2024.5.30 闲话 DLC

因为好像不小心做了一些不那么平凡的事情,于是追加一个 DLC,抢先发布这个内容 .

问题:

\[\sum_{k=0}^n\dbinom nkH_k=2^n\left(H_n-\sum_{i=1}^n\dfrac1{i2^i}\right) \]

证明:

写为生成函数的形式,首先有调和级数的生成函数:

\[\mathcal H(x)=\sum_{n\ge0}H_nx^n=-\dfrac{\ln(1-x)}{1-x} \]

则:

\[\begin{aligned}\mathrm{LHS}&=[x^n]\dfrac1{1-x}\mathcal H\left(\dfrac x{1-x}\right)\\&=-[x^n]\dfrac1{1-x}\cdot\dfrac{\ln \frac{1-2x}{1-x}}{\frac{1-2x}{1-x}}\\&=[x^n]\left(\dfrac{\ln(1-x)}{1-2x}-\dfrac{\ln(1-2x)}{1-2x}\right)\\&=2^nH_n-2^n\sum_{i=1}^n\dfrac1{i2^i}\\&=\mathrm{RHS}\end{aligned} \]

证明完毕 .

挑战一波更难的问题:

Generalized Mneimneh's Identity

\[\sum_{k=0}^n\dbinom nk x^ky^{n-k}H_k=(x+y)^n\left(H_n-\sum_{i=1}^n\dfrac{y^i(x+y)^{-i}}i\right) \]

其实做法基本没啥区别:

\[\begin{aligned}\mathrm{LHS}&=-[z^n]\dfrac1{1-yz}\left(\dfrac{\ln(1-xz)}{1-xz}\circ\dfrac z{1-yz}\right)\\&=-[z^n]\dfrac1{1-yz}\cdot\dfrac{\ln(\frac{1-(x+y)z}{1-yz})}{\frac{1-(x+y)z}{1-yz}}\\&=[z^n]\left(\dfrac{\ln(1-yz)}{1-(x+y)z}-\dfrac{\ln(1-(x+y)z)}{1-(x+y)z}\right)\\&=(x+y)^nH_n-(x+y)^n\sum_{i=1}^n\dfrac1{i(\frac{x+y}y)^i}\\&=\mathrm{RHS}\end{aligned} \]

因为最近(?)随便开了一篇论文 General Mneimneh-type Binomial Sum involving Harmonic Numbers,内容好像有点深刻,证明上来就一排积分糊脸好像不太可做,所以没咋仔细看,不过自己推了一下好像对于里面的所有式子上面那个做法都能跑(免责声明:没有真的对每个式子都从头到尾推一遍,不一定对

如果有什么不对的地方评论区 D 一下(

话说好久没一日双更学术了啊 .

posted @ 2024-05-30 17:20  Jijidawang  阅读(167)  评论(2编辑  收藏  举报
😅​