Question

We obtained (19.10) for${\mathit{J}}_{\mathbf{p}}\left(x\right)\mathbf{,}\mathit{p}\mathbf{\ge }\mathbf{0}\mathbf{.}$It is, however, valid for$\mathit{p}\mathbf{\ge }\mathbf{-}\mathbf{1}$, that is for${\mathit{N}}_{\mathbf{p}}\left(x\right)\mathbf{,}\mathbf{0}\mathbf{\le }\mathit{p}\mathbf{<}\mathbf{1}$. The difficulty in the proof occurs just after (19.7); we said that $\mathit{u}\mathbf{,}\mathit{v}\mathbf{,}\mathit{u}\mathbf{\text{'}}\mathbf{,}\mathit{v}\mathbf{\text{'}}$are finite at $\mathit{x}\mathbf{=}\mathbf{0}$which is not true for${\mathit{N}}_{\mathbf{p}}\left(x\right)$.

However, the negative powers of x cancel if$\mathit{p}\mathbf{<}\mathbf{1}$. Show this for $\mathit{p}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$by using two terms of the power series (12.9) or (13.1) for the function ${\mathit{N}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{-}{\mathit{J}}_{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}\left(x\right)$ [see (13.3)].

Short answer

Answer

The answer is stated below.

Step-by-step solution

The concept of the power series

In mathematics, a power series (in one variable) is an infinite series of the form

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}{\left(x-c\right)}^{\mathbf{n}}\mathbf{=}{\mathit{a}}_{\mathbf{0}}\mathbf{+}{\mathit{a}}_{\mathbf{1}}\left(x-c\right)\mathbf{+}{\mathit{a}}_{\mathbf{2}}{\left(x-c\right)}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}$

Where,represents the coefficient of theterm and is a constant.

Use the power series for calculation

The differential equation satisfied by, and the differential equation satisfied by.

Recall that,.

Then:

$$\begin{gathered} N_{1/2}\left(x\right)=\frac{\cos \left(\mathrm{\pi }{\displaystyle \frac{1}{2}}\right)j_{1/2}\left(x\right)-J_{-1/2}\left(x\right)}{\sin \left(\mathrm{\pi }{\displaystyle \frac{1}{2}}\right)} \\ =\frac{0-J_{1/2}\left(x\right)-J_{-12}\left(x\right)}{1} \\ =-J_{-1/2}\left(x\right) \\ {J}_{-1/2}\left(x\right)=-N_{1/2}\left(x\right) \end{gathered}$$

Assume. The differential equations satisfied by these two functions are:

Recall and integrate

Recall that $J_p\left(x\right)=\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{\left(n+1\right)\left(n+1+p\right)}{\left(\frac{x}{2}\right)}^{2n+p}$and integrate this equation to get:

Recall that, becauseis root of. Then, the above equation becomes:

Then,.

Now, let.

Then,becomes the indeterminate form.

So, obtain:

Now, it can be shown as:

Therefore, it can be written as: