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hasan_researc
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Homework Statement
This is the how the question begins.
1. Bessel's equation is [tex]z^{2}\frac{d^{2}y}{dz^{2}} + z\frac{dy}{dz} + \left(z^{2}- p^{2}\right)y = 0[/tex].
For the case [tex]p^{2} = \frac{1}{4}[/tex], the equation has two series solutions which (unusually) may be expressed in terms of elementary functions:
[tex]J_{1/2} = \left(\frac{2}{\pi z}\right)^{1/2} sin z[/tex]
[tex]J_{-1/2} = \left(\frac{2}{\pi z}\right)^{1/2} cos z[/tex]
[ The factors [tex]\left(\frac{2}{\pi z}\right)^{1/2}[/tex] are supefluous, but are included by convention, for reasons that are not relevant to the present purposes.]
Clearly [tex]J_{-1/2}[/tex] is singular at z = 0. Show that [tex]J_{1/2}(0) = 0[/tex].
2......(for later)
Homework Equations
The Attempt at a Solution
I am going to assume the solutions [tex]J_{1/2}[/tex] and [tex]J_{-1/2}[/tex] without worrying about why/how they come about.
Obviously, when z = 0, [tex]cos z \neq 0 [/tex]. Therefore, [tex]\left(\frac{2}{\pi z}\right)^{1/2}[/tex] blows up and [tex]J_{-1/2}[/tex] is singular at z = 0.
On the other hand, if we draw separately the graphs of [tex]\left(\frac{2}{\pi z}\right)^{1/2}[/tex] and [tex] sin z [/tex] and then combine the two in a single graph of [tex]J_{1/2}[/tex], we find that it is sinusoidal with an amplitude given by [tex]\left(\frac{2}{\pi z}\right)^{1/2}[/tex]. This means that the curve oscillates as it moves towards z = 0 with an amplitude that tends to infinity as z tends to 0. How do I conclude from this that [tex]J_{1/2}(0) = 0[/tex].