$(1+x)^n$的展开式中, $n$的幂的系数?

hbghlyj

在$\displaystyle\sum_{i=0}^∞\frac{n(n-1)\cdots(n-i+1)}{i!}x^i$中, $n$的幂的系数?
取$i≤2$展开:
\[1+n \left(x-\frac{x^2}{2}\right)+\frac{n^2 x^2}{2}\]
取$i≤20$展开看看:

1. CoefficientList[Sum[Product[n - k + 1, {k, 1, i}]/i! x^i, {i, 0, 20}], n]

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次数	系数
0	\[1\]
1	\[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\to\log(1+x)\]
2	\[\frac{x^2}{2}-\frac{x^3}{2}+\frac{11 x^4}{24}-\frac{5 x^5}{12}+\cdots\to\frac{1}{2} \log ^2(1+x)\]
3	\[\frac{x^3}{6}-\frac{x^4}{4}+\frac{7 x^5}{24}-\frac{5 x^6}{16}+\cdots\to\frac{1}{6} \log ^3(1+x)\]
$k$	$$\frac{x^k}k-\cdots\to\frac{\log^k(1+x)}{k!}\quad\text?$$

相关帖子:二项式级数

hbghlyj

$(1+x)^n$看作$n$的函数$f(n)=\exp(n\log(1+x))$, 根据exp的展开式,
$$f(n)=\sum_{i=0}^∞\frac{n(n-1)\cdots(n-i+1)}{i!}x^i=\sum_{k=0}^∞\frac{(\log(1+x))^k}{k!}n^k\quad\forall|x|<1$$
当$n\to\infty$, 两边的$n^k$的系数相等, 因为都等于$\displaystyle\lim_{n→∞}\frac{f(n)}{n^k}$.
可以使用这种方法由二项级数得到对数级数

hbghlyj

planetmath.org/stirlingnumbersofthefirstkind
Generating Function.
There is also a strong connection with the generalized binomial formula, which furnishes us with the following generating function:$$(1+t)^{x}=\sum_{n=0}^{\infty} \sum_{k=1}^{n} s(n, k) x^{k} \frac{t^{n}}{n !}$$