A real sequence $\mathbf{x}=(x_k)_{k\in\mathbb{N}_0}$ is of the form $$ x_k=\alpha r_k,\quad \alpha\in\mathbb{R}\backslash\mathbb{Q},\quad r_k\in\mathbb{Q},\tag{*} $$ if and only if all of its terms are irrational and the successive ratios $x_{k+1}/x_k$ are rational numbers. In this case, both $\alpha$ and the $r_k$ are uniquely determined up to a common rational factor.

The set $\left\{\alpha r: r\in\mathbb{Q}\right\}$ can be interpreted as a $\mathbb{Q}$-vector space with basis $\alpha$. Any element of the sequence $\mathbf{x}$ of the form (*) is a basis for this vector space as well. In particular, if we define $\tilde\alpha=x_0$ and $\tilde r_k=x_k/x_0\in\mathbb{Q}$, then $x_k=\tilde\alpha\tilde r_k$ for $k=0,1,\ldots$.

Consider now a real, irrational sequence $\mathbf{y}=(y_k)_{k\in\mathbb{N}_0}$ of the form $$ y_k=\alpha r_k+\beta s_k,\quad \alpha, \beta\in\mathbb{R}\backslash\mathbb{Q},\quad r_k, s_k\in\mathbb{Q},\tag{**} $$ where $\alpha$ and $\beta$ are linearly independent over $\mathbb{Q}$. This assumption implies that, again up to rational factors, the numbers $\alpha, \beta, r_k, s_k$ are uniquely determined by the $y_k$.

Question: Given the numerical values of a sequence $\mathbf{y}$ of the form (**), what is a practical way of computing irrational numbers $\tilde\alpha, \tilde\beta$ and rational numbers $\tilde r_k, \tilde s_k$ such that $y_k=\tilde a\tilde r_k+\tilde \beta\tilde s_k$ for $k=0,1,\ldots$?


1 Answer 1


Setting $\tilde\alpha=y_0$, $\tilde \beta=y_1$, and using an integer relations algorithm such as Mathematica's FindIntegerNullVector to find integers $n_k,m_k,p_k$ such that $$ n_k y_k = m_k y_0 + p_k y_1,\quad k=2,3,\ldots $$ solves the problem.


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