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${\mathbf{h}}_{\mathbf{n}}{\mathbf{\left(}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\right)}}^{\mathbf{"}}\mathbf{\left[}{\mathbf{x}}^{\mathbf{-}\mathbf{n}}{\mathbf{j}}_{\mathbf{n}}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^n\mathbf{\right]}\mathbf{"}\mathbf{=}{\mathbf{x}}^{\mathbf{-}\mathbf{n}}{\mathbf{j}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Short answer

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

Step-by-step solution

Concept of Spherical Bessel functions:

Spherical Bessel function is given as follows:

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

With:

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

Formula used:

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

Consider jn(x)=∑r=0∞(-1)rΓ(r+1)Γ(r+1+n)(x2)2r+n and simplify it       :

Consider and simplify the equation to obtain:

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

Put value of r - 1 = t in the equation:

Substitute for (r- 1) in equation to obtain:

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^{-\mathrm{n}}{\mathrm{j}}_{\mathrm{n}}\left(\mathrm{x}\right)\right)=\frac{1}{{\mathrm{x}}_{\mathrm{n}}}\sum _{\mathrm{t}=0}^{\infty }\frac{{(-1)}^{\mathrm{t}+1}\left(\mathrm{t}+1\right)}{\mathrm{\Gamma }(\mathrm{t}+1+1)\mathrm{\Gamma }(\mathrm{t}+1+1+\mathrm{n})}{\left(\frac{1}{2}\right)}^{2\mathrm{t}+\mathrm{n}+1}{\left(\mathrm{x}\right)}^{2\mathrm{t}+\mathrm{n}+1} \\ =\frac{1}{{\mathrm{x}}_{\mathrm{n}}}\sum _{\mathrm{t}=0}^{\infty }\frac{{(-1)}^{\mathrm{t}+1}\left(\mathrm{t}+1\right)}{(\mathrm{t}+1)\mathrm{\Gamma }(\mathrm{t}+1)\mathrm{\Gamma }(\mathrm{t}+1+1+\mathrm{n})}{\left(\frac{1}{2}\right)}^{2\mathrm{t}+\mathrm{n}+1}{\left(\mathrm{x}\right)}^{2\mathrm{t}+\mathrm{n}+1} \\ =-{\mathrm{x}}^{-\mathrm{n}}\sum _{\mathrm{t}=0}^{\infty }\frac{{(-1)}^{\mathrm{t}}}{\mathrm{\Gamma }(\mathrm{t}+1)\mathrm{\Gamma }(\mathrm{t}+1+1+\mathrm{n})}{\left(\frac{1}{2}\right)}^{2\mathrm{t}+\mathrm{n}+1}{\left(\mathrm{x}\right)}^{2\mathrm{t}+\mathrm{n}+1} \\ =-{\mathrm{x}}^{-\mathrm{n}}{\mathrm{j}}_{\mathrm{n}+1}\left(\mathrm{x}\right) \end{gathered}$$

Hence, it is proved.