五十一(C)
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設 \(f_{t,i}\) 表示第 \(t\) 輪 \(x=i\) 的機率。有
\[f_{t,i}=\sum_{j=i}^n\frac1{j+1}f_{t-1,j}
\]
設 \(F_t(x)=\sum_{i=0}^nf_{t,i}x^i\)。有
\[\begin{aligned}
F_t(x)&=\sum_{i=0}^nf_{t,i}x^i\\
&=\sum_{i=0}^n\sum_{j=i}^n\frac1{j+1}f_{t-1,j}x^i\\
&=\sum_{j=0}^n\frac1{j+1}f_{t-1,j}\sum_{i=0}^jx^i\\
&=\sum_{j=0}^n\frac1{j+1}f_{t-1,j}\frac{x^{j+1}-1}{x-1}\\
\end{aligned}\]
利用定積分展開 \(\frac{x^{j+1}-1}{j+1}\):
\[\begin{aligned}
F_t(x)&=\frac1{x-1}\sum_{j=0}^nf_{t-1,j}\int_{1}^xy^j\text{d}y\\
&=\frac1{x-1}\int_{1}^x\text{d}y\sum_{j=0}^nf_{t-1,j}y^j\\
&=\frac1{x-1}\int_{1}^xF_{t-1}(y)\text{d}y\\
\end{aligned}\]
\(1\) 到 \(x\) 難以處理,設
\[\begin{aligned}
G_t(x)&=F_t(x+1)\\
&=\frac1{x}\int_{1}^{x+1}F_{t-1}(y)\text{d}y\\
&=\frac1{x}\int_{0}^{x}G_{t-1}(y)\text{d}y\\
\end{aligned}\]
若設 \(G_t(x)=\sum_{i=0}^ng_{t,i}x^i\),則有
\[\begin{aligned}
G_t(x)&=\frac1{x}\int_{0}^{x}\sum_{i=0}^ng_{t-1,i}y^i\text{d}y\\
G_t(x)&=\frac1{x}\sum_{i=0}^ng_{t-1,i}\frac{x^{i+1}}{i+1}\\
G_t(x)&=\sum_{i=0}^ng_{t-1,i}\frac{x^i}{i+1}\\
\end{aligned}\]
因此有 \(g_{t,i}=\frac1{i+1}g_{t-1,i}\),即 \(g_{t,i}=\frac{g_{0,i}}{(i+1)^t}\)。
考慮求 \(g_0\),有
\[G_t(x)=F_t(x+1)
\]
於是
\[\sum_{i=0}^ng_{t,i}x^i=\sum_{i=0}^nf_{t,i}(x+1)^i
\]
展開
\[\begin{aligned}
\sum_{i=0}^ng_{t,i}x^i
&=\sum_{i=0}^nf_{t,i}\sum_{j=0}^ix^j\binom{i}{j}\\
&=\sum_{j=0}^nx^j\sum_{i=j}^nf_{t,i}\binom ij\\
\end{aligned}\]
因此
\[g_{t,i}=\sum_{j=i}^nf_{t,j}\binom ji
\]
可以把組合數拆開,差卷積計算 \(g_0\)。
求得 \(g_m\) 之後還要求 \(f_m\),有二項式反演
\[f_{t,i}=\sum_{j=i}^ng_{t,j}\binom ji(-1)^{j-i}
\]
仍然是差卷積。