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Question Number 213007 by issac last updated on 30/Oct/24

{can}^{'} t find coefficient f^{(n)} (α) of Y_{ν} (z)

formal power series of Y_{ν} (z) is

Y_{ν} (z) = \underset{h = 0}{\sum^{\infty}} \frac{Y_{ν}^{(h)} (α)}{h!} {(z - α)}^{h}

B u t . . {can}^{'} t generalize coeff Y_{ν}^{(h)} (α)

series representation ↓

Y_{ν} (z) = - \frac{1}{π} {(\frac{2}{z})}^{ν} \cdot \underset{h = 0}{\sum^{ν - 1}} \frac{Γ (ν - h)}{h!} {(\frac{z}{2})}^{2 h} + \frac{2}{π} J_{ν} (z) \ln (\frac{1}{2} z) - \frac{1}{π} {(\frac{z}{2})}^{ν} \cdot \underset{h = 0}{\sum^{\infty}} \frac{{(- 1)}^{h} (ψ^{(0)} (h + 1) - ψ^{(0)} (h + ν + 1))}{h! (h + ν)!} {(\frac{z}{2})}^{2 h}

Answered by MrGaster last updated on 03/Nov/24

Y_{ν} (z) = \frac{1}{π} {(\frac{2}{z})}^{ν} \cdot \underset{h = 0}{\sum^{ν - 1}} \frac{Γ (ν - h)}{h!} {(\frac{z}{2})}^{2 h} + \frac{2}{π} J_{ν} (z) \ln (\frac{1}{z} z) - \frac{1}{π} {(\frac{z}{2})}^{ν} \cdot \underset{h = 0}{\sum^{\infty}} \frac{{(- 1)}^{h} (ψ^{(0)} (h + 1) - ψ^{(0)} (h + ν + 1)}{h! (h + ν)!} {(\frac{z}{2})}^{2 h}

Y_{ν} (z) \underset{h = 0}{\sum^{\infty}} \frac{Y_{ν}^{(h)} (α)}{h!} {(z - α)}^{h}

Y_{ν}^{(h)} (α) = {\begin{cases} - \frac{1}{π} {(\frac{2}{α})}^{ν} \frac{Γ (ν - h)}{h!} {(\frac{α}{2})}^{2 h}, h = 0.1, \dots, ν - 1 \\ \frac{2}{π} J_{ν}^{(h - ν)} (α) \ln (\frac{1}{2} α) - \frac{1}{π} {(\frac{α}{2})}^{ν} \frac{{(- 1)}^{h - ν} (ψ^{(0)} (h - ν + 1) - ψ^{(0)} (h + 1))}{(h - ν)! (h + ν)!} {(\frac{α}{2})}^{2 (h - ν)}, h = ν, ν + 1, \dots \end{cases}

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