2023.11.5 闲话

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听 Sonnety 说春卷饭又发新歌了
😭回来吧 3579😭
🌟我最骄傲的信仰🌟
⚡️历历在目的 GF⚡️
😭眼泪莫名在流淌😭
💥依稀忆阅读随笔💥
👍还有给力的讲解👍
⚡️把题目都给打退⚡️
✨通宵熬夜都不累✨

不过 joke3579 问那个分治好像就直接高低位就完了，就考察复合 \(f\circ g\) 的话假设 \(f\) 可以被分成 \(f(z)=f_0(z)+z^{\lfloor n/2\rfloor}f_1(z)\)，那么 \(f(g(z)) = f_0(g(z)) + g(z)^{\lfloor n/2\rfloor}f_1(g(z))\) .

每层一次多项式乘法，因为 \(\deg g=\Theta(1)\) 所以是 \(\Theta(\mathsf M(n)\log n)\) .

Bones - Imagine Dragons

Gimme, gimme, gimme some time to think
I'm in the bathroom looking at me
Face in the mirror is all I need
Wait until the reaper takes my life
Never gonna get me out alive
I will live a thousand million lives

My patience is waning
Is this entertaining?
My patience is waning
Is this entertaining?

I got this feeling, yeah, you know
Where I'm losing all control
'Cause there's magic in my bones
I got this feeling in my soul
Go ahead and throw your stones
'Cause there's magic in my bones

Playing with a stick of dynamite
There was never grey in black and white
There was never wrong till there was right
Feeling like a boulder hurdling
Seeing all the vultures circling
Burning in the flames I'm working in
Turning in a bed that's darkening

My patience is waning
Is this entertaining?
My patience is waning
Is this entertaining?

I got this feeling, yeah, you know
Where I'm losing all control
Cause there's magic in my bones
I got this feeling in my soul
Go ahead and throw your stones
'Cause there's magic in my bones
'Cause there's magic in my bones

Look in the mirror of my mind
Turning the pages of my life
Walking the path so many paced a million times
Drown out the voices in the air
Leaving the ones that never cared
Picking the pieces up and building to the sky

My patience is waning
Is this entertaining?
My patience is waning
Is this entertaining?

I got this feeling yeah you know
Where I'm losing all control
'Cause there's magic in my bones
I got this feeling in my soul
Go ahead and throw your stones
'Cause there's magic in my bones

There goes my mind
Don't mind
There goes my mind
There goes my mind
Don't mind
There goes my mind
'Cause there's magic in my bones

---

APJ 看了 (number: 1) 眼 ABC279H，不禁感叹：「五边形数定理」.

五边形数定理

\[\prod_{k\ge1}(1-z^k)=\sum_{k=-\infty}^{\infty}(-1)^kz^{k(3k+1)/2} \]

后文求和号下标不加限制就是取全体整数 .

要证明五边形数定理，先考虑证明 Jacobi 三重积：

\[\prod_{r\ge1}(1-q^{2r})(1-q^{2r-1}z)(1+q^{2r-1}z^{-1})=\sum_kq^{k^2}z^k \]

首先有 \(q\)-二项式定理的两个式子（一个是正指数一个是负指数）：

\[\begin{aligned}&\prod_{i=0}^{n-1}(1+q^iz)=\sum_{i=0}^nq^{i(i-1)/2}\dbinom ni_qz^i\\&\prod_{i=0}^{n-1}\dfrac1{1-q^iz}=\sum_{k\ge0}\dbinom{k+n-1}k_qz^k\end{aligned} \]

取 \(n\to\infty\) 即得：

\[\begin{aligned}&\prod_{i\ge 0}(1+q^iz)=\sum_{i\ge 0}\dfrac{q^{i(i-1)/2}\cdot z^i}{(1-q^i)\cdots(1-q)}\\&\prod_{i\ge 0}\dfrac1{1-q^iz}=\sum_{i\ge 0}\dfrac{z^i}{(1-q^i)\cdots(1-q)}\end{aligned} \]

那么进而可以知道：

\[\begin{aligned}\prod_{k\ge0}(1+q^{2k-1}z)&=\prod_{k\ge 0}(1+z^{2n}(qz))\\&=\sum_{k\ge 0}\dfrac{z^{k(k-1)}(qz)^k}{(1-q^{2k})\cdots(1-q^2)}\\&=\dfrac1{\prod_{j\ge 0}(1-q^{2j+2})}\sum_{k\ge 0}q^{k^2}z^k\prod_{j\ge0}(1-q^{2n+2+2j})\\&=\dfrac1{\prod_{j\ge 0}(1-q^{2j+2})}\sum_{k}q^{k^2}z^k\prod_{j\ge0}(1+q^{2j}(-q^{2n+2}))\\&=\dfrac1{\prod_{j\ge 0}(1-q^{2j+2})}\sum_{k}q^{k^2}z^k\prod_{l\ge0}\dfrac{q^{l(l-1)}(-q^{2n+2})^l}{(1-q^{2l})\cdots(1-q^2)}\\&=\dfrac1{\prod_{j\ge 0}(1-q^{2j+2})}\sum_{k}q^{k^2}z^k\prod_{l\ge0}\dfrac{(-1)^l\cdot q^{l^2+l+2kl}}{(1-q^{2l})\cdots(1-q^2)}\\&=\dfrac1{\prod_{j\ge 0}(1-q^{2j+2})}\sum_{k}q^{k^2}z^k\prod_{l\ge0}\dfrac{(-1)^l\cdot q^{l^2+l+2kl}z^{-m}z^m}{(1-q^{2l})\cdots(1-q^2)}\\&=\dfrac1{\prod_{j\ge 0}(1-q^{2j+2})}\sum_{l\ge 0}\dfrac{(-1)^l\cdot(qz^{-1})^l}{(1-q^{2l})\cdots(1-q^2)}\prod_{k}q^{(k+l)^2}z^{k+l}\\&=\dfrac1{\prod_{j\ge 0}(1-q^{2j+2})}\prod_{k\ge 0}\dfrac1{1+q^{2k+1}z^{-1}}\sum_{k}q^{k^2}z^{k}\end{aligned} \]

移项即得 .

\[\tag*{$\square$} \]

那么在 Jacobi 三重积中取 \(z=t^{\frac 12},\,q=-t^{\frac 32}\) 则有：

\[\begin{aligned}\sum_k(-1)^kt^{k(3k+1)/2}&=\prod_{k\ge0}(1-t^{3k})(1-t^{1/2}t^{3k-3/2})(1-t^{-1/2}t^{3k-3/2})\\&=\prod_{k\ge 0}(1-t^{3k})(1-t^{3k-1})(1-t^{3k-2})\end{aligned} \]

类似裂项可以得到等号右侧等于 \(\prod_{k\ge 1}(1-z^k)\)，从而证得五边形数定理 . 这种做法可能比较正常 .

我逐渐理解一切.jpg

Reference.

• qwaszx《生成函数选讲》.
• Jacobi三重积与欧拉五边形数定理 - TravorLZH .
• ［整理］Jacobi三乘积恒等式的一个简单证明 - Trebor .

UPD. 似乎有更简单的证法 .