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Yapay Zeka Hızlı Bakış & Çıkarımlar
AI Insight⚡ TL;DR(1 Cümlede Hızlı Çıkarım)
Next, we prove that if $f$ is a positive harmonic function on the tree and $x$, $y$ are two vertices with $d(x,y)=2$, then $f(x)/(qr)\leq f(y)\leq qr\cdot f(x)$.
📌 3 Maddede Bu Makale
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Amaç & Hipotez:On a biregular tree of degrees $q+1$ and $r+1$, we study the growth of two classes of harmonic functions.
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Temel Bulgular & Veri:First, we prove that if $f$ is a bounded harmonic function on the tree and $x$, $y$ are two adjacent vertices, then $|f(x)-f(y)|\leq 2 (qr-1)\|f\|_\infty/((q+1)(r+1))$, thus generalizing a result of Cohen and Colonna for regular trees.
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Sonuç & Katkı:Next, we prove that if $f$ is a positive harmonic function on the tree and $x$, $y$ are two vertices with $d(x,y)=2$, then $f(x)/(qr)\leq f(y)\leq qr\cdot f(x)$.
🔬Tür:Araştırma Makalesi
📍Kapsam:Genişletilmiş Akademik Örneklem
⚙️Değişkenler:Biregular tree · Growth · Harmonic function
Abstract
On a biregular tree of degrees $q+1$ and $r+1$, we study the growth of two classes of harmonic functions. First, we prove that if $f$ is a bounded harmonic function on the tree and $x$, $y$ are two adjacent vertices, then $|f(x)-f(y)|\leq 2 (qr-1)\|f\|_\infty/((q+1)(r+1))$, thus generalizing a result of Cohen and Colonna for regular trees. Next, we prove that if $f$ is a positive harmonic function on the tree and $x$, $y$ are two vertices with $d(x,y)=2$, then $f(x)/(qr)\leq f(y)\leq qr\cdot f(x)$.
Anahtar kelimeler
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Bu Makaleye Atıf Yap
VİELİ, F. J. G. (2022). Growth of harmonic functions on biregular trees. Journal of Algebra Combinatorics Discrete Structures and Applications, Cilt: 9 - Sayı: 2, 1–8. https://ajindex.com/articles/861484