周期6 线性递推$a_n=a_{n-1}-a_{n-2}$

hbghlyj

Renascence_5 发表于 2016-5-21 07:23
for the first one,give you a hint. Using $$\frac{1}{x^2+x+1}=\frac{1-x}{1-x^3}$$ and then the geometric series.maybe you will get a series related to the digamma function or polygarithm.

分圆多项式 $\Phi_6(x)=x^2-x+1$ 的倒数的级数
$$\frac{1}{\Phi_6(x)}=1+x-x^{3}-x^{4}+x^{6}+x^{7}-x^{9}-x^{10}+\ldots$$的系数 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, ... (OEIS A010892) 满足递推式 $a(n)=a(n-1)-a(n-2)$
在 A New Kind of Science 第128页 例子(e)

1. Clear[f];Column[Labeled[Row[{Spacer[10], #[[1]], Grid[{#[[2]]}, Background->GrayLevel[0.9], Frame->All, FrameStyle->Gray], "⋯"}, Spacer[0.5]], #[[3]], {{Top, Left}}]&/@{{"(a)", RecurrenceTable[{f[n]==1 + f[n - 1], f[1]==1}, f, {n, 38}], "f[n]=1+f[n-1], f[1]=1"}, {"(b)", RecurrenceTable[{f[n]==1 - f[n - 1], f[1]==1}, f, {n, 48}], "f[n]=1-f[n-1], f[1]=1"},
2. {"(c)", RecurrenceTable[{f[n]==2f[n - 1], f[1]==1}, f, {n, 22}], "f[n]=2f[n-1], f[1]=1"},
3. {"(d)", RecurrenceTable[{f[n]==f[n - 1] + f[n - 2], f[1]==f[2]==1}, f, {n, 26}], "f[n]=f[n-1]+f[n-2], f[1]=f[2]=1"},
4. {"(e)", RecurrenceTable[{f[n]==f[n - 1] - f[n - 2], f[1]==f[2]==1}, f, {n, 44}], "f[n]=f[n-1]-f[n-2], f[1]=f[2]=1"},
5. {"(f)", RecurrenceTable[{f[n]== - f[n - 1] + f[n - 2], f[1]==f[2]==1}, f, {n, 27}], "f[n]=-f[n-1]+f[n-2], f[1]=f[2]=1"}}
6. , Spacings->1.5, BaseStyle->{FontFamily->"NewScienceSans", Italic, GrayLevel[0.2]}]

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symbolic graphics

1. CloudGet["https://wolfr.am/nks-page0128a-graphic"]

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给出了表达式$a(n)={\sin\left[(n+1)\frac{\pi}{3}\right]\over\sin\frac\pi3}=U_n(\frac12)$, $U_n$ is the Chebyshev polynomial of the second kind.
线性递推$b(n)=b(n-1)-b(n-2)$的通解为$b(n)=b(0) a(n)+[b(1)-b(0)] a(n-1)$

hbghlyj

MathWorld
generating function of $U_n(x)$
\[\frac{1}{1-2 x t+t^{2}} =\sum_{n=0}^{\infty} U_{n}(x) t^{n}\]
Gegenbauer polynomials一般形式
\[{\frac {1}{(1-2xt+t^{2})^{\alpha }}}=\sum _{n=0}^{\infty }C_{n}^{(\alpha )}(x)t^{n}\qquad (0\leq |x|<1,|t|\leq 1,\alpha >0)\]
Wolfram functions series representations
\begin{align}
U_n(\frac{1}{2}) &=2^{2-n} \sum_{k=0}^{\lfloor\frac{n+1}{2}\rfloor} (-3)^{k-1} \left(\frac{2k}{n+1}-\frac{3}{4}\right) \binom{n+1}{2k}\label1\\
&= \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n-k}k\label2
\end{align}
(1) 参见 周期3 线性递推$a_n=a_{n-1}+a_{n-2}$
(2) 参见 3#

hbghlyj

To prove that $a_n=a_{n-1}-a_{n-2}$, use Pascal's identity$$\binom{n-k}{k}=\binom{n-k-1}{k}+\binom{n-k-1}{k-1}\quad(n\ge2k+1,k\ge1)$$to rewrite the sum:
\begin{align*}
a_n &=1+\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n-k}{k}\\
&=\color{red}{1+\sum_{k=1}^{\lfloor\frac{n-1}{2}\rfloor}(-1)^k\binom{n-k-1}{k}}
+\color{blue}{\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}(-1)^k\binom{n-k-1}{k-1}}\\
&=\color{red}{\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}(-1)^k\binom{n-k-1}{k}}
\color{blue}{-\sum_{k=0}^{\lfloor\frac{n-2}{2}\rfloor}(-1)^k\binom{n-k-2}{k}}\\
&= a_{n-1}-a_{n-2},
\end{align*}

• For the first step, the term $k=0$ of the sum is $1$. For the term $k= \lfloor\frac n2\rfloor$:
  If $n$ is odd, $k=\frac{n-1}2$, so $\lfloor\frac{n-1}{2}\rfloor=k$, by Pascal's identity $\binom{n-k}{k}=\binom{n-k-1}{k}+\binom{n-k-1}{k-1}$.
  If $n$ is even, $k=\frac n2$, so $\lfloor\frac{n-1}{2}\rfloor=k-1$, $\binom nk=\binom{n-k-1}{k-1}$.
• For the red part, the term $k=0$ of the sum is $1$.
• For the blue part, substitute $k$ by $k+1$ and use $\lfloor\frac{n-2}{2}\rfloor=\lfloor\frac n2\rfloor-1$

hbghlyj

如何证明分圆多项式 $\Phi_n(x),n=1,2,\cdots$ 的倒数的级数的系数周期是$n$?

Czhang271828

hbghlyj 发表于 2023-4-20 17:37
如何证明分圆多项式 $\Phi_n(x),n=1,2,\cdots$ 的倒数的级数的系数周期是$n$?

当然以 $n$ 为周期, 并且个人猜想最小正周期其实 $n$ (应该很好证明?)

根据 $\displaystyle \dfrac{1}{x^n-1}=-\sum_{k\geq 0}x^{kn}$, 其各级系数对应周期为 $n$ 的数列
\[
-1,0,\ldots,0,-1,0,\ldots, 0,-1,\ldots.
\]
可以发现, 对任意周期为 $n$ 的数列乘上 $x^{d}-1$ ($d$ 为 $n$ 的因子), 得到的新数列仍是周期 $n$ 的. 例如原数列 $\{a_k\}_{k\in \mathbb N}$ 中的 $a_l=a_{l+n}$ 对应新数列 $\{a_k'\}_{k\in \mathbb N}$ 中的等式
\[
a_{l}'=a_{l-d}-a_l=a_{l-d+n}-a_{l+n}=a_{l+n}'.
\]