicki already pointed out above, Polya-Szegö (Problems and Theorems in Analysis II. Theory of Functions. Zeros. Polynomials. Determinants. Number Theory. Geometry.) is a good reference for this particular result.

See §6, Problem 49 in that book: By the Fundamental Theorem of Algebra you can restrict to quadratic polynomials. For lines it is trivial. For parabolas opening down it is easy. The only nontrivial case happens for parabolas opening up.

Now determine a polynomial $h \in \mathbb Z[K,L,A,B,C]_ 4$ such that for all $a,b,c\in\mathbb R$ and all $\ell\in\mathbb N_{\ge2}$ one has $$aX^2+bX+c=\sum_{k=0}^\ell\frac{(\ell-2)!}{k!(\ell-k)!}h(k,\ell,a,b,c)X^k(1-X)^{\ell-k}.$$

Show that each $f\in\mathbb R[X]_ 2$ with $f>0$ on $\mathbb R$ has the desired representation by comparing the discriminants of $f=aX^2+bX+c\in\mathbb R[X]_ 2$ and $h(K,\ell,a,b,c)\in\mathbb Z[K]_ 2$ for $a,b,c\in\mathbb R$ and large $\ell\in\mathbb N_{\ge2}$.

This approach unfortunately does not carry over for the numerous generalizations of the theorem.