乘积$(1+a x)(1+a^2 x)(1+a^3 x) \dots(1+a^m x)$按照 $x$ 的升幂排列

nttz

\begin{gathered}
(1+a x)(1+a^2 x) \\
(1+a x)(1+a^2 x)(1+a^3 x)
\end{gathered}更一般地，求出乘积
\[
(1+a x)(1+a^2 x)(1+a^3 x) \dots(1+a^m x)
\]
并把乘积按照 $x$ 的升幂排列。求出展开式里 $x^p$ 的系数的一般表达式。当 $a$趋近于 1 时，这些系数怎样变化？由此推出
\[
(1+x)^m
\]
按照 $x$ 的升幂的展开式．

hbghlyj

$$f(x)=(1+ax)(1+{{a}^{2}}x)(1+{{a}^{3}}x)....(1+{{a}^{n}}x)$$

$x^n$的系数$\displaystyle=\frac{f^{(n)}(0)}{n!}={{a}^{\frac{n(n+1)}{2}}}$
$x^{n-1}$的系数$\displaystyle={f^{(n-1)}(0)\over(n-1)!}={{a}^{\frac{n(n+1)}{2}}}\sum_{k=1}^{n}{{\frac1{{a}^{k}}}}=a^{n(n-1)\over2}\frac{1-a^n}{1-a},$
$x^{n-2}$的系数$\displaystyle=\frac{f^{(n-2)}(0)}{(n-2)!}={{a}^{\frac{n(n+1)}{2}}}\sum_{1≤i<j≤n}{{\frac1{{a}^{i+j}}}}=a^{n(n-3)\over2}\frac{\left(1-a^n\right) \left(a-a^n\right)}{(1-a)^2 (1+a)},$
$\vdots$
$\vdots$
$x$的系数$\displaystyle=f'(0)=a\frac{1-a^n}{1-a}$

nttz

回复 2# hbghlyj


    这道题是初等数学教材上的习题内容，能否不用导数去解决？

hbghlyj

回复 3# nttz

q-Pochhammer symbol
    ${\displaystyle (a;q)_{n}=\prod _{k=0}^{n-1}(1-aq^{k})=(1-a)(1-aq)(1-aq^{2})\cdots (1-aq^{n-1})}$
WHAT IS A $q$-SERIES?
What is a $q$-series? The simplest and most manifestly useless definition would be a series with $q$'s in the summands. Having begun this essay with meaningless drivel, we next admit that there does not exist a "good" definition of a $q$-series. We might define a $q$-series to be one with summands containing expressions of the type
$$
(a)_{n}:=(a ; q)_{n}:=(1-a)(1-a q) \cdots\left(1-a q^{n-1}\right), \quad n \geq 0,\tag{1.1}
$$
where we interpret $(a ; q)_{0}=1$. If the base $q$ is understood, we often use the notation at the far left-hand side of (1.1), but in this paper, since we are going to use this notation for rising factorials or shifted factorials, to avoid confusion, we shall always write $(a ; q)_{n}$. This definition of a $q$-series is also not completely satisfactory, because often in the theory of $q$-series, we let parameters in the summands tend to 0 or to $\infty$, and consequently it may happen that no factors of the type $(a ; q)_{n}$ remain in the summands. In some cases, what may remain is a theta function. Following the lead of Ramanujan, we shall define a general theta function $f(a, b)$ by
$$
f(a, b):=\sum_{n=-\infty}^{\infty} a^{n(n+1) / 2} b^{n(n-1) / 2}, \quad|a b|<1 .\tag{1.2}
$$
In the theory of $q$-series, theta functions also frequently arise in identities satisfied by series with products (1.1) in their summands. Thus, for these reasons, theta functions are an integral part of the theory and are also considered to be $q$-series. As we shall see in the sequel, infinite $q$-products
$$
(a ; q)_{\infty}:=\lim _{n \rightarrow \infty}(a ; q)_{n}, \quad|q|<1,
$$
arise both in product representations of theta functions and, more generally, in identities for $q$-series.

hbghlyj

en.wikipedia.org/wiki/Q-Pochhammer_symbol

By removing a triangular partition with $n − 1$ parts from such a partition, we are left with an arbitrary partition with at most $n$ parts. This gives a weight-preserving bijection between the set of partitions into $n$ or $n − 1$ distinct parts and the set of pairs consisting of a triangular partition having $n − 1$ parts and a partition with at most $n$ parts. By identifying generating series, this leads to the identity:
${\displaystyle (-a;q)_{\infty }=\prod _{k=0}^{\infty }(1+aq^{k})=\sum _{k=0}^{\infty }\left(q^{k \choose 2}\prod _{j=1}^{k}{\frac {1}{1-q^{j}}}\right)a^{k}=\sum _{k=0}^{\infty }{\frac {q^{k \choose 2}}{(q;q)_{k}}}a^{k}}$

nttz

回复 11# hbghlyj


    能不能说出每个系数怎么算的过程呢

nttz

回复 16# hbghlyj
韦达定理本身就是比较两边的系数，但是如果是x的次数是j，那么它的系数怎么求呢
a^(m1+m2+m3+..+mj)   m1,m2...mj都不同，范围都是1到n，然后累加

hbghlyj

Pentagonal number theorem

${\displaystyle \prod _{n=1}^{\infty }\left(1-x^{n}\right)=\sum _{k=-\infty }^{\infty }\left(-1\right)^{k}x^{k\left(3k-1\right)/2}=1+\sum _{k=1}^{\infty }(-1)^{k}\left(x^{k(3k+1)/2}+x^{k(3k-1)/2}\right).}$
In other words,
${\displaystyle (1-x)(1-x^{2})(1-x^{3})\cdots =1-x-x^{2}+x^{5}+x^{7}-x^{12}-x^{15}+x^{22}+x^{26}-\cdots .}$

hbghlyj

$$(1+x)(1+x^2)…(1+x^{2^n})=\frac{1-x^{2^{n+1}}}{1-x}$$
$\prod_{n=1}^\infty(1+x^n)$ is the generating series for the partitions of integers into distinct parts.

1, 1, 2, 2, 3, 4, 5, 6, 8, 10, ... (OEIS A000009).

hbghlyj

回复 18# nttz
问n-m的话只能递推了吧...
$x^p$的系数是$a,a^2,⋯,a^n$的$p$元组乘积之和。
把$a,a^2,⋯,a^n$的$p$元组乘积之和记为$σ(n,p)$.显然$σ(n,0)=1,σ(n,n)=a^{n(n+1)\over2}$.
考虑$σ(n+1,p+1)$中的每个$p+1$元组乘积,若$p+1$元组中含有$a^{n+1}$,剩下的是$a,a^2,⋯,a^n$的$p$元组;若$p+1$元组中不含$a^{n+1}$,剩下的是$a,a^2,⋯,a^n$的$p+1$元组,因此
$$σ(n+1,p+1)=a^{n+1}σ(n,p)+σ(n,p+1)\tag1$$
若$p+1$元组中含有$a$,剩下的是$a^2,a^3,⋯,a^{n+1}$的$p$元组,每个提取出$a$,剩下的是$a,a^2,⋯,a^n$的$p$元组;若$p+1$元组中不含$a$,剩下的是$a^2,a^3,⋯,a^{n+1}$的$p+1$元组,每个提取出$a$,剩下的是$a,a^2,⋯,a^n$的$p$元组,因此
$$σ(n+1,p+1)=a^{p+1}σ(n,p)+a^{p+1}σ(n,p+1)\tag2$$

hbghlyj

把$p=n-2$代入(1)得到$σ(n + 1, n - 1) =  a^{n+1}σ(n, n - 2) + σ(n , n - 1)$
令$g(n)=σ(n,n-2)$,得到递推式$$g(n + 1) =  a^{n+1}g(n) +\frac{a^{\frac{1}{2} n (n-1)} \left(1-a^n\right)}{1-a},\;g(2)=1$$

1. g[n] /. RSolve[
2.    g[n + 1] == a^(n + 1) g[n] + a^(n (n - 1)/2) (1 - a^n)/(1 - a) &&
3.     g[2] == 1, g[n], n][[1]]

Copy the Code

输出
$$\frac{a^{-\frac{3n}{2}+\frac{n^2}{2}} \left(-1+a^n\right) \left(-a+a^n\right)}{(-1+a)^2 (1+a)}$$

hbghlyj

回复 20# hbghlyj
像这样可以继续做下去求出n-m次项的系数:
把$p=n-3$代入(1)得到$σ(n+1,n-2)=a^{n+1}σ(n,n-3)+σ(n,n-2)$
令$h(n)=σ(n,n-3)$,得到递推式
$$h(n + 1) =  a^{n+1}h(n) +g(n),\;h(3)=1$$

1. h[n] /. RSolve[
2.    h[n + 1] ==
3.      a^(n + 1) h[n] + (a^(-((3 n)/2) + n^2/2) (-1 + a^n) (-a + a^n))/((-1 + a)^2 (1 + a)) && h[3] == 1,
4.    h[n], n][[1]]

Copy the Code

输出$$\frac{a^{-\frac{5n}{2}+\frac{n^2}{2}} \left(-1+a^n\right) \left(a^3+a^{2
   n}-a^{1+n}-a^{2+n}\right)}{(-1+a)^3 (1+a) \left(1+a+a^2\right)}$$
验证一下:

1. Table[FullSimplify[Coefficient[Product[1 + a^k x, {k, 1, n}], x^(n - 3)] - (
2.   a^(-((5 n)/2) + n^2/
3.     2) (-1 + a^n) (a^3 + a^(2 n) - a^(1 + n) - a^(2 + n)))/((-1 +
4.      a)^3 (1 + a) (1 + a + a^2))],{n,4,10}]

Copy the Code

输出{0, 0, 0, 0, 0, 0, 0}
所以$h(n)$的表达式应该是正确的.

tommywong

會唔會係留$\sum$符號嘅表達式啊
用佢嚟推導$(1+x)^m$嘅展開式點都係牛刀殺雞架喎

$(1+ax)(1+a^2 x)=1+(a+a^2)x+a^3 x^2$
$(1+ax)(1+a^2 x)(1+a^3 x)=1+(a+a^2+a^3)x+(a^3+a^4+a^5) x^2+a^6 x^3$
$(1+ax)(1+a^2 x)(1+a^3 x)(1+a^4 x)$
$=1+(a+a^2+a^3+a^4)x+(a^3+a^4+2a^5+a^6+a^7) x^2+(a^6+a^7+a^8+a^9) x^3+a^{10} x^4$

$\displaystyle [x^p]\prod_{j=1}^m (1+a^j x)=\sum_{1\le j_1<j_2<\cdots<j_p\le m}  a^{j_1+j_2+\cdots+j_p}$

$\displaystyle [x^p]\prod_{j=1}^m (1+x)=\sum_{1\le j_1<j_2<\cdots<j_p\le m}1=\binom{m}{p}$

nttz

回复 22# tommywong
这个表达很简洁，但是没有具体的关于j，n的表达式

nttz

还只是递推$σ(n,n-2)$ 和 $σ(n,n-3)$ 的结果，要求的是 $σ(n,p)$

kuing

在知乎上有相关帖子：
【如何证明 q 级数中的柯西二项式定理？】
zhihu.com/question/427390721/answer/1549407155
（我看不懂，纯搬运）

其实论坛上也有：
【一个没见过的 Cauchy 恒等式】
forum.php?mod=viewthread&tid=4862#pid39254

另外，这帖的标题起得不够好，建议改一下。

Czhang271828

回复 17# kuing

个人之前写过一个 triple produit de Jacobi 的组合学推导, 以及其他


论坛 Banana space (不是 Banach space) 上有 triple produit de Jacobi 的复变推导, 记得 hbghlyj mark 过.