Talk:Confluent hypergeometric function

Whittaker function

Is U(a,b,z) the Whittaker function? (anon, Oct 2006)

I don't know, that's not what A&S calls them.linas 00:43, 11 December 2006 (UTC)

I am certainly not an expert, but I now know a bit about Kummer/Whittaker functions. Enough to find severe discrepancies between A&S and maple. Anybody have an opinion about whether I should tack some things up on the main page?

Kummer's function

I am interested in the real part of Kummmer's function in the case a=2n+1, b=a+1 (real part of incomplete gamma). From a numerical point of view, which is cheaper to approximate, what is the convergence like for each and what methods are used? (anon, Nov 2006)

a sub n is defined in this article, but what is b sub n? — Preceding unsigned comment added by 213.122.105.23 (talk) 12:11, 29 April 2015 (UTC)

continuous fraction for e^z

The original text used to say

by setting b = 0 and c = 1

It is hard to tell what it meant because there was no c around.

M(1, 2, z)⁄M(0, 1, z)
= 1/
1 − 1⁄2 z/
1 + 1⁄6 z/
1 − 2⁄12 z/
1 + 2⁄20 z/
…
1 − k⁄(2 k − 1) (2 k) z/
1 + k⁄(2 k) (2 k + 1) z/
…
= 1 + 1/ 1 − 1⁄2 z/
1 + 1⁄6 z/
1 − 1⁄6 z/
1 + 1⁄10 z/
…
1 − 1⁄2 (2 k − 1) z/
1 + 1⁄2 (2 k + 1) z/
…

Transforming this fraction with the sequence (1, 2, 3, 2, …, 2 k + 1, 2, …) gives

1/
1 − z/
2 + z/
3 − z/
2 + z/
…
(2 k − 1) − z/
2 + z/
…
= (e^z − 1)⁄z

which is not quite what was postulated.

--Yecril (talk) 13:47, 3 October 2008 (UTC)

Formal power series?

The following is simply too cryptic for inclusion as it stands

Moreover,

$U(a,b,z)=z^{-a}\cdot \,_{2}F_{0}\left(a,1+a-b;\,;-{\frac {1}{z}}\right),$

where the hypergeometric series $_{2}F_{0}(\cdot ,\cdot ;;z)$ degenerates to a formal power series in z (which converges nowhere).

Please explain precisely what it is that this is supposed to convey, including a reference. Sławomir Biały (talk) 18:37, 3 July 2009 (UTC)

Addendum: Presumably this is supposed to hold as an asymptotic series as z→0 in the right half-plane. But a reference (or at least a clarification) is needed to establish this. Sławomir Biały (talk) 19:05, 3 July 2009 (UTC)

Referring to @book{andrews2000special,

 title=Template:Special functions,
 author={Andrews, G.E. and Askey, R. and Roy, R.},
 year={2000},
 publisher={Cambridge Univ Pr}

} Page 189 They agree, the formal form above diverges and they provide a convergent alternative solution by taking limits on 2F1.

${\frac {1}{\Gamma (a)}}\int _{0}^{\infty }e^{-xt}t^{a-1}(1+t)^{b-a-1}dt,$

Rrogers314 (talk) 20:53, 16 July 2009 (UTC)

No one is disagreeing that the "formal form" diverges. The question is, what exactly is intended by the string of symbols

$U(a,b,z)=z^{-a}\cdot \,_{2}F_{0}\left(a,1+a-b;\,;-{\frac {1}{z}}\right).$

Because a power series it most certainly is not. Sławomir Biały (talk) 03:03, 21 July 2009 (UTC)

It's the result of various transformations and limits giving a asymptotic series for x "large". The above reference covers this and computes R_n(x) as O(1/x^n) . If you would like I could try to capture the reasoning or result. To give credit; how much can I quote before violating copyright? The book is succinct and I have a tendency to wander off; this means that quoting is probably preferred in some instatnces. My guess about your request is:

1) How does this form, both as symbols and series, come about

2) The effectiveness as a asymptotic series.

3) Skipping the actual intermediate details

?? Rrogers314 (talk) 15:17, 18 August 2009 (UTC)

Clarification request

This section seems confusing

... .Similarly

$U(a,2a,x)={\frac {e^{\frac {x}{2}}}{\sqrt {\pi }}}x^{{\tfrac {1}{2}}-a}K_{a-{\tfrac {1}{2}}}\left({\tfrac {x}{2}}\right),$

When a is a non-positive integer, this equals $2^{-a}\theta _{-a}\left({\tfrac {x}{2}}\right)$ where θ is a Bessel polynomial.

It isn't clear what the function $K_{a-{\tfrac {1}{2}}}\left({\tfrac {x}{2}}\right)$ is supposed to be. The preceding text would incline me guess at Kelvin function, but it really shouldn't have to be a guess. Could somebody please add an appropriate definition? Thank you.