Talk:Bessel function

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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 the 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)

I've got another issue. A quick calculation using the first identity (sum of two modified bessel functions of the first kind), and then the corresponding expression for the modified bessel function of the 2nd kind (using alpha=1/2), gives an exponential that is exp(z/2), but isn't it exp(-z/2)? We can do a direct calculation of the integral of K_{1/2}(z), and we see the exponential is indeed negative. —Preceding unsigned comment added by 72.229.19.115 (talk) 15:09, 11 September 2009 (UTC)

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)

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 $J_\alpha$ 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 $J_\alpha$.
Similarly, if you compare the first two terms in the power series for $Y_\alpha$ you find the requirement that $x \ll \sqrt{|\alpha-1|}$ ... however, $Y_\alpha$ 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)

I notice that the given asymptotic expression for I has a dependency on alpha, but that for K doesn't. In fact the next term in the asymptotic expansion for K looks like that for I, but with the sign reversed. If I knew how to edit the main article I'd change it, else the impression is given that the asymptotic form for K is different, in some fundamental way, to that of the other Bessel functions. — Preceding unsigned comment added by 134.225.3.100 (talk) 14:14, 16 June 2011 (UTC)

I don't think it makes sense to give the next-order term for the I expression, but not for the J, K, or Y expressions. Especially as, in the text, the I asymptotic form is explained as following from the J asymptotic form (since I(x) ~ J(ix)). Especially since no reference is given for the higher-order terms. So, I've removed these terms from the I formula. — Steven G. Johnson (talk)
Regarding a reference for higher order terms for I, see A&S p377 9.7.1 and for K A&S p378 9.7.2. The higher order terms for I have just been restored by A. Pichler - I shall add the terms for K and also the reference. --catslash (talk) 16:31, 19 June 2011 (UTC)
If A&S has the higher-order formulas, then by all means add them with a reference for J, Y, K, and I.
(A. Pichler has consistently refused to see the need for adding references for formulas, unfortunately. I have had to revert a lot of his additions as a consequence of his ignoring WP:CITE policy. In the early days of Wikipedia when we didn't have much content and didn't have a good citation mechanism, adding unreferenced equations was excusable, but now it is not really justifiable.) — Steven G. Johnson (talk) 16:38, 19 June 2011 (UTC)

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 $\Gamma(m+\alpha+1)$, 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)

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)

The description of the plot does not match the plot. I think it should read alpha=0..2,

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).

Recurrence formulas for Spherical Bessel functions

I miss some recurrence relations for spherical bessel functions, like these:

The functions jα, yα, hα(1), and hα(2) satisfy the recurrence relations:
$\frac{2\alpha+1}{x} z_\alpha(x) = z_{\alpha-1}(x) + z_{\alpha+1}(x)$
$(2\alpha+1)\frac{dz_\alpha}{dx} = \alpha z_{\alpha-1}(x) - (\alpha+1)z_{\alpha+1}(x)$
$\frac{dz_\alpha}{dx} = z_{\alpha-1}(x) - \frac{\alpha+1}{x}z_{\alpha}(x)$
$\frac{dz_\alpha}{dx} = \frac{\alpha}{x}z_{\alpha}(x) - z_{\alpha+1}(x)$
where z denotes j, y, h(1), or h(2).

(or at least the first two relations to be consistent with relations specified for other functions).

--Mojca Miklavec (talk) 09:51, 26 June 2009 (UTC)

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)

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)

Compare the signs in the recurrence relations with the ones used in Abramowitz's book. They look wong. —Preceding unsigned comment added by 195.169.141.54 (talk) 13:45, 23 May 2008 (UTC)

No, the formulas as written here are correct. (You can easily check them numerically with Matlab.) There is no inconsistency with Abramowitz and Stegun: the formulas on page 361 are for J and Y Bessel functions, not I and K. —Steven G. Johnson (talk) 14:53, 18 June 2008 (UTC)

Could someone elaborate on what happens to the Bessel functions for negative real arguments? Would I be right in guessing that for even orders, these functions are even; for odd orders they are odd; for non-integer orders, they are complex? There's no mention of any of this in the article; not even the graphs show this.Toolnut (talk) 00:17, 24 August 2011 (UTC)

/* potential error */

In Properties , For integer order α = n, Jn is often defined via a Laurent series for a generating function: <formula>

the formula on the left has no dependency on the variable x while the rigth has one.
In general this section lacks some developpement (or links to other wiki pages).
I mean skipping all the background to have series of bessel functions is a little disapointing for a reader like me, like the after taste after eating a salad without a dressing ...(e.g. not using for example the 20th century approach of hilbert space with the discrete and/or continous function bases to show where it comes from ...) — Preceding unsigned comment added by 91.179.53.212 (talk) 08:07, 18 May 2014 (UTC)

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)

closure equation

I'm wondering how one would go about proving the closure equation that is cited in the article:

$\int_0^\infty x J_\alpha(ux) J_\alpha(vx) dx = \frac{1}{u} \delta(u - v)$

Does anyone know how, or can they give me a reference? Thanks --Lavaka 05:35, 13 April 2007 (UTC)

I seem to recall that equation being from the Bowman book; perhaps it is proved there. —Steven G. Johnson 16:55, 13 April 2007 (UTC)
It's also found in Arfken and weber. They write: This may be proved by the use of Hankel transforms in section 15.1. An alternate approach, starting from a relation similar to (9.82), is given by Morse and Feshbach (Section 6.3). (9.82 is the equation for the Green's function of the Helmholtz equation.) —Steven G. Johnson 16:59, 13 April 2007 (UTC)
Thanks Steven, I'll try those --Lavaka 17:26, 13 April 2007 (UTC)

It's not in Bowman, but I found the same passage you did in Arfken and Weber, which isn't especially enlightening. I believe that the Hankel Transform, with respect to some variable $u$, of $J_0(vx)$ is indeed $\frac{1}{u} \delta (u - v)$, but I'm a bit unsure if the integral formulation is true. When we extend something like the Fourier Transform to include tempered distributions, no longer can we write certain integral formulations, even if the objects have a Fourier Transform. I don't know if Arfken and Weber are correct, but I haven't been able to find any other good sources. --Lavaka 20:56, 13 April 2007 (UTC)

PuZHANG (talk) 11:49, 14 November 2010 (UTC) I saw the passage in Arfken and Weber. They say people can prove the closure relation with Hankel transform. However, the Hankel transform pair is again based on this closure relation according to the item Hankel Transform in wikipedia. Actually to prove this relation is part of a problem in Jackson's Classical Electrodynamics (problem 3.16 in the 3rd ed.), where some hint is given. The proof should not be difficult following the hint.

What does this integral mean? Surely the integrand is not Lebesgue integrable because the integrand would have oscillations of approximately constant size out to infinity! Or more precisely, the integrand is like the product of two sine waves of different frequencies. It wouldn't even be a definable improper integral – the integral out to some X would have oscillations as a function of X that do not decay to zero amplitude. Eric Kvaalen (talk) 13:22, 29 July 2014 (UTC)

I have figured out the answer to my own question. If one defines a boxcar function of x that depends on a small parameter ε as:

$f_\epsilon(x)=\epsilon\ \mathrm{rect}\left(\frac{x-1}\epsilon\right)$

(where rect() is the rectangle function) then the Hankel transform of it (of any given order α), gε(k), approaches Jα(k) as ε approaches zero, for any given k. Conversely, the Hankel transform (of the same order) of gε(k) is fε(x):

$\int_0^\infty k J_\alpha(kx) g_\epsilon(k) dk = f_\epsilon(x)$

which is zero everywhere except near 1. (This can be proved, at least for α zero, from Fourier analysis of a circularly symmetrical function, so it does not require the "closure" equation that we are discussing.) As ε approaches zero, this integral approaches δ(x−1). So by abuse of language (or "formally"), one might say that

$\int_0^\infty k J_\alpha(kx) J_\alpha(k) dk = \delta(x-1)$

even though the integral is not actually defined. A change of variables then yields the formula in question. Eric Kvaalen (talk) 09:23, 11 August 2014 (UTC)

This article was severely lacking differentiation! I had to search forever around the net to find a reference. Hopefully this helps out a lot of people.

Aaron —Preceding unsigned comment added by 128.163.172.117 (talk) 00:15, 5 December 2007 (UTC)

Independency of Bessel function of first and second kind

I am missing a formula describing this. Will somebody add it?

TomyDuby (talk) 06:39, 5 October 2008 (UTC)

Bourget's hypothesis

I am puzzled by what the article says: the problem is open in the case of integers n and n+m, except when m=1,2,3,4. I read somewhere that Watson's treatise mentions that the answer is known for n rational and m positive integer. It has something to do with a theorem by Siegel that zeroes are transcendental numbers. Some help please! --Bdmy (talk) 10:33, 21 March 2009 (UTC)

I will fix it now Stefán (talk) 15:05, 7 October 2009 (UTC)

Bessel function of the second kind (again)

you have mentioned in your article that Y-n(x)=(-1)nYn(x),but by comparing with the definition of Y, I suspect that the relation should be Y-n(x)=(-1)n-1Yn(x).Am I correct about this?

You are probably looking at the definition
$Y_\alpha(x) = \frac{J_\alpha(x) \cos(\alpha\pi) - J_{-\alpha}(x)}{\sin(\alpha\pi)}.$
It's true that if J had a certain parity with respect to the order α, then Y would have the opposite parity, because sin(απ) is odd in α (and because cos(απ) is even). However, note that:
• Jn only has a definite parity when the order n is an integer
• The above definition is only valid for non-integer order α (it's indeterminate otherwise)
• Yn only has a definite parity when the order n is an integer

So you have to use the integral definition for Yn. The integrand in the second term seems to have the correct behaviour to get Y-n(x)=(-1)nYn(x), but I have to admit the first term defeats me for the moment. --catslash (talk) 11:30, 10 August 2009 (UTC)

Integral representation of Bessel functions of the second kind

How to derive the Integral representation of Bessel functions of the second kind from its definition Y(x)={Jn(x)cos(n times pi)-J-n(x)}/sin(n times pi) with n tends to a integer ? I eager to know the proof because the Integral representation explain the asymptotic behaviour of Y with large x. —Preceding unsigned comment added by 61.18.170.29 (talkcontribs)

Number of squiggles

A single squiggle is the usual notation for asymptotic to, isn't it? It is on the asymptotic analysis page. The asymptotic expressions are more than approximations; they give the limit. (Also what is asymptotically proportional to?). --catslash (talk) 15:17, 12 August 2009 (UTC)

By "asymptotically proportional to," I mean that I've seen $f(x) \sim g(x)\!$ mean $f(x) \in \Theta(g(x))$ (see Big O notation)...that is, the single tilde in my experience sometimes ignores constant factors. $\approx$ seems less likely to be misinterpreted. — Steven G. Johnson (talk) 22:45, 12 August 2009 (UTC)
Would anyone else find the order of the error helpful in the asymptotic expansions? If I'm far from my copy of Bender and Orzag or Hinch then this page is invaluable, but could be slightly even more valuable still... 7daysahead (talk) 20:04, 31 July 2010 (UTC)
I've never seen $\approx$ being used as meaning "asymptotically equal", to be honest. IMO this is therefore more likely to be misinterpreted than using a single tilde - it's in fact quite clear that it doesn't mean "up to constant factors" here, since explicit constant factors are already included in many of the asymptotics. --Roentgenium111 (talk) 21:40, 11 September 2012 (UTC)
Who said it meant "up to constant factors"? $\approx$ typically means "approximately equal", including constant factors. — Steven G. Johnson (talk) 01:08, 12 September 2012 (UTC)

Plot of spherical Bessel functions

A simple question. When I go to the limes $x \to 0$, I should get $j_0 \to 1, j_2 \to -1$, correct? My problem is that the plot of the spherical Bessel functions has $j_2 \to 0$. Am I wrong or is the plot incorrect? 130.225.67.200 (talk) 14:19, 27 August 2009 (UTC)

Plot and statements in the text are correct, as $j_n(x)=O(x^n)$. 18:09, 27 August 2009 (UTC) —Preceding unsigned comment added by A. Pichler (talkcontribs)

multiple error of definition ?

• in the Relation to Laguerre polynomials

where the $t$ comes from ? is an integral missing somewhere or what ?

• Hankel function

It is stated that for interger alpha, the limit must be calculated... but can someone write the result ?

• Modified Bessel functions
• Again, the relation between the 1st kind and the laguerre polynomials displays a $t$ without further explanations.
• Again, the definition of the 2nd kind holds only for non-interger alpha, but this time no mention of it is noted, neither any hint has to get the version with interger alpha (which is funy in a sense since only version with interger alpha are displayed on the graph).
• the last formula, that provide a relation between the modified bessel function of second kind and laguerre polynomials displays TWO new parameters : $z$ and $a$, again, what are they ?
• on a side note, I don't know if we got the same notation rules for series or not, but for me $\sum_{k=0}$ is a function (like $I_n$) so either $J_\alpha(x)$ is a function from $\N$ into $\R$ either a superscript is missing in the formula (and there is a difference between a serie and the limit of the serie).

141.30.111.12 (talk) 07:03, 7 October 2009 (UTC)

Reference style

There is consensus for the change from <references /> to {{Reflist}}. Two of the main contributors (User:Michael Hardy, User:Paul August) support the move, as does User:MuDavid who initially added <references /> to the article. Therefore I'm change the reference style now. --bender235 (talk) 17:32, 8 July 2010 (UTC)

unsourced integral representations

The following information was added by User:Toolnut (with the comment Have tried to convince non-believers of the truth of the integral form):

There exist many integral representations of these functions. The following for Kα(x), is useful for the calculus of the Feynman propagator in field theory:
$K_\alpha(x) = \frac{1}{2} \int_{-\infty}^{+\infty} e^{-ix\sinh t -\alpha |t|} \, dt =\int_0^\infty e^{-\alpha t}\cos{(x \sinh{t})} \, dt \text{ , for } \alpha >1 \, ;$
this can be verified to be a solution to the modified Bessel equation (see below), given $\alpha>1$. The way to convince ourselves of the truth of the definite integral form
$y(x)=\int_a ^b u(x,t)\,dt$
of any function of the variable x purported to be a solution to a differential equation in x is by substitution and application of Leibniz integral rule, i.e. taking the integration w.r.t. the independent integration variable t outside the differentiation w.r.t. x.

Whether or not these are true, they don't belong in the article without a citation to a reputable source. Wikipedia is not the forum to "convince non-believers". See WP:V.

— Steven G. Johnson (talk) 18:37, 22 August 2011 (UTC)

I was not the first person who put it there: I tested the original posting and found it deficient and inaccurate and have corrected it beyond a shadow of a doubt. The original contributor cites the "Feynman propagator" as the source and there is a link to it on Wikipedia. The integral forms of the other Bessel functions are also in the article: I only wish to see them to be convincingly true. These integrals do occur in real-life applications, such as the one I was working on, and I was happy to have found their solutions on Wikipedia, though slightly off.
Toolnut (talk) 19:37, 22 August 2011 (UTC)
Thanks for clarifying, but they still should not be included without a reputable source. If you have seen them in published applications, perhaps you are familiar with a suitable source? — Steven G. Johnson (talk) 19:45, 22 August 2011 (UTC)
Wikipedia was my source on this one, together with basic calculus; there are other sources, but kind of hard to find and not too elaborate: I'll see what I can do. The integral form of the Bessel function of the second kind is also unsourced in this article. Although these solutions to the differential equations are not unique (they could possibly be linear combinations of the first and second kinds of linearly independendent solutions) and so, to be sure that they represent their respective, specific, and established Bessel functions, I have tested these representations with numerical integration and compared them to their respective programmed Bessel functions found in applications like Matlab and Excel: I am thorough that way.Toolnut (talk) 20:07, 22 August 2011 (UTC)
Unfortunately, derivations and checks by Wikipedia users are not sufficient for Wikipedia; see WP:OR. I'll look at the second-kind integral formula to see if I can find a reference or otherwise fix it. — Steven G. Johnson (talk) 21:37, 22 August 2011 (UTC)
I found two integral formulas in the Watson text (a different formulation than yours) and added them with the cite. (The one for K in the Watson text actually seems nicer than yours in some ways, since it is non-oscillatory for real x and the integrand goes to zero as a double exponential in t.) — Steven G. Johnson (talk) 21:50, 22 August 2011 (UTC)
I had lied: I hadn't checked the correspondence between the besselk function and that integral until now and have found that they are not the same, although my integral is still a solution to the modified Bessel equation, so it must be expressible as a linear combination of besselk and besseli. Thanks for finding the right integrals: the one that I corrected had nearly the same form as mine, but evaluated to a complex number for a real argument. I'm glad these discussions are leading towards the truth. I will check your integrals and see how a linear combination of those two could evaluate into the solution I had.Toolnut (talk) 22:53, 22 August 2011 (UTC)

I see now what had happened: the only error in the original integral form was that it had an errant i in the exponent of e, so the form that you've uncovered may equally be written as the corrected original, integrating from -∞ to +∞:

$K_\alpha(x)=\frac{1}{2}\int_{-\infty}^\infty e^{-x\cosh{t}-\alpha t}dt,\ \Re\{x\}>0;$

instead, I had corrected it by choosing to leave the complex exponential alone and making the sign of αt unchanging to ensure that it converged and stayed real. We should also point out that the integral forms of the Hankel functions, introduced in the preceding section, would also diverge unless $\Im\{x\}<0$ for $H_\alpha^{(1)}(x)$, and $\Im\{x\}>0$ for $H_\alpha^{(2)}(x)$: you may want to add those caviats in there, as you have done for $I_\alpha(x) \ \&\ K_\alpha(x)$.Toolnut (talk) 08:04, 23 August 2011 (UTC)

The integral representations in the Hankel section didn't match the cited reference (A&S). (The ones in the reference converge for Re x > 0, which is obviously much more useful and matches those for J and Y.) — Steven G. Johnson (talk) 17:52, 23 August 2011 (UTC)

Removal of +1

I undid this diff as the "+1" appears to have been around for awhile. As I don't understand the math, would someone who does please verify that I did the right thing? Thanks Jim1138 (talk) 22:45, 8 June 2012 (UTC)

You are right, at least per the cited source. Also the test case of J0(0)=1 requires a +1. Sławomir Biały (talk) 00:38, 9 June 2012 (UTC)
The source Abramowitz and Stegun [1] agrees with the article as presently written, as does Wolfram alpha [2]. Furthermore, it is clear for the following reason. We have $J_0(0)=1$. Without the "+1" in the Gamma function, we would instead compute $J_0(0)=0$. Sławomir Biały (talk) 11:42, 10 July 2012 (UTC)
Also see this: http://www.wolframalpha.com/input/?i=Sum[%28-1%29^m%2F%28m!+Gamma[m+%2B+\[Alpha]+%2B+1]%29+%28x%2F2%29^%282+m+%2B+\[Alpha]%29%2C+{m%2C+0%2C++++Infinity}] Sławomir Biały (talk) 17:19, 12 July 2012 (UTC)

Astronomy

The applications of Bessel functions to astronomy, by Bessel himself, could be mentioned. — Preceding unsigned comment added by 78.105.0.33 (talk) 14:00, 19 July 2012 (UTC)

The article on Friedrich Bessel mentions the use of Bessel functions in gravity. — Preceding unsigned comment added by Siberian Patriot (talkcontribs)

Error in spherical Bessel functions?

I believe that the second equality of the statement of y_n near the beginning of the Spherical Bessel functions section is incorrect. It does not appear in the citation given (Abramowitz & Stegun), and I cannot derive it. As far as I can tell, it should read

$y_{n}(x) = \sqrt{\frac{\pi}{2x}} Y_{n+1/2}(x) = (-1)^{n+1} \sqrt{\frac{\pi}{2x}} J_{-n-1/2}(x).$

That is, a 'J' instead of a 'Y.' It seems like a simple typo. Can an expert confirm? — Preceding unsigned comment added by 129.74.141.120 (talk) 16:47, 12 September 2012 (UTC)

Well
$\mathrm{Y}_{\alpha}(z) = \frac{\mathrm{J}_{\alpha}(z) \cos(\alpha \pi) - \mathrm{J}_{- \alpha}(z)}{\sin(\alpha \pi)}$
(A&S 9.1.2 p358), which for $\scriptstyle{\alpha = n + \frac{1}{2}}$ reduces to
$\mathrm{Y}_{n + \frac{1}{2}}(z) = - \frac{\mathrm{J}_{- n - \frac{1}{2}}(z)}{\cos(\pi n)} = (- 1)^{n + 1} \mathrm{J}_{- n - \frac{1}{2}}(z)$
so it's certainly correct as you have written it. Equivalently
$\mathrm{y}_{n}(z) = (- 1)^{n + 1} \mathrm{j}_{- n - 1}(z)$
for integer $\scriptstyle{n}$ (A&S 10.1.15 p439), so again
$\mathrm{y}_{n}(z) = (- 1)^{n + 1} \sqrt{\frac{\pi}{2 z}} \mathrm{J}_{- n - \frac{1}{2}}(z)$
I shall change the second $\scriptstyle{Y}$ to a $\scriptstyle{J}$. --catslash (talk) 23:30, 12 September 2012 (UTC)

Clarification needed?

This: {{clarify||First eqn. would give J_0(x)~1 (constant), which does not seem to agree with the graph given above|date=September 2012}} is unexpected.

The graph does show ${\scriptstyle J_0(x)\;\sim\;1\;\;(x \to 0)}$ and the curve is shown flat (${\scriptstyle J_0 '(0)\;=\;0}$) at this point, since it is a local maximum. Nevertheless, something was not clear enough; is it that ${\scriptstyle (x \to 0)}$ is not explicitly stated after each formula? I shall remove the {{clarify}} until it is clear what is not clear. --catslash (talk) 22:21, 12 September 2012 (UTC)

Laypersons

I have no beef with the article, because Bessel functions are notoriously difficult, but I'm wondering if the lede could somehow give an indication to the non-mathematical layperson about what Bessel functions are all about. Probably that is impossible, but I'm putting the idea out there as a challenge.77Mike77 (talk) 00:05, 16 March 2013 (UTC)

gamma in Modified Bessel Function of the Second Kind Asymptotic Form

The little gamma in the asymptotic form for K0(z) for small z should be clarified. It is the Euler–Mascheroni constant. — Preceding unsigned comment added by 128.208.188.101 (talk) 23:26, 16 July 2013 (UTC)

Differential equation verification - what am I missing?

What am I missing - why can't I verify the differential equation that generates the stated solution? For instance, take the differential equation and then divide it by $x^2$:

$x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + x^2 y = 0$
$\frac{d^2 y}{dx^2} + \frac{1}{x} \frac{dy}{dx} + y = 0$

This should have as solution a Bessel equation of the first kind of order 0. Using the Taylor series expansion for the solution we can readily take the first and second derivative as the solution is a linear combination of continous functions of x, so we get:

$y = \sum_{m=0}^\infty \frac{(-1)^m}{{(m!)}^2} {\left(\frac{x}{2}\right)}^{2m}$
$\frac{dy}{dx} = \sum_{m=0}^\infty \frac{(-1)^m}{{(m!)}^2} \frac{2m}{2} {\left(\frac{x}{2}\right)}^{2m - 1}$
$\frac{d^2 y}{dx^2} = \sum_{m=0}^\infty \frac{(-1)^m}{{(m!)}^2} \frac{2m(2m - 1)}{4} {\left(\frac{x}{2}\right)}^{2m - 2}$

By combining the terms in the initial differential equation and taking the common factor we get:

$\frac{d^2 y}{dx^2} + \frac{1}{x} \frac{dy}{dx} + y = \sum_{m=0}^\infty \frac{(-1)^m}{{(m!)}^2} {\left(\frac{x}{2}\right)}^{2m - 2} {\left( \frac{2m(2m - 1)}{4} + \frac{1}{x} \frac{2m}{2} \frac{x}{2} + {\left(\frac{x}{2}\right)}^2 \right)}$
$\frac{d^2 y}{dx^2} + \frac{1}{x} \frac{dy}{dx} + y = \sum_{m=0}^\infty \frac{(-1)^m}{{(m!)}^2} {\left(\frac{x}{2}\right)}^{2m - 2} \frac{1}{2} {\left( m (2m - 1) + m + \frac{x^2}{2} \right)}$
$\frac{d^2 y}{dx^2} + \frac{1}{x} \frac{dy}{dx} + y = \sum_{m=0}^\infty \frac{(-1)^m}{{(m!)}^2} {\left(\frac{x}{2}\right)}^{2m - 2} \frac{1}{2} {\left( 2 {m}^2 + \frac{x^2}{2} \right)}$

Now what I don't get is how this final result gives a zero for all m and x values? Why I can't verify the initial differential equation using one of the stated solutions for the equation? If the objection is that the initial equation is not defined in the origin because of $x^{-1}$ as a multiplier for the first derivative, it's possible to repeat the demostration without reducing the differential equation to this form and you'll still end up with a paranthesis that it's not zero for all m or x values.

Is my assumtion that the solution for the initial equation is the Bessel function of the first kind wrong, or is something missing from the text (i.e. like the solution for the stated dif equation is a linear combination of first and second kind Bessel functions, or something in the like)? 89.120.104.106 (talk) 10:52, 3 October 2013 (UTC)Apass

Your working is correct. The series in the last line is zero because the $\scriptstyle{\frac{x^{2}}{2}}$ bit of each term in the summation is cancelled by the $\scriptstyle{2 m^{2}}$ bit of the next term in the series. The trick to see this is: (1) Split the summation into two series, one just including the $\scriptstyle{2 m^{2}}$ bit and one just the $\scriptstyle{\frac{x^{2}}{2}}$ bit, (2) Reparameterize the first of these, replacing $\scriptstyle{m}$ by $\scriptstyle{m + 1}$. This makes the terms in the two series line up; the same $\scriptstyle{m}$ corresponds to the same power of $\scriptstyle{x}$ (i.e. $\scriptstyle{\left( \frac{x}{2}\right)^{2 m}}$) in each one. (3) Zip the series back together - and voila: nothing! --catslash (talk) 18:27, 3 October 2013 (UTC)