"Quasipolynomial"의 두 판 사이의 차이

수학노트
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imported>Pythagoras0
imported>Pythagoras0
1번째 줄: 1번째 줄:
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==example==
 
* assume $a_n = \left((-1)^n+1\right)+\left((-1)^n+3\right) n$
 
* assume $a_n = \left((-1)^n+1\right)+\left((-1)^n+3\right) n$
 
* then
 
* then
5번째 줄: 6번째 줄:
 
$$
 
$$
  
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==some results==
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;thm
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If $P$ is a rational convex $d$-polytope, then $L_{P}(t)$ is a quasipolynomial in $t$ of degree $d$. Its period divides the least common multiple of the denominator of the coordinates of the vertices of $P$.
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;lemma
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If $\sum_{t \ge 0} f(t)z^t = \frac{g(z)}{h(z)}$, then $f$ is a quasipolynomial of degree $d$ with period $p$ if and only if $g$ and $h$ are polynomials such that $\deg(g)<\deg(h)$, all roots of $h$ are $p$-th roots of unity of multiplicity at most $d+1$, and there is a root of multiplicity equal to $d+1$ (all of this assuming that $g/h$ has been reduced to lowest terms.
  
 
;prop
 
;prop
 
Let $f$ be a quasipolynomial of degree $d$. If $f(n+1)\geq f(n)$ for all $n\in \mathbb{N}$, then the top degree coefficient of $f$ must be constant.
 
Let $f$ be a quasipolynomial of degree $d$. If $f(n+1)\geq f(n)$ for all $n\in \mathbb{N}$, then the top degree coefficient of $f$ must be constant.

2016년 8월 28일 (일) 20:04 판

example

  • assume $a_n = \left((-1)^n+1\right)+\left((-1)^n+3\right) n$
  • then

$$ \sum_{n=0}^{\infty}a_nt^n = \frac{2 \left(t^3+3 t^2+t+1\right)}{(1-t)^2 (t+1)^2} $$


some results

thm

If $P$ is a rational convex $d$-polytope, then $L_{P}(t)$ is a quasipolynomial in $t$ of degree $d$. Its period divides the least common multiple of the denominator of the coordinates of the vertices of $P$.

lemma

If $\sum_{t \ge 0} f(t)z^t = \frac{g(z)}{h(z)}$, then $f$ is a quasipolynomial of degree $d$ with period $p$ if and only if $g$ and $h$ are polynomials such that $\deg(g)<\deg(h)$, all roots of $h$ are $p$-th roots of unity of multiplicity at most $d+1$, and there is a root of multiplicity equal to $d+1$ (all of this assuming that $g/h$ has been reduced to lowest terms.

prop

Let $f$ be a quasipolynomial of degree $d$. If $f(n+1)\geq f(n)$ for all $n\in \mathbb{N}$, then the top degree coefficient of $f$ must be constant.