To show the first few terms of ${\mathbf{J}}_{\mathbf{0}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{J}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{J}}_{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{J}}_{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{J}}_{\mathbf{-}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$. Show that${\mathbf{J}}_{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathbf{J}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}{\mathbf{andJ}}_{\mathbf{-}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{J}}_{\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}$.

The Bessel function:

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

The given relations to prove:

$$\begin{gathered} {\mathrm{J}}_{-1}\left(\mathrm{x}\right)=-{\mathrm{J}}_1\left(\mathrm{x}\right) \\ {\mathrm{J}}_{-2}\left(\mathrm{x}\right)=-{\mathrm{J}}_2\left(\mathrm{x}\right) \end{gathered}$$

According to the Bessel function:
localid="1659262746953" $$\begin{gathered} {\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^{2\mathrm{n}+\mathrm{p}}}{⌜(\mathrm{n}+\mathrm{p}+1)\mathrm{n}!}{\mathrm{J}}_{\mathrm{o}}\left(\mathrm{x}\right) \\ =\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{n}}}{{(\mathrm{n}!)}^2}{\mathrm{J}}_0\left(\mathrm{x}\right) \\ =\left\{1-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^2}{{(1!)}^2}+\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^4}{{(2!)}^2}-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^6}{{(3!)}^2}+..........\right\} \end{gathered}$$

${\mathrm{J}}_1\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^{2\mathrm{n}+1}}{┌(\mathrm{n}+2)\mathrm{n}!}$

Therefore,

localid="1659262971772" $J_1\left(x\right)=\frac{x}{2}\left\{1-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^2}{1!2!}+\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^4}{2!3!}-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^6}{3!4!}+..........\right\}$

Similarly,

${\mathrm{J}}_1\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^{2\mathrm{n}+1}}{┌(\mathrm{n}+2)\mathrm{n}!}$

$=\frac{x}{2}\left\{1-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^2}{1!2!}+\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^4}{2!3!}-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^6}{3!4!}+..........\right\}$

Calculate to obtain ${\mathrm{J}}_{-1}\left(\mathrm{x}\right)=-{\mathrm{J}}_1\left(\mathrm{x}\right)\mathrm{and}{\mathrm{J}}_{-2}\left(\mathrm{x}\right)=-{\mathrm{J}}_2\left(\mathrm{x}\right)$

${\mathrm{J}}_{-\mathrm{p}}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^{2\mathrm{n}+\mathrm{p}}}{┌(\mathrm{n}-\mathrm{p}+1)\mathrm{n}!}$

$$\begin{gathered} {\mathrm{J}}_{-1}\left(\mathrm{x}\right)=-\left\{\frac{\mathrm{x}}{2}-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^3}{1!2!}+\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^4}{2!3!}-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^6}{3!4!}+..........\right\} \\ {\mathrm{J}}_{-1}\left(\mathrm{x}\right)=-{\mathrm{J}}_1\left(\mathrm{x}\right) \end{gathered}$$

Similarly, simplify further as follows:

${\mathrm{J}}_{-2}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^{2\mathrm{n}-2}}{\mathrm{n}!(\mathrm{n}-2)\mathrm{n}!}$

$=-\frac{4}{x^2}\left\{\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^4}{1!2!}+\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^6}{1!3!}-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^8}{1!4!}+..........\right\}$

$J_{-2}\left(x\right)={\left(\frac{x}{2}\right)}^2\left\{\frac{1}{2}-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^2}{1!3!}+\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^4}{2!4!}-\frac{{\left({\displaystyle \frac{\mathrm{x}}{2}}\right)}^6}{3!5!}+..........\right\}$

${\mathrm{J}}_{-2}\left(\mathrm{x}\right)=-{\mathrm{J}}_2\left(\mathrm{x}\right)$

Hence, it is proved that $J_{-1}\left(x\right)=-J_1\left(x\right)\mathrm{and}{\mathrm{J}}_{-2}\left(\mathrm{x}\right)=-{\mathrm{J}}_2\left(\mathrm{x}\right)$.