2020-10-06 11:56:04

$$\sum_n \mathrm{ans}_{n,n+k} \frac{x^n}{n!(n+k)!} = (1-x)^{-1/2} \exp\left(\frac{x(1+x)}{2-2x}\right) \frac {\left(\dfrac{2-x}{2-2x}\right)^k}{k!} \cdot { {}_0 F_1\left(;k+1; x\left(\frac{2-x}{2-2x}\right)^2\right) }$$

#### 广义超几何函数微分有限

$$f = {}_p F_q(a_1,\dots a_p; b_1, \dots, b_q;x) = \sum_n \frac{a_1^{\overline n} \cdots a_p^{\overline n}}{b_1^{\overline n} \cdots b_q^{\overline n}n!}x^n$$

$$(\vartheta + a_1) \cdots (\vartheta + a_p)f = D (\vartheta + b_1 - 1) \cdots (\vartheta + b_q - 1)f$$

1. 若 $f,g$ 微分有限，那么 $f+g,fg$ 微分有限。
2. 若 $f$ 微分有限，$g$ 是代数的，那么 $f\circ g$ 微分有限。

#### 求 $f+g$

$$f^{(n)} = \sum_{0\le k<n} p_k f^{(k)}$$

$$(f+g)^{(k)} = \sum_{i<n} a_i f^{(i)} + \sum_j b_jg^{(j)}$$

\begin{aligned} (f+g)^{(k+1)} &= \left(\sum_{i<n} a_i f^{(i)} + \sum_j b_jg^{(j)}\right)'\\ &= \sum_{i<n} \left(a_i' f^{(i)} + a_i f^{(i+1)} \right) + \sum_j \left( b_j'g^{(j)} + b_j g^{(j+1)}\right) \end{aligned}

#### 求 $fg$

$$(fg)^{(k)} = \sum_{i<n,j<m} a_{i,j} f^{(i)}g^{(j)}$$

#### 求 $f\circ g$

$$(f\circ g)^{(k)} = \sum_j a_{k,j} (f^{(k)} \circ g)$$

$$(\mathbf v \circ g) \mathbf A^{-1} \mathbf h = 0$$

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