四元四次多项式非负性证书的高效求和平方方法
An efficient sum of squares nonnegativity certificate for quaternary quartic
AI总结：
研究四元四次非负多项式通过二次形式分解为平方和，利用Hilbert定理并探讨单个二次形式是否足够分解的问题，发现非负多项式分解可能需更优方法。
摘要：
对于任何四元四次非负多项式f≥0，存在二次形式q和q'，使得qq'f为四次平方和。通过将f表示为f=au²+2bu+c，并利用Hilbert(1893)定理，指出对于三元六次非负多项式h≥0，存在二次形式q''使得q''h为四次平方和。为确定是否仅需一个q即可使qf为平方和，给出非平方和的f=au²+2bu+c≥0的例子，但所有例子中af均为平方和。尽管单个q是否足够仍待验证，但描述了q的存在与Δ的特定平方和分解的关系。
英文摘要：
For any 4-variate quartic form $f\geq 0$ (i.e. $f$ nonnegative, homogeneous polynomial of degree $4$ with real coefficients) there exist quadratic forms $q$ and $q'$ so that $qq'f$ is a sum of squares (s.o.s.) of quartics, by reducing to the case of $f=au^2+2bu+c$ with $a$, $b$, $c$ $3$-variate forms of degrees 2, 3, 4, respectively, and invoking on its discriminant $Δ=ac-b^2$ a theorem by Hilbert (1893) asserting that for any ternary sextic $h\geq 0$ there exists a quadric $q''$ so that $q''h$ is s.o.s. of quartics. Towards deciding whether just one $q$ always suffices to make $qf$ a s.o.s, we give explicit examples of non-s.o.s. $f=au^2+2bu+c\geq 0$ with non-s.o.s. $Δ$. However, in all these examples $af$ are s.o.s. That is, the straightforward s.o.s. decomposition via Hilbert (1893) need not be the best possible. While it remains open whether one $q$ always suffices (and we conjecture that $q=a$ suffices), we describe how the existence of such $q$ is related to particular types of s.o.s. decompositions for $Δ$.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
