Bessel's equation of the second kind with integer order.

[yungman]

This is the equation given for the Y.

Y_{p}=\frac{J_{p}(x)cos(p\pi)-P_{-p}(x)}{sin(p\pi)}

In many books, if p is an integer n, they just said Y_{n}=lim(p\rightarrow n) Y_{p}

J_{p}(x)=\sum^{k=0}_{\infty}\frac{(-1)^{k}}{k!\Gamma(k+p+1)}(\frac{x}{2})^{2k+p} which give

J_{n}(x)=\sum^{k=0}_{\infty}\frac{(-1)^{k}}{k!\Gamma(k+n+1)}(\frac{x}{2})^{2k+n}=\sum^{k=0}_{\infty}\frac{(-1)^{k}}{k!(k+n)!}(\frac{x}{2})^{2k+n}

For p=integer n, J_{-n}(x)=(-1)^{n}J_{n}(x) \Rightarrow J_{-n}(x)=J_{n}(x),n=0,2,4...

and J_{-n}(x)=-J_{n}(x), n=1,3,5...

No books that I have gave the answer for Y when p is an integer. This is what I am doing so far and I cannot get the right answer:

\stackrel{lim}{p \rightarrow n} Y_{p}=\frac{J_{p}(x)cos(p\pi)-P_{-p}(x)}{sin(p\pi)}=J_{n}(x)\frac{cos(p\pi)-1}{sin(p\pi)}, n=0,2,4,6...

=J_{n}(x)\frac{cos(p\pi)+1}{sin(p\pi)}, n=1,3,5...

So I can evaluate the lim by L'Hopital that \frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}

\frac{cos(p\pi)-1}{sin(p\pi)}=\frac{[cos(p\pi)-1]'}{[sin(p\pi)]'}=\frac{-\pi sin(p\pi)}{\pi cos(p\pi)}

If we take p to an integer, the whole thing become zero! I can't get Y_{n}


also what is \stackrel{lim}{x\rightarrow 0} x^{0}?

Thanks for your time

Alan

 

[yungman]

To suppliment to my question, the reason I asked the question is because I don't see from my approach, I can get the graph of the Y function which go to negative infinity when x goes to zero.

What I did is basically separate the J_{n}(x) out and look at the \frac{cos(p\pi)+1}{sin(p\pi)} alone.

J_{n}(x) equal to 1 for n=0 and 0 for n>0. So I cannot prove the Y functions go to negative infinity as x approach zero.


I've gone through at least 6 books and most of them have different formulas for Y. None explain how they come up with the formulas.

Two books suggested using the formula used in Cauchy Euler equation to get the second solution:

y_{2}(x)=J_{n}(x)ln(x)+x^{n}\sum^{\infty}_{1}b_{k}x^{k}

That was actually the reason I asked the question about the Euler equation yesterday to really understand how the equation come from and not just copy and use it.

Thanks for your time

Alan

 

[yungman]

Anybody please?
Thanks
Alan

 

[Mute]

I think the problem is that you're factoring out J_n(x) when you're not really allowed to do that. You can't generally selectively apply the limiting procedure to different parts of the function.

\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x\rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}

provided a) the top and bottom limits exist and b) the limit of the denominator is not zero. Because the denominator is zero in this case, you can't split the limit over top and bottom like that - you'll have to take the Bessel functions into account when you do L'Hopital's rule. Be careful, because you'll be taking derivatives with respect to the order of the Bessel function, not its argument. You can see the derivatives on this page

http://www.math.sfu.ca/~cbm/aands/page_362.htm

Note that it involves the digamma function.

 

[yungman]

This is the equation given for the Y.

Y_{p}=\frac{J_{p}(x)cos(p\pi)-J_{-p}(x)}{sin(p\pi)}

In many books, if p is an integer n, they just said Y_{n}=lim(p\rightarrow n) Y_{p}

J_{p}(x)=\sum^{k=0}_{\infty}\frac{(-1)^{k}}{k!\Gamma(k+p+1)}(\frac{x}{2})^{2k+p} which give

J_{n}(x)=\sum^{k=0}_{\infty}\frac{(-1)^{k}}{k!\Gamma(k+n+1)}(\frac{x}{2})^{2k+n}=\sum^{k=0}_{\infty}\frac{(-1)^{k}}{k!(k+n)!}(\frac{x}{2})^{2k+n}

For p=integer n, J_{-n}(x)=(-1)^{n}J_{n}(x) \Rightarrow J_{-n}(x)=J_{n}(x),n=0,2,4...

and J_{-n}(x)=-J_{n}(x), n=1,3,5...

No books that I have gave the answer for Y when p is an integer. This is what I am doing so far and I cannot get the right answer:

\stackrel{lim}{p \rightarrow n} Y_{p}=\frac{J_{p}(x)cos(p\pi)-J_{-p}(x)}{sin(p\pi)}=J_{n}(x)\frac{cos(p\pi)-1}{sin(p\pi)}, n=0,2,4,6...

=J_{n}(x)\frac{cos(p\pi)+1}{sin(p\pi)}, n=1,3,5...

So I can evaluate the lim by L'Hopital that \frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}

\frac{cos(p\pi)-1}{sin(p\pi)}=\frac{[cos(p\pi)-1]'}{[sin(p\pi)]'}=\frac{-\pi sin(p\pi)}{\pi cos(p\pi)}

If we take p to an integer, the whole thing become zero! I can't get Y_{n}


also what is \stackrel{lim}{x\rightarrow 0} x^{0}?

Thanks for your time

Alan

I type it wrong, there is no P_{-p}(x), it is J_{-p}(x).

 

[yungman]

I think the problem is that you're factoring out J_n(x) when you're not really allowed to do that. You can't generally selectively apply the limiting procedure to different parts of the function.

\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x\rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}

provided a) the top and bottom limits exist and b) the limit of the denominator is not zero. Because the denominator is zero in this case, you can't split the limit over top and bottom like that - you'll have to take the Bessel functions into account when you do L'Hopital's rule. Be careful, because you'll be taking derivatives with respect to the order of the Bessel function, not its argument. You can see the derivatives on this page

http://www.math.sfu.ca/~cbm/aands/page_362.htm

Note that it involves the digamma function.

I think in this case, the limit p approaching an integer, the limit of Y is zero/zero. that is the reason I use L'Hopital.

Do you mean applying L'Hopital of the whole equation by differentiate the top and bottom respect to p, not respect to x? This is what I am trying to do, is this correct? I am looking at the equations you posted and take the dirivative of J_p respect to p.

Y_{p}=\frac{J_{p}(x)cos(p\pi)-J_{-p}(x)}{sin(p\pi)}

J_{p}(x)=\sum^{k=0}_{\infty}\frac{(-1)^{k}}{k!\Gamma(k+p+1)}(\frac{x}{2})^{2k+p} and J_{-p}(x)=\sum^{k=0}_{\infty}\frac{(-1)^{k}}{k!\Gamma(k-p+1)}(\frac{x}{2})^{2k-p}


L'Hopital rule stated that:

\stackrel{lim}{p \rightarrow n}Y_{p}=\stackrel{lim}{p \rightarrow n}\frac{J_{p}(x)cos(p\pi)-J_{-p}(x)}{sin(p\pi)}=\stackrel{lim}{p \rightarrow n}\frac{\frac{d}{dp}[J_{p}(x)cos(p\pi)-J_{-p}(x)]}{\frac{d}{dp}sin(p\pi)}

then substitute n into p. I'll work on this tomorrow with your equation.
Thanks
Alan