Question

Question: Use the Section 15 recursion relations and (17.4) to obtain the following recursion relations for spherical Bessel functions. We have written them for${\mathbf{j}}_{\mathbf{n}}$, but they are valid for${\mathbf{y}}_{\mathbf{n}}$and for the${\mathbf{h}}_{\mathbf{n}}\left(\frac{d}{\mathrm{dx}}\right)\mathbf{"}\left["x^{n+1}{j}_n{\left(x\right)}^{"}\right]\mathbf{=}{\mathbf{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{j}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\left(x\right)$

Short answer

The resultant answer is $\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}}\left(\mathrm{x}\right)\right)=\left({\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}-1}\left(\mathrm{x}\right)\right)$.

Step-by-step solution

Concept of Spherical Bessel functions:

The 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:

localid="1659265307597" $$\begin{gathered} {\mathbf{j}}_{\mathbf{n}}\left(x\right)\mathbf{=}{\mathbf{x}}^{\mathbf{n}}{\left(-\frac{1}{x}\frac{d}{dx}\right)}^{\mathbf{n}}\left(\frac{sin\left(x\right)}{x}\right) \\ {\mathbf{y}}_{\mathbf{n}}\left(x\right)\mathbf{=}\mathbf{-}{\mathbf{x}}^{\mathbf{n}}{\left(-\frac{1}{x}\frac{d}{\mathrm{dx}}\right)}^{\mathbf{n}}\left(\frac{\cos \left(x\right)}{x}\right) \end{gathered}$$

Formula used:

${\mathbf{j}}_{\mathbf{n}}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\left(-1\right)}^{\mathbf{r}}}{\mathbf{\Gamma }\left(r+1\right)\mathbf{\Gamma }\left(r+1+n\right)}{\left(\frac{x}{2}\right)}^{\mathbf{2}\mathbf{r}\mathbf{+}\mathbf{n}}$

Use the Spherical Bessel function to calculate the function:

Using the spherical Bessel function to calculate, obtain:

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}}\left(\mathrm{x}\right)\right]={\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}-1}\left(\mathrm{x}\right)\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}}\left(\mathrm{x}\right)\right) \\ =\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}+1}\sqrt{\frac{\mathrm{\pi }}{2\mathrm{x}}}{\mathrm{J}}_{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}\right)\right) \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}}\left(\mathrm{x}\right)\right]=\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}+\frac{1}{2}}\sqrt{\frac{\mathrm{\pi }}{2}}{\mathrm{J}}_{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}\right)\right) \\ =\sqrt{\frac{\mathrm{\pi }}{2}}\left({\mathrm{x}}^{\mathrm{n}+\frac{1}{2}}{\mathrm{J}}_{\mathrm{n}+\frac{1}{2}-1}\left(\mathrm{x}\right)\right) \end{gathered}$$

As known that,

$\left(\frac{d}{dx}\left(x^p{J}_p\left(x\right)\right)=x^p{J}_{p-1}\left(x\right)\right)$.

Therefore,

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}}\left(\mathrm{x}\right)\right]=\sqrt{\frac{\mathrm{\pi }}{2x}}\left({\mathrm{x}}^{\mathrm{n}+1}{\mathrm{J}}_{\mathrm{n}+\frac{1}{2}-1}\left(\mathrm{x}\right)\right) \\ =\left[{\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}-1}\left(\mathrm{x}\right)\right] \end{gathered}$$

Hence, it is proved that $\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}}\left(\mathrm{x}\right)\right)=\left({\mathrm{x}}^{\mathrm{n}+1}{\mathrm{j}}_{\mathrm{n}-1}\left(\mathrm{x}\right)\right)$.