2025.10.26 闲话

新年的军队里面涉及到的反演：

\[\begin{aligned}&\sum_{i = 0}^{n - 1} a_ix^i = \sum_{r = 0}^{n - 1}b_r \frac{x^{r}(1 - x)^{n - r - 1}}{r!(n - r - 1)!}\\\iff&\sum_{r = 0}^{n - 1} \frac{b_r}{r!(n - r - 1)!}x^{r} = \sum_{k = 0}^{n - 1} a_k x^k(1 - x)^{n - k - 1}\end{aligned} \]

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\[\begin{aligned}&\sum_{i=0}^{n-1}a_ix^i=\sum_{r=0}^{n-1}b_r\dfrac{x^r(1-x)^{n-r-1}}{r!(n-r-1)!}\\\iff&\sum_{i=0}^{n-1}a_ix^i=(1-x)^{n-1}\sum_{r=0}^{n-1}(-1)^rb_r\dfrac{x^r}{(x-1)^r}\dfrac1{r!(n-r-1)!}\\\iff&(1-t)^{n-1}\sum_{i=0}^{n-1}a_i\dfrac{t^i}{(t-1)^i}=\sum_{r=0}^{n-1}(-1)^rb_r\dfrac{t^r}{r!(n-r-1)!}&\left(\text{subs }t=\dfrac x{x-1}\right)\\\iff&\sum_{i=0}^{n-1}(-1)^ia_it^i(1-t)^{n-i-1}=\sum_{r=0}^{n-1}(-1)^rb_r\dfrac{t^r}{r!(n-r-1)!}\\\iff&\sum_{i=0}^{n-1}a_it^i(1-t)^{n-i-1}=\sum_{r=0}^{n-1}b_r\dfrac{t^r}{r!(n-r-1)!}\end{aligned} \]

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