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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)=Σ_(h=0) ^∞ ((Y_ν ^((h)) (α))/(h!))(z−α)^h But.. can′t generalize coeff Y_ν ^((h)) (α) series representation ↓ Y_ν (z)=−(1/π)((2/z))^ν ∙Σ_(h=0) ^(ν−1) ((𝚪(ν−h))/(h!))((z/2))^(2h) +(2/π)J_ν (z)ln((1/2)z)−(1/π)((z/2))^ν ∙Σ_(h=0) ^∞ (((−1)^h (ψ^((0)) (h+1)−ψ^((0)) (h+ν+1)))/(h!(h+ν)!))((z/2))^(2h)

$$\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\:\mathrm{coefficient}\:{f}^{\left({n}\right)} \left(\alpha\right)\:\mathrm{of}\:{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{formal}\:\mathrm{power}\:\mathrm{series}\:\mathrm{of}\:{Y}_{\nu} \left({z}\right)\:\mathrm{is} \\ $$$${Y}_{\nu} \left({z}\right)=\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{Y}_{\nu} ^{\left({h}\right)} \left(\alpha\right)}{{h}!}\left({z}−\alpha\right)^{{h}} \\ $$$${But}..\:\mathrm{can}'\mathrm{t}\:\mathrm{generalize}\:\mathrm{coeff}\:{Y}_{\nu} ^{\left({h}\right)} \left(\alpha\right) \\ $$$$\mathrm{series}\:\mathrm{representation}\:\downarrow \\ $$$${Y}_{\nu} \left({z}\right)=−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{h}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\:\frac{\boldsymbol{\Gamma}\left(\nu−{h}\right)}{{h}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{h}} +\frac{\mathrm{2}}{\pi}{J}_{\nu} \left({z}\right)\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \centerdot\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{h}} \left(\psi^{\left(\mathrm{0}\right)} \left({h}+\mathrm{1}\right)−\psi^{\left(\mathrm{0}\right)} \left({h}+\nu+\mathrm{1}\right)\right)}{{h}!\left({h}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{h}} \\ $$

Answered by MrGaster last updated on 03/Nov/24

Y_ν (z)=(1/π)((2/z))^ν ∙Σ_(h=0) ^(ν−1) ((Γ(ν−h))/(h!))((z/2))^(2h) +(2/π)J_ν (z)ln((1/z)z)−(1/π)((z/2))^ν ∙Σ_(h=0) ^∞ (((−1)^h (ψ^((0)) (h+1)−ψ^((0)) (h+ν+1))/(h!(h+ν)!))((z/2))^(2h) Y_ν (z)Σ_(h=0) ^∞ ((Y_ν ^((h)) (α))/(h!))(z−α)^h Y_ν ^((h)) (α)= { ((−(1/π)((2/α))^ν ((Γ(ν−h))/(h!))((α/2))^(2h) ,h=0.1,…,ν−1)),(((2/π)J_ν ^((h−ν)) (α)ln((1/2)α)−(1/π)((α/2))^ν (((−1)^(h−ν) (ψ^((0)) (h−ν+1)−ψ^((0)) (h+1)))/((h−ν)!(h+ν)!))((α/2))^(2(h−ν)) ,h=ν,ν+1,…)) :}

$${Y}_{\nu} \left({z}\right)=\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{h}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\frac{\Gamma\left(\nu−{h}\right)}{{h}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{h}} +\frac{\mathrm{2}}{\pi}{J}_{\nu} \left({z}\right)\mathrm{ln}\left(\frac{\mathrm{1}}{{z}}{z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \centerdot\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{h}} \left(\psi^{\left(\mathrm{0}\right)} \left({h}+\mathrm{1}\right)−\psi^{\left(\mathrm{0}\right)} \left({h}+\nu+\mathrm{1}\right)\right.}{{h}!\left({h}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{h}} \\ $$$${Y}_{\nu} \left({z}\right)\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{Y}_{\nu} ^{\left({h}\right)} \left(\alpha\right)}{{h}!}\left({z}−\alpha\right)^{{h}} \\ $$$${Y}_{\nu} ^{\left({h}\right)} \left(\alpha\right)=\begin{cases}{−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{\alpha}\right)^{\nu} \frac{\Gamma\left(\nu−{h}\right)}{{h}!}\left(\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}{h}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:,{h}=\mathrm{0}.\mathrm{1},\ldots,\nu−\mathrm{1}}\\{\frac{\mathrm{2}}{\pi}{J}_{\nu} ^{\left({h}−\nu\right)} \left(\alpha\right)\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\alpha\right)−\frac{\mathrm{1}}{\pi}\left(\frac{\alpha}{\mathrm{2}}\right)^{\nu} \frac{\left(−\mathrm{1}\right)^{{h}−\nu} \left(\psi^{\left(\mathrm{0}\right)} \left({h}−\nu+\mathrm{1}\right)−\psi^{\left(\mathrm{0}\right)} \left({h}+\mathrm{1}\right)\right)}{\left({h}−\nu\right)!\left({h}+\nu\right)!}\left(\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}\left({h}−\nu\right)} \:,{h}=\nu,\nu+\mathrm{1},\ldots}\end{cases} \\ $$

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