Question

Show that${\mathbf{J}}_{\mathbf{P}}\left(x\right)\mathbf{N}{\mathbf{\text{'}}}_{\mathbf{P}}\left(x\right)\mathbf{-}\mathbf{J}{\mathbf{\text{'}}}_{\mathbf{P}}\left(x\right){\mathbf{N}}_{\mathbf{P}}\left(x\right)\mathbf{=}\frac{\mathbf{J}{\mathbf{\text{'}}}_{\mathbf{P}}\left(x\right){\mathbf{J}}_{\mathbf{-}\mathbf{p}}\left(x\right)\mathbf{-}{\mathbf{J}}_{\mathbf{p}}\left(x\right)\mathbf{j}{\mathbf{\text{'}}}_{\mathbf{-}\mathbf{p}}\left(x\right)}{\mathbf{sinp\pi }}\mathbf{=}\frac{\mathbf{2}}{\mathbf{\pi x}}$.

Short answer

The resultant answer is${\mathrm{J}}_{\mathrm{P}}\left(\mathrm{x}\right)\mathrm{N}{\text{'}}_{\mathrm{P}}\left(\mathrm{x}\right)-\mathrm{J}{\text{'}}_{\mathrm{P}}\left(\mathrm{x}\right){\mathrm{N}}_{\mathrm{P}}\left(\mathrm{x}\right)=\frac{2}{\mathrm{\pi x}}$ .

Step-by-step solution

Concept of Equation (13.3):

Equation (13.3):

$$\begin{gathered} {\mathbf{N}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{cos}\mathbf{}{\mathbf{\pi pJ}}_{\mathbf{P}}\left(x\right)\mathbf{-}{\mathbf{J}}_{\mathbf{-}\mathbf{p}}\left(x\right)}{\mathbf{sin\pi p}} \\ \mathbf{N}{\mathbf{\text{'}}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{cos}\mathbf{}\mathbf{\pi pJ}{\mathbf{\text{'}}}_{\mathbf{P}}\left(x\right)\mathbf{-}\mathbf{J}{\mathbf{\text{'}}}_{\mathbf{-}\mathbf{p}}\left(x\right)}{\mathbf{sin\pi p}} \end{gathered}$$

Consider the equation and simplify it:

By considering the equation and simplifying it, obtain:

$$\begin{gathered} {\mathrm{J}}_{\mathrm{P}}\left(\mathrm{x}\right)\mathrm{N}{\text{'}}_{\mathrm{P}}\left(\mathrm{x}\right)-\mathrm{J}{\text{'}}_{\mathrm{P}}\left(\mathrm{x}\right){\mathrm{N}}_{\mathrm{P}}\left(\mathrm{x}\right)=\frac{\cos {\mathrm{\pi pJ}}_{\mathrm{P}}\left(\mathrm{x}\right)\mathrm{J}{\text{'}}_{\mathrm{p}}\left(\mathrm{x}\right)-{\mathrm{J}}_{\mathrm{P}}\left(\mathrm{x}\right)\mathrm{J}{\text{'}}_{-\mathrm{p}}\left(\mathrm{x}\right)}{\mathrm{sin\pi p}}-\frac{\cos \mathrm{\pi pJ}{\text{'}}_{\mathrm{P}}\left(\mathrm{x}\right){\mathrm{J}}_{\mathrm{p}}\left(\mathrm{x}\right)-\mathrm{J}{\text{'}}_{\mathrm{P}}\left(\mathrm{x}\right){\mathrm{J}}_{-\mathrm{p}}\left(\mathrm{x}\right)}{\mathrm{sin\pi p}} \\ =\frac{\mathrm{J}{\text{'}}_{\mathrm{P}}{\mathrm{J}}_{-\mathrm{P}}-\mathrm{J}{\text{'}}_{-\mathrm{P}}{\mathrm{J}}_{\mathrm{P}}}{\mathrm{sin\pi p}} \\ =\frac{{\displaystyle \frac{2\sin xp}{xx}}}{\mathrm{\pi x}} \end{gathered}$$Hence, it is proved that${\mathrm{J}}_{\mathrm{P}}\left(\mathrm{x}\right)\mathrm{N}{\text{'}}_{\mathrm{P}}\left(\mathrm{x}\right)-\mathrm{J}{\text{'}}_{\mathrm{P}}\left(\mathrm{x}\right){\mathrm{N}}_{\mathrm{P}}\left(\mathrm{x}\right)=\frac{2}{\mathrm{\pi x}}$.