Question

Prove the given system of equations ${\mathit{I}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\surd }\mathbf{\left(}\frac{\mathbf{2}}{\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{\right(}\mathit{x}\mathbf{)}$and${\mathit{K}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\surd }\mathbf{(}\frac{\mathbf{\pi }}{\mathbf{2}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathit{e}}^{\mathbf{(}\mathbf{-}\mathbf{x}\mathbf{)}}\mathbf{.}$

Short answer

The given system of equations ${\mathrm{K}}_{1/2}\left(\mathrm{x}\right)=\surd (\frac{\mathrm{\pi }}{2\left(\mathrm{x}\right)}{\mathrm{e}}^{(-\mathrm{x})}$ and $I_{1/2}\left(x\right)=\surd \left(\frac{2}{\mathrm{\pi }\left(\mathrm{x}\right)}sinh\right(x)$ are proved.

Step-by-step solution

Concept of recursion relation:

The recursion relation is given as follows:

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(x^{-p}{J}_p\left(x\right)\right)\mathbf{=}\mathbf{-}{\mathit{x}}^{\mathbf{-}\mathbf{1}}{\mathit{J}}_{\mathbf{p}}\mathbf{+}\mathbf{1}\mathbf{\left(}\mathit{z}\mathbf{\right)}$

Prove the given system of equations 1/2(x)=√(2π(x)sinh(x)  and K1/2(x)=√(π2(x)e(-x).

Consider the given equations as below.

$K_{1/2}\left(x\right)=\surd (\frac{\mathrm{\pi }}{2\left(\mathrm{x}\right)}{e}^{(-\mathrm{x})}.$

And

$I_{1/2}\left(x\right)=\surd \left(\frac{2}{\mathrm{\pi }\left(\mathrm{x}\right)}sinh\right(x)$

Compare the given equation with standard equation as follows:

$$\begin{gathered} K\left(x\right)={\left(i\right)}^{-p}{J}_P\left(ix\right) \\ =\frac{\pi }{2}{\left(i\right)}^{p+1}{H}_P\left(ix\right) \\ =\frac{\pi }{2}{\left(i\right)}^{p+1}\left[J_P\left(ix\right)+i{N}_P\left(IX\right)\right] \end{gathered}$$

Simplify further as follows:

role="math" localid="1659266495025" $$\begin{gathered} l_{1l2}\left(x\right)=i^{\left(-\frac{1}{2}\right)}{l}_{1l2}\left(ix\right) \\ =i^{\left(-\frac{1}{2}\right)}\sqrt{\frac{2}{\pi i\left(x\right)}}\sin \left(ix\right) \\ =\sqrt{\frac{2}{\pi i\left(x\right)}}\sin \left(ix\right) \\ {k}_{\frac{1}{2}}\left(x\right)=\frac{\pi }{2}{\left(i\right)}^{\frac{3}{2}}\left(\sqrt{\frac{2}{\pi i\left(x\right)}}\sin \left(x\right)-\left(i\right){\left(i\right)}^{-\frac{1}{2}}\sqrt{\frac{2}{\pi i\left(x\right)}}\cos i\left(x\right)\right) \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} k_{\frac{1}{2}}\left(x\right)=\sqrt{\frac{\pi }{2\left(x\right)}}\left(\cos h\left(x\right)-\sin h\left(x\right)\right) \\ =\sqrt{\frac{\pi }{2\left(x\right)}}{\left(e\right)}^{-x}{J}_{\frac{1}{2}}\left(ix\right)+i{Y}_{\frac{1}{2}}\left(ix\right))k_{\frac{1}{2}}\left(x\right) \\ =\sqrt{\frac{\pi }{2\left(x\right)}}{\left(e\right)}^{-x} \end{gathered}$$

Hence, it proved.