Question

Using (17.3) and (15.1) to (15.5), find the recursion relations for${\mathit{I}}_{\mathbf{p}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ . In particular, show that${\mathit{I}}_{\mathbf{0}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{I}}_{\mathbf{1}}$ .

Short answer

The equation $I_0^{\text{'}}\left(x\right)=I_1\left(x\right)$ is proved.

Step-by-step solution

A concept of recursion relation:

A recursion relation is an equation that defines a sequence based on a rule that gives the next term as a function of the previous terms for some function.

Compare the given equation:

On comparing the given equation, obtain:

$$\begin{gathered} \frac{d}{dx}{\left(x\right)}^p{l}_p\left(x\right)={\left(i\right)}^{-p}{J}_P\left(ix\right) \\ =\sum _{n-0}^{\infty }\frac{1}{\sqrt{\overline{)n+1}n+1+p}}{\left(\frac{x}{2}\right)}^{2n+p} \\ \frac{d}{dx}{\left(x\right)}^p{l}_p\left(x\right)=\frac{d}{dx}\sum _{n-0}^{\infty }\frac{{\left(x\right)}^{2n+p}}{\sqrt{\overline{)n+1}n+1+p}}\left(\frac{1}{2^{2n-p}}\right) \\ =\sum _{n-0}^{\infty }\frac{\left(n+p\right){\left(x\right)}^{2n+p-}}{\sqrt{n+1}\ln +1+p}\left(\frac{1}{2^{2n-p}}\right) \\ ={\left(x\right)}^p\sum _{n-0}^{\infty }\frac{\left(n+p\right){\left(x\right)}^{2n+p-}}{\sqrt{\sqrt{n+1}\ln +p-1+1}}{\left(\frac{x}{2}\right)}^{2n-p} \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} \frac{d}{dx}\left(x{)}^p{I}_p\right(x)=(x{)}^p{I}_{p-1}\left(x\right) \\ \frac{d}{dx}\left(x{)}^p{I}_p\right(x)=(x{)}^{-p}{I}_{p+1}\left(x\right) \\ \frac{d}{dx}\left(x{)}^p{I}_p\right(x)=(x{)}^{-p}{I}_{p-1}\left(x\right) \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} \left(x{)}^p{I}_p^{\text{'}}\right(x)+p(x{)}^{p-1}{I}_p\left(x\right)=\left(x{)}^p{I}_{p-1}\right(x\left) \\ {I}_p^{\text{'}}\right(x)=I_{p-1}(x)-\left(\frac{p}{X}\right)I_p(x) \\ \frac{d}{DX}/(x{)}^{-P}{I}_p\left(x\right)=(x{)}^{-P}-I_{P+}(x) \\ {\left(x\right)}^{-p}{I}_p\left(x\right)+p{\left(x\right)}^{-p-1}{I}_p\left(x\right)={\left(x\right)}^{-p}{I}_{p+1}\left(x\right) \end{gathered}$$ .....(1)

Therefore,

$I_p^{\text{'}}\left(x\right)=I_{p+1}\left(x\right)+\left(\frac{p}{x}\right)I_p\left(x\right)$ …… (2)

Add and subtract equation (1) and (2):

Adding equation (1) and (2) as follows:

$l_{p+1}^{\text{'}}\left(x\right)=I_{p-1}\left(x\right)-2\left(\frac{p}{x}\right)I_p\left(x\right)$

Subtracting equation (1) and (2) as follows:

$l_{p+1}^{\text{'}}\left(x\right)=I_{p-1}\left(x\right)-2\left(\frac{p}{x}\right)I_p\left(x\right)$

Put   in equation (2):

Putting $p=0$ in equation (2) as follows:

$$\begin{gathered} I_0^{\text{'}}\left(x\right)-I_1\left(x\right)=0 \\ {I}_0^{\text{'}}\left(x\right)=I_1\left(x\right) \end{gathered}$$

Hence, proved.