- #1

yungman

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This is the equation given for the Y.

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

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

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

[tex]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}[/tex]

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

and [tex]J_{-n}(x)=-J_{n}(x), n=1,3,5...[/tex]

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:

[tex]\stackrel{lim}{p \rightarrow n}[/tex] [tex] 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...[/tex]

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

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

[tex]\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)}[/tex]

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

also what is [tex]\stackrel{lim}{x\rightarrow 0} x^{0}?[/tex]

Thanks for your time

Alan

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

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

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

[tex]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}[/tex]

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

and [tex]J_{-n}(x)=-J_{n}(x), n=1,3,5...[/tex]

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:

[tex]\stackrel{lim}{p \rightarrow n}[/tex] [tex] 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...[/tex]

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

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

[tex]\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)}[/tex]

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

also what is [tex]\stackrel{lim}{x\rightarrow 0} x^{0}?[/tex]

Thanks for your time

Alan

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