Question

To show that$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}{\mathbf{x}}^{\mathbf{-}\mathbf{3}\mathbf{/}\mathbf{2}}{\mathbf{J}}_{\mathbf{3}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{3}}^{\mathbf{-}\mathbf{1}}\sqrt{\mathbf{2}\mathbf{/}\mathbf{\pi }}$ .

Short answer

Hence,$\underset{\mathrm{x}\to 0}{\lim }{\mathrm{x}}^{-3/2}{\mathrm{J}}_{3/2}\left(\mathrm{x}\right)=3^{-1}\sqrt{\frac{J}{\mathrm{\tau \tau }}}$

Step-by-step solution

Concept of the Property of Jp(x)

Infinite series:

${\mathbf{J}}_{\mathbf{p}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}}{\mathbf{⌜}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{⌜}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{p}\mathbf{)}}{\mathbf{\left(}\frac{\mathbf{x}}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{p}}$

Calculation of the function limx→0x-3/2J3/2(x) :

Considering the infinite series, ${\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜(\mathrm{n}+1)⌜(\mathrm{n}+1+\mathrm{p})}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}+\mathrm{p}}$.

Let $p=\frac{3}{2}$in the above recursion relation.

Therefore, simplify as follows:

localid="1659257652571" $$\begin{gathered} J_{\frac{3}{2}}\left(x\right)={\left(\frac{x}{2}\right)}^{\frac{3}{2}}\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜(\mathrm{n}+1)⌜(\mathrm{n}+{\displaystyle \frac{5}{2}})}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}+\mathrm{p}} \\ ={\left(\frac{x}{2}\right)}^{\frac{3}{2}}\left\{\frac{1}{⌜\left({\displaystyle \frac{5}{2}}\right)}-\frac{{\left({\displaystyle \frac{x}{2}}\right)}^2}{{\displaystyle \frac{5}{2}}⌜\left({\displaystyle \frac{5}{2}}\right)}+\frac{{\left({\displaystyle \frac{x}{2}}\right)}^4}{2\times {\displaystyle \frac{7}{2}}\times {\displaystyle \frac{5}{2}}⌜\left({\displaystyle \frac{5}{2}}\right)}.............................\right\} \end{gathered}$$

localid="1659258104658" $$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }{\mathrm{x}}^{\frac{3}{2}}{\mathrm{J}}_{\frac{3}{2}}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{\lim }{\left(\frac{1}{2}\right)}^{\frac{3}{2}}={\left(\frac{x}{2}\right)}^{\frac{3}{2}}\left\{\frac{1}{⌜\left({\displaystyle \frac{5}{2}}\right)}-\frac{{\left({\displaystyle \frac{x}{2}}\right)}^2}{{\displaystyle \frac{5}{2}⌜\left(\frac{5}{2}\right)}}+\frac{{\left({\displaystyle \frac{x}{2}}\right)}^4}{2\times {\displaystyle \frac{7}{2}}\times {\displaystyle \frac{5}{2}}⌜\left({\displaystyle \frac{5}{2}}\right)}.............................\right\} \\ =\frac{{\left({\displaystyle \frac{1}{2}}\right)}^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}\times {\displaystyle \frac{1}{2}}\times \sqrt{\mathrm{\tau \tau }}} \\ =\frac{\left(4\right)}{{\displaystyle 3{\left(8\right)}^{\frac{1}{2}}\times \sqrt{\mathrm{\tau \tau }}}} \\ \underset{\mathrm{x}\to 0}{\lim }{\mathrm{x}}^{\frac{3}{2}}{\mathrm{J}}_{\frac{3}{2}}\left(\mathrm{x}\right)=\frac{\left(4\right)}{{\displaystyle 3\times 2{\left(2\right)}^{\frac{1}{2}}\times \sqrt{\mathrm{\tau \tau }}}} \\ =3^{-1}\sqrt{\frac{2}{\mathrm{\tau \tau }}} \end{gathered}$$

Hence, $\underset{\mathrm{x}\to 0}{\lim }{\mathrm{x}}^{\frac{3}{2}}{\mathrm{J}}_{\frac{3}{2}}\left(\mathrm{x}\right)=3^{-1}\sqrt{\frac{2}{\mathrm{\tau \tau }}}$ is proved.