Question

To show that, $\sqrt{\mathbf{\tau \tau x}\mathbf{/}}\mathbf{2}{\mathbf{J}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{sin}\mathbf{}\mathbf{x}$.

Short answer

Hence, $\sqrt{\frac{\mathrm{\pi x}}{2}}{\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{x}\right)=\sin \mathrm{x}$is proved.

Step-by-step solution

Concept of the Infinite series:

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 ττx/2J1/2(x) :

Considering the provided statement to show that the provided infinite series for ${\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)$converges for all the values of x by the ratio test.

The infinite series is,

$\sqrt{\frac{\pi x}{2}}{J}_{\frac{1}{2}}\left(x\right)=\sin x$

Substituting the series for $\frac{1}{2}$ for p in the equation (1) as follows:

$$\begin{gathered} \sqrt{\frac{\mathrm{\pi x}}{2}}{\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{\pi x}}{2}}\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜(\mathrm{n}+1)⌜(\mathrm{n}+{\displaystyle 1+\frac{1}{2}})}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{\pi }+\frac{1}{2}} \\ =\sqrt{\frac{\mathrm{\pi x}}{2}}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{{\displaystyle \mathrm{n}!\left(\mathrm{n}+\frac{1}{2}\right)\left(\mathrm{n}+\frac{{\displaystyle 1}}{{\displaystyle 2}}-1\right)..........\frac{1}{2}⌜\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)2^{\mathrm{\pi }}\sqrt{2}}}{\mathrm{x}}^{2\mathrm{n}} \\ =\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{\left(2\mathrm{n}+1\right)!}{\mathrm{x}}^{2\mathrm{n}+1} \\ =\sin \mathrm{x} \end{gathered}$$

$$\begin{gathered} \sqrt{\frac{\mathrm{\pi x}}{2}}{\mathrm{J}}_{\frac{1}{2}}\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{\pi x}}{2}}\sum _{\mathrm{n}=0}^{\infty }\frac{{(-1)}^{\mathrm{n}}}{⌜(\mathrm{n}+1)⌜(\mathrm{n}+{\displaystyle 1+\frac{1}{2}})}{\left(\frac{\mathrm{x}}{2}\right)}^{2\mathrm{\pi }+\frac{1}{2}} \\ =\sqrt{\frac{\mathrm{\pi x}}{2}}\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{{\displaystyle \mathrm{n}!\left(\mathrm{n}+\frac{1}{2}\right)\left(\mathrm{n}+\frac{{\displaystyle 1}}{{\displaystyle 2}}-1\right)..........\frac{1}{2}⌜\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)2^{\mathrm{\pi }}\sqrt{2}}}{\mathrm{x}}^{2\mathrm{n}} \\ =\sum _{\mathrm{n}=0}^{\infty }\frac{{\left(-1\right)}^{\mathrm{n}}}{\left(2\mathrm{n}+1\right)!}{\mathrm{x}}^{2\mathrm{n}+1} \\ =\sin \mathrm{x} \end{gathered}$$

Hence,$\sqrt{\frac{\pi x}{2}}{J}_{\frac{1}{2}}\left(x\right)=\sin x\mathrm{is}\mathrm{proved}$