Question

Use the table above and the definitions in Section 17 to find approximate formulas for large x for $h_n^{\left(2\right)}\left(ix\right)$:

Short answer

The approximate value for large value of x is $i^n{x}^{-1}{e}^x$.

Step-by-step solution

Take out the equations.

The given term is $h_n^{\left(2\right)}\left(ix\right)$.

$h_n^{\left(2\right)}\left(x\right)=j_n\left(x\right)-i{y}_n\left(x\right)$

As, $h_n^{\left(2\right)}\left(x\right)=j_n\left(x\right)-i{y}_n\left(x\right)$.

Therefore, $h_n^{\left(2\right)}\left(ix\right)=j_n\left(ix\right)-i{y}_n\left(ix\right)$ …… (1)

For larger values of x as follows:

$j_n\left(x\right)=\frac{1}{x}\sin \left(x-\frac{n\pi }{2}\right)+O\left(x^{-2}\right)$ …… (2)

$y_n\left(x\right)=-\frac{1}{x}\cos \left(x-\frac{n\pi }{2}\right)+O\left(x^{-2}\right)$ …… (3)

Use equations (1), (2), and (3) for calculation.

Using equation (1), (2), and (3) as follows:,

$$\begin{gathered} h_n^{\left(2\right)}\left(ix\right)=j_n\left(ix\right)-i{y}_n\left(ix\right) \\ =\frac{1}{ix}\sin \left(ix-\frac{n\pi }{2}\right)-i\left(-\frac{1}{ix}\cos \left(ix-\frac{n\pi }{2}\right)\right) \\ =\frac{1}{ix}\sin \left(ix-\frac{n\pi }{2}\right)+i\frac{1}{ix}\cos \left(ix-\frac{n\pi }{2}\right) \\ =\frac{-i^2}{ix}\sin \left(ix-\frac{n\pi }{2}\right)+\frac{1}{x}\cos \left(ix-\frac{n\pi }{2}\right) \end{gathered}$$

Simplify further as follows:

$$\begin{gathered} h_n^{\left(2\right)}\left(ix\right)=\frac{-i}{x}\sin \left(ix-\frac{n\pi }{2}\right)+\frac{1}{x}\cos \left(ix-\frac{n\pi }{2}\right) \\ =\frac{1}{x}\left[\cos \left(ix-\frac{n\pi }{2}\right)-i\sin \left(ix-\frac{n\pi }{2}\right)\right] \\ =x^{-1}\left[\cos \left(ix-\frac{n\pi }{2}\right)-i\sin \left(ix-\frac{n\pi }{2}\right)\right] \end{gathered}$$

As,$e^{-i\theta }=\cos \theta -i\sin \theta$

Therefore, calculate:

$$\begin{gathered} h_n^{\left(2\right)}\left(ix\right)=x^{-1}[\left.\cos \left(ix-\frac{n\pi }{2}\right)-i\sin \left(ix-\frac{n\pi }{2}\right)\right] \\ =x^{-1}{e}^{-i\left(ix-\frac{n\pi }{2}\right)} \\ =x^{-1}{e}^{\left(-i^{2}x+i\frac{n\pi }{2}\right)} \\ =x^{-1}{e}^{\left(x+i\frac{n\pi }{2}\right)}=x^{-1}{e}^x\left(e^{i\frac{n\pi }{2}}\right) \end{gathered}$$

Simplifying further to get:

$$\begin{gathered} h_n^{\left(2\right)}\left(ix\right)=x^{-1}{e}^x(i{)}^n \\ =i^n{x}^{-1}{e}^x \end{gathered}$$