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Productly linearly independent sequences - MaRDI portal

Productly linearly independent sequences (Q2834198)

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scientific article; zbMATH DE number 6656726
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Productly linearly independent sequences
scientific article; zbMATH DE number 6656726

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    Productly linearly independent sequences (English)
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    25 November 2016
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    linear independence
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    infinite product
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    The notion in the title is defined as follows. \(K\) sequences \((a_{k,n})_{n=1}^\infty \; (k=1,\ldots,K)\) of positive real numbers are said to be \textit{productly linearly independent}, if the \(K+1\) numbers \(1\) and \(\prod_{n=1}^\infty\big(1+1/(a_{k,n}c_n)\big),\; k=1,\ldots,K\), are \(\mathbb{Q}\)-linearly independent for every sequence \((c_n)_{n=1}^\infty\) of positive integers. In the particular case \(K=1\), a sequence \((a_n)_{n=1}^\infty\) of positive real numbers is said to be \textit{productly irrational}, if \(\prod_{n=1}^\infty\big(1+1/(a_nc_n)\big)\) is irrational for any sequence \((c_n)\) as above. (Note that, in Definition 2, \(\sum_{n=1}^\infty\) has to be replaced by \(\prod_{n=1}^\infty\).) For both notions, very technical criteria are given, too cumbersome to be quoted here.NEWLINENEWLINEMoreover, the authors show the irrationality of \((\ast)\!\!: \prod_{n=1}^\infty(1-J^{-2^n})^{-1}\) for each integer \(J>1\). It should be pointed out that, using Mahler's transcendence method, much more can be proved on the infinite product \(\prod_{n\geq1}\big(1+(1-a)z^{d^n}\big)/(1-az^{d^n})\) defining, for \((a,d)\in\overline{\mathbb{D}}\times\mathbb{Z}_{\geq2}\), a function \(P_{a,d}(z)\) being holomorphic in the open unit disk \(\mathbb{D}\). Namely, if \(a\in\overline{\mathbb{D}}\) is algebraic and \(d\in\mathbb{Z}_{\geq2}\) satisfy \((a,d)\neq(0,2)\), then \(P_{a,d}(\alpha)\) is transcendental for any non-zero algebraic \(\alpha\in\mathbb{D}\), for which \((\ast\ast)\!\!:1+(1-a)\alpha^{d^n}\neq0\) holds for any \(n\in\mathbb{Z}_{\geq1}\). Note that both conditions, \((\ast\ast)\) and \((a,d)\neq(0,2)\), are necessary for transcendence, the second one since \(P_{0,2}(z)=(1-z^2)^{-1}\). Note also that \(P_{1,2}(1/J)\) is just the authors' product \((\ast)\).
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