Question

From (17.4), show that,${\mathbf{h}}_{\mathbf{n}}^{\left(1\right)}\left(\mathrm{ix}\right)\mathbf{=}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}\mathbf{/}\mathbf{x}$ .

Short answer

The figures show the behavior of Bessel functions of different kinds.

Step-by-step solution

Concept of Spherical Bessel functions:

Spherical Bessel function is given as follows:

${\mathbf{h}}_{\mathbf{n}}^{\left(1\right)}\left(x\right)\mathbf{=}{\mathbf{j}}_{\mathbf{n}}\left(x\right)\mathbf{+}{\mathbf{iy}}_{\mathbf{n}}\left(x\right)$

With:

$$\begin{gathered} {\mathbf{j}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}{\mathbf{x}}^{\mathbf{n}}{\left(-\frac{1}{x}\frac{d}{\mathrm{dx}}\right)}^{\mathbf{n}}\left(\frac{\sin \left(x\right)}{x}\right) \\ {\mathbf{y}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{n}}{\mathbf{(}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{)}}^{\mathbf{n}}\mathbf{\left(}\frac{\mathbf{cos}\left(x\right)}{\mathbf{x}}\mathbf{\right)} \\ \mathbf{Formula}\mathbf{}\mathbf{used}\mathbf{:} \\ {\mathbf{j}}_{\mathbf{n}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\left(-1\right)}^{\mathbf{r}}}{\mathbf{\Gamma }\left(r+1\right)\mathbf{\Gamma }\left(r+1+n\right)}{\left(\frac{x}{2}\right)}^{\mathbf{2}\mathbf{r}\mathbf{+}\mathbf{n}} \end{gathered}$$

Calculate the equation by the Spherical Bessel function:

Using the Spherical Bessel function to obtain:

$$\begin{gathered} {\mathrm{h}}_{\mathrm{n}}^{\left(1\right)}\left(\mathrm{x}\right)={\mathrm{j}}_{\mathrm{n}}\left(\mathrm{x}\right)+{\mathrm{iy}}_{\mathrm{n}}\left(\mathrm{x}\right) \\ ={\mathrm{h}}_{\mathrm{n}}^{\left(1\right)}\left(\mathrm{ix}\right) \\ ={\mathrm{j}}_{\mathrm{n}}\left(\mathrm{ix}\right)+{\mathrm{iy}}_{\mathrm{n}}\left(\mathrm{ix}\right) \\ ={\mathrm{h}}_0^{\left(1\right)}\left(\mathrm{ix}\right) \\ \mathrm{Simplify}\mathrm{further}\mathrm{as}\mathrm{follows}: \\ {\mathrm{h}}_{\mathrm{n}}^{\left(1\right)}\left(\mathrm{x}\right)={\mathrm{j}}_{\mathrm{n}}\left(\mathrm{ix}\right)+{\mathrm{iy}}_0\left(\mathrm{ix}\right) \\ =\left[\frac{\sin \mathrm{ix}}{\mathrm{ix}}-\mathrm{i}\frac{\cos \mathrm{ix}}{\mathrm{ix}}\right] \\ =\left[\frac{\sinh \mathrm{x}}{\mathrm{x}}-\mathrm{i}\frac{\cosh \mathrm{x}}{\mathrm{x}}\right] \\ =-\frac{{\mathrm{e}}^{-\mathrm{x}}}{\mathrm{x}} \\ \mathrm{Hence},{\mathrm{h}}_0^{\left(1\right)}\left(\mathrm{ix}\right)=-\frac{{\mathrm{e}}^{-\mathrm{x}}}{\mathrm{x}}. \end{gathered}$$