Question Number 213007 by issac last updated on 29/Oct/24
$$\mathrm{can}'\mathrm{t}\:\mathrm{find}\:\:\mathrm{coefficient}\:{f}^{\left({n}\right)} \left(\alpha\right)\:\mathrm{of}\:{Y}_{\nu} \left({z}\right) \\ $$$$\mathrm{formal}\:\mathrm{power}\:\mathrm{series}\:\mathrm{of}\:{Y}_{\nu} \left({z}\right)\:\mathrm{is} \\ $$$${Y}_{\nu} \left({z}\right)=\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{{Y}_{\nu} ^{\left({h}\right)} \left(\alpha\right)}{{h}!}\left({z}−\alpha\right)^{{h}} \\ $$$${But}..\:\mathrm{can}'\mathrm{t}\:\mathrm{generalize}\:\mathrm{coeff}\:{Y}_{\nu} ^{\left({h}\right)} \left(\alpha\right) \\ $$$$\mathrm{series}\:\mathrm{representation}\:\downarrow \\ $$$${Y}_{\nu} \left({z}\right)=−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{h}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\:\frac{\boldsymbol{\Gamma}\left(\nu−{h}\right)}{{h}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{h}} +\frac{\mathrm{2}}{\pi}{J}_{\nu} \left({z}\right)\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \centerdot\underset{{h}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{h}} \left(\psi^{\left(\mathrm{0}\right)} \left({h}+\mathrm{1}\right)−\psi^{\left(\mathrm{0}\right)} \left({h}+\nu+\mathrm{1}\right)\right)}{{h}!\left({h}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{h}} \\ $$