Talk:Bessel function

From Wikipedia, the free encyclopedia

You have new messages (last change).
Other languages WikiProject Echo has identified Bessel function as a foreign language featured article. You may be able to improve this article with information from the Chinese language Wikipedia.

Contents

[edit] Error in Definition?

I noticed that Bessel's function of the first kind defined alpha as having to be integer or non-negative while Bessel's function of teh second kind does not mention any restriction about alpha. I just checked with Mathworld and it seems they don't mention that "integer or nonnegative" part about teh parameter alpha for functions of the first kind. Where does that restriction even come from? I'm not so sure about that part of the definition for the functions of the first kind.

—The preceding unsigned comment was added by 128.32.146.75 (talk) 23:52, 18 January 2007 (UTC).

The article does not say that J is only defined for alpha an integer or nonnegative. It says that J is only finite at x=0 for alpha an integer or nonnegative, which is true. I've changed the article in an attempt to clarify this. —Steven G. Johnson 03:12, 19 January 2007 (UTC)

[edit] cylindrical alpha and spherical alpha

I am a student of Quantum Physics and I believe the article is wrong hwne it asserts that alpha is an integer for cylindrical problems and a half integer (n + 1/2) for spherical problems. In fact, in the homework problem I am doing right now, I have derived that for cylindrical problems alpha is a half integer, and in "Introduction to Quantum Mechanics" by David J. Griffiths it proves that spherical problems involve integer alphas. DavidGrayson 03:22, 5 May 2006 (UTC)

You might want to re-check your homework problem before you turn it in. —Steven G. Johnson 03:34, 5 May 2006 (UTC)

[edit] asymptotic conditions

Nice article. I think there's a small problem in the section on asymptotic forms. Currently we say that the solutions are defined for 0 < x << 1 and x >> 1, but I believe this should be 0 < x << α and x >> α?

I don't think so. The notation x<<1 just means a "small value, much smaller than one"; the value doesn't need to be smaller than α even if α is small. FOr the other limit, its possible that maybe one might need to have x be larger than α when alpha is very large ... not clear, can you supply an argument to defend this? linas 01:34, 2 Apr 2005 (UTC)

Hmmm, I think if it's valid for x >> α, it should be valid for 0 < x << α too. Because it might not have to be less than one if alpha is large. I believe what put me on to this was: http://www.library.cornell.edu/nr/bookcpdf/c6-5.pdf --Chinasaur 03:49, 12 Apr 2005 (UTC)

Actually, looking at that .pdf it seems pretty clear that it should be relative to alpha, so I'm going to go ahead and change it. Sorry to be unbold, but I don't really know what I'm talking about here. But if the book says so it's probably right. You can change it back if somehow the book is wrong... --Chinasaur 03:52, 12 Apr 2005 (UTC)

Hi Chinasaur, the reference you cite is incorrect. Here's a homework problem: Let α=1000000 and let x=1000 Then clearly one has that x is much much less than α, i.e. x << α However, the asymptotic formula does not hold in this case. Look at the reference you just cited. Look at equation 6.5.1, and plug in these numbers; you will not get 6.5.3 in this case. Clearly, that reference is incorrect; the author or typesetter made a mistake there. I'm going to revert this edit.linas 14:37, 13 Apr 2005 (UTC)

Very good to know, and thanks for checking it for us since I won't have time to become comfortable with this math for at least a month. I'm surprised that Numerical Recipes would have this wrong, but I'll check it against Abramowitz and Stegun when I get the chance; refering to text books can be my contribution for now... --Chinasaur 05:04, 20 Apr 2005 (UTC)

I respectfully submit that I believe Chinasaur, and The Numerical Recipes book (Press, Teukolsky, Vetterling, and Flannery), are correct. So I have set the switchover point back to α. I hope I'm right and am not merely engaging in a reversion war. My understanding is that Jα behaves like xα out to about α, and then switches over to be a slowly damped (1/sqrt(x)) sine/cosine function after that. So J1000000 is very flat out to about 1000000, and then suddenly rises and turns into the sine/cosine form. I haven't found a really clear statement of this, but Abramowitz and Stegun indicate that the first zero (for large α) is somewhere after x=α, and Hildebrand Advanced Calculus for Applications makes an argument for the sine/cosine form for x >> α. Now all of this assumes that α is large -- it doesn't make a lot of sense to say x << α when α is 2. Maybe there should be a note clarifying this. I also made a similar change for the modified Bessel functions I and K, though I'm not very familiar with them. --William Ackerman 18:12, 31 January 2006 (UTC)

These limits are clearly wrong as α approaches zero. See also linas' counterexample above for the small-x, large-α case. For the large-x large-α case, I did some numerical checking with Matlab and the bounds of validity clearly seem to go as α2 rather than as α; this confirms the Arfken & Weber reference below.
If you look in Arfken and Weber, the condition that they give for Jα is that x \ll \sqrt{\alpha+1}. (Actually, they give the analogous formula for spherical Bessel functions, but the J formula is equivalent by a shift of 1/2 in α.) You can actually derive this condition quite easily by comparing the first two terms in the power-series expansion for Jα.
Similarly, if you compare the first two terms in the power series for Yα you find the requirement that x \ll \sqrt{|\alpha-1|} ... however, Yα is a bit screwy because you have to handle α near 1 specially. In particular, from the series expansion it looks like there is the additional condition that x^{2\alpha} \ln x \ll \Gamma(\alpha) \Gamma(\alpha+1) that becomes significant near α=1. Combining these two, it looks like it is sufficient to use x \ll \sqrt{\alpha+1} everywhere. (References would be appreciated, however.)
Conversely, Arken and Weber explicitly give the condition for the large-x expansion of J to be correct, and what they give is 8x \gg |4\alpha^2 - 1|, which could be simplified to x \gg |\alpha^2-1/4|. In other words, the condition for the validity of the large-x expansion is not simply the reverse of the condition for the small-x expansion. The large-x conditions on the other functions follow from the same considerations and are identical as far as I can tell.
In other words, both previous versions of the article seem wrong; I've edited the article with the above bounds. —Steven G. Johnson 23:57, 31 January 2006 (UTC)
I did some MatLab fiddling and also decided that both old suggestions are wrong. Sounds like you have gotten to the bottom of it. --Chinasaur 01:37, 22 March 2006 (UTC)

[edit] Taylor Lambda function

Hi, slight consern over the Taylor series expansion equation. I was attempting to find it (so that I could use the lower terms for a partial convolution), but the equation this page has has a Lambda(m+alpha+1) in it without explaining what the Lambda function. After a lot of searching, I found the same equation, but the "Lambda(m+alpha+1)" term was replaced with "(m+alpha)!". If anyone can find a source for this equation and update the picture, it would be helpful for people in the future. I would do it, but I don't know how to upload pics.--Rayc 23:28, 15 November 2005 (UTC)

I suspect you are referring to Γ(m + α + 1), the gamma function, which is a generalization of the factorial. I added a link in the article. linas 23:51, 15 November 2005 (UTC)
Thanks. I really should of known that, seeing as how this is for my master's degree. Though, that means that a normal person without 6 years of math would be even more confused.--Rayc 23:57, 15 November 2005 (UTC)

[edit] I_n plot

The I_n plot lines are incorrectly labelled as J_n (I think!) ... but I can't fix it since I don't have matlab (and don't have time to implement it fully in gnuplot). --Russell E 02:53, 25 January 2006 (UTC)

I put a request on the author's talk page: User talk:Alejo2083. —Steven G. Johnson 03:15, 25 January 2006 (UTC)
I fixed it :-) Alessio Damato 17:37, 26 January 2006 (UTC)

[edit] Recurrence formulas

It would be nice if somebody add paragraph about recurrence formulas for Bessel polynomials —The preceding unsigned comment was added by 129.2.175.16 (talk) 22:07, 6 December 2006 (UTC).

[edit] Modified bessel function of the second kind

I think the modified bessel function of the second kind was lacking an e^(-i*alpha*pi) in fornt of the I_alpha(ix), so I changed it. Is that correct? Margarin 15:54, 17 January 2007 (UTC)

No, the previous version was correct:
K_n(x) = \frac{\pi}{2} i^{n+1} H^{(1)}_n(ix)
= \frac{\pi}{2} i^{n+1} \left[ J_n(ix) + i Y_n(ix) \right]
= \frac{\pi}{2} i^{n+1} \left[ J_n(ix) + i \frac{J_n(ix) \cos(n\pi) - J_{-n}(ix)}{\sin(n\pi)} \right]
= \frac{\pi}{2 \sin(n\pi)} i^{n+1} \left[ J_n(ix) (\sin(n\pi) + i\cos(n\pi)) - i J_{-n}(ix) \right]
= \frac{\pi}{2 \sin(n\pi)} i^{n+1} \left[ J_n(ix) i e^{-in\pi} - i J_{-n}(ix) \right]
= \frac{\pi}{2 \sin(n\pi)} i^{n+1} \left[ i^{n+1} I_n(x) e^{-in\pi} + (-i)^{n+1} I_{-n}(x) \right]
= \frac{\pi}{2 \sin(n\pi)} \left[ i^{2(n+1)} I_n(x) e^{-in\pi} + I_{-n}(x) \right]
= \frac{\pi}{2 } \frac{I_{-n}(x) - I_n(x)}{\sin(n\pi)}
The key point was that, in the last step, i^{2(n+1)} e^{-in\pi} = e^{i\pi (n+1 - n)} = -1 \,
—Steven G. Johnson 18:36, 17 January 2007 (UTC)

[edit] Properties

Regarding the recurrence relations for the modified Bessel functions, one has

C_{\alpha-1}(x) - C_{\alpha+1}(x) = \frac{2\alpha}{x} C_\alpha(x)
C_{\alpha-1}(x) + C_{\alpha+1}(x) = 2\frac{dC_\alpha}{dx}

where Cα denotes Iα or eαπiKα (and not Kα as in the article).

DAKAgreg 14:32, 5 February 2007 (UTC)

[edit] Bessel functions of the second kind

The formula seems to be wrong, as it's stated as Y_{\alpha} and the formula uses n. Albmont 17:53, 27 March 2007 (UTC)

I assume you are referring to the integral form. You're right, it should be fixed now, and matches Abramowitz and Stegun p. 360. (I'm not sure this is the most useful formula in the world, on the other hand.) —Steven G. Johnson 20:25, 27 March 2007 (UTC)
Retrieved from "http://en.wikipedia.org../../../b/e/s/Talk%7EBessel_function_a23b.html"