The Alternating Compositions of Weighted Differential Operators Yield The Weights' Wronskian With Which Constant?
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arXiv
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| Natura: | Preprint |
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2026
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| _version_ | 1866911672230215680 |
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| author | Shah, Kian C. Kiselev, Arthemy V. |
| author_facet | Shah, Kian C. Kiselev, Arthemy V. |
| contents | The alternated composition of $N=2p$ differential operators $ w_j(x)\,\partial_x^p$ of strict order $p$ on the line $\mathbb{R}\ni x$ is again a differential operator of strict order $p$; its coefficient is the constant $\mathrm{const}(p)$, depending only on the arity $N$, times the Wronskian determinant of the originally taken coefficients $w_1$, $\ldots$, $w_N$. The case $p=1$ of the Lie bracket for two vector fields fixes $\mathrm{const}(1)=1$. When $p=2$, finding $\mathrm{const}(2)=2$ is easy; we obtain $\mathrm{const}(3)=90$. The problem is to know $\mathrm{const}(p\geqslant 4)$. We express the formula of $\mathrm{const}(p)$ in terms of the sum with signs over the much smaller set of 'late-growing' permutations, thus reaching the exact values $c(p=4)= 586\,656$, $c(p=5)\approx 1.9\cdot 10^{12}$, and $c(p=6)\approx 7.9\cdot 10^{21}$; the positive integer sequence $\mathrm{const}(p)$ seems to be new. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2605_11137 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | The Alternating Compositions of Weighted Differential Operators Yield The Weights' Wronskian With Which Constant? Shah, Kian C. Kiselev, Arthemy V. Combinatorics 15A15, 05E18, 05A10 (Primary) 05E16, 05A16 (Secondary) The alternated composition of $N=2p$ differential operators $ w_j(x)\,\partial_x^p$ of strict order $p$ on the line $\mathbb{R}\ni x$ is again a differential operator of strict order $p$; its coefficient is the constant $\mathrm{const}(p)$, depending only on the arity $N$, times the Wronskian determinant of the originally taken coefficients $w_1$, $\ldots$, $w_N$. The case $p=1$ of the Lie bracket for two vector fields fixes $\mathrm{const}(1)=1$. When $p=2$, finding $\mathrm{const}(2)=2$ is easy; we obtain $\mathrm{const}(3)=90$. The problem is to know $\mathrm{const}(p\geqslant 4)$. We express the formula of $\mathrm{const}(p)$ in terms of the sum with signs over the much smaller set of 'late-growing' permutations, thus reaching the exact values $c(p=4)= 586\,656$, $c(p=5)\approx 1.9\cdot 10^{12}$, and $c(p=6)\approx 7.9\cdot 10^{21}$; the positive integer sequence $\mathrm{const}(p)$ seems to be new. |
| title | The Alternating Compositions of Weighted Differential Operators Yield The Weights' Wronskian With Which Constant? |
| topic | Combinatorics 15A15, 05E18, 05A10 (Primary) 05E16, 05A16 (Secondary) |
| url | https://arxiv.org/abs/2605.11137 |