We obtained (19.10) for${\mathit{J}}_{\mathbf{p}}\left(x\right)\mathbf{,}\mathit{p}\mathbf{\ge }\mathbf{0}\mathbf{.}$It is, however, valid for$\mathit{p}\mathbf{\ge }\mathbf{-}\mathbf{1}$, that is for${\mathit{N}}_{\mathbf{p}}\left(x\right)\mathbf{,}\mathbf{0}\mathbf{\le }\mathit{p}\mathbf{<}\mathbf{1}$. The difficulty in the proof occurs just after (19.7); we said that $\mathit{u}\mathbf{,}\mathit{v}\mathbf{,}\mathit{u}\mathbf{\text{'}}\mathbf{,}\mathit{v}\mathbf{\text{'}}$are finite at $\mathit{x}\mathbf{=}\mathbf{0}$which is not true for${\mathit{N}}_{\mathbf{p}}\left(x\right)$.

However, the negative powers of x cancel if$\mathit{p}\mathbf{<}\mathbf{1}$. Show this for $\mathit{p}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$by using two terms of the power series (12.9) or (13.1) for the function ${\mathit{N}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{-}{\mathit{J}}_{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}\left(x\right)$ [see (13.3)].

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