Question

Use the series in (7)to verify that$I_v\left(x\right)=i^{-v}{J}_V\left(ix\right)$is a real function.

Short answer

$I_v\left(x\right)=i^{-v}{J}_V\left(ix\right)$ is a real function is verified.

Step-by-step solution

Define Bessel’s equation.

Let the Bessel equation be${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\left(x^2-v^2\right)\mathit{y}\mathbf{=}\mathbf{0}$. A Bessel equation of the first kind, by using the coefficients c₂₀ and r = v, indicated by J_n(x) , is,

$J_v\left(x\right)=\sum _{n=0}^x\frac{{(-1)}^n}{n!\Gamma (1+v+n)}{\left(\frac{x}{2}\right)}^{2n+v}$

Derive a required function.

To verify the formula, $I_v\left(x\right)=i^{-v}{J}_V\left(ix\right)$…(1)

Which does not involve complex variables but gives a real function. Then, the series is,

$$\begin{gathered} J_v\left(x\right)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!\Gamma (1+v+n)}{\left(\frac{x}{2}\right)}^{2n+v} \\ {J}_v\left(ix\right)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!\Gamma (1+v+n)}{\left(\frac{ix}{2}\right)}^{2n+v}.......2 \end{gathered}$$

Derive for the real function.

Substitute the equation (2) into (1) which yields,

$$\begin{gathered} I_v\left(x\right)=i^{-v}\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!\Gamma (1+v+n)}{\left(\frac{ix}{2}\right)}^{2n+v} \\ =i^{-v}\sum _{n=0}^{\infty }\frac{{(-1)}^n\times i^{2n+v}}{n!\Gamma (1+v+n)}{\left(\frac{x}{2}\right)}^{2n+v} \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n\times i^{2n+x-x}}{n!\Gamma (1+v+n)}{\left(\frac{x}{2}\right)}^{2n+v} \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n\times {(-1)}^n}{n!\Gamma (1+v+n)}{\left(\frac{x}{2}\right)}^{2n+v}\left[\because i^2=-1\right] \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^{2n}}{n!\Gamma (1+v+n)}{\left(\frac{x}{2}\right)}^{2n+v} \end{gathered}$$

This implies that, $I_v\left(x\right)$is a real function of all values of n. Hence verified.