Talk:Binomial series

message: This page should discuss the general form of the binomial series given by  (x + y)^{\alpha} = \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^{\alpha - k} y^k . This is because in the series the roles of x and y are not symmetric. Currently this page discusses only the special case x = 1, and ignores the other important special case y = 1. The general form unifies both, and for example the convergence criteria need then to be stated only once. --Kaba3 (talk) 22:25, 20 September 2014 (UTC)
thread_title:  Missing cases 

message: : {\alpha \choose k} = \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!}=\frac{(-1)^k}{k!}(-\alpha)_k,
seems to be inconsistent... According to Pochhammer symbol, there shouldn't be any negative signs on the RHS... Specifically, if we define
:(x)_n=\frac{\Gamma(x+1)}{\Gamma(x-n+1)}.
then according to Mathematica, the consistent RHS is
:\frac{1}{k!}(\alpha)_k,. I think someone may have used the mathworld article which only uses one type of Pochhammer symbol.--Lionelbrits 19:40, 30 March 2007 (UTC)
thread_title: Inconsistency

message: It seems worth enlarging the discussion to the complex values of α.
I added a few facts on the binomial coefficients and a sketch of the proof of
the elementary results about convergence.PMajer 11:18, 28 April 2007 (UTC)
thread_title: Complex values of α

message: Is it not easier to understand “when to stop” if you write : \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-(k-1))}{k!}? Anders Ytterström (talk) 12:16, 14 June 2008 (UTC)
thread_title: Easier

message: The example at the end seems out of place, possibly a case of self-promotion.  I do not know any textbook coverage of the binomial series that mentions such an example.  If there is such a textbook, it should be cited; otherwise I propose to remove the example. WardenWalk (talk) 18:42, 1 August 2009 (UTC)
thread_title:  2-ray ground reflect model 

message: Well, when I provided the first proof of convergence, my idea was to make it as short as possible, so I chose the solution of deriving the asymptotics from the product formula of the Gamma function. Today, having gathered some small experience of wikipedia, it seems to me that after all, even if this is an encyclopaedia, short self-contained proofs are a kind of information of interest for many readers. Accordingly, I've added another small section to get the bounds on the binomial coefficients via basic inequalities, which puts the whole thing on a more elementary level. Of course, if anybody finds it too long, we may make it shorter. I'm not sure about where to put them, BTW. For now I'm trying the bottom --pma (talk) 18:18, 7 September 2009 (UTC)
thread_title:  Asymptotics on the coefficients 

message: In the section on convergence it says:
(i)    If |x| < 1, the series converges for any complex number α.
In the sketch of the proof it says:
To prove (i), apply the ratio test and use formula (2) above to show that whenever α is not a nonnegative integer, the radius of convergence is exactly 1.
Seems to me it doesn't matter weather or α is "not a nonnegative integer.  You'll always end up with the limit of:
 \frac{n+1}{\alpha-n} = \frac{n}{n} \frac{1 + \frac{1}{n}}{\frac{\alpha}{n} - 1} = \frac{1}{-1} 
whose absolute value is 1, no matter what complex number α we have.  In anycase the point was to proof (i) for all complex numbers α.  Am I missing something?--85.220.91.106 (talk) 15:44, 30 September 2011 (UTC)
replies: text: What you are missing is that the terms of the series ultimately become zero when \alpha\in\N (the ordinary binomial theorem) so the series converges regardless, and has infinite ratio of convergence. So (i) is satisfied but the ratio of convergence statement is not. The ratio of successive coefficients you are taking the limit of becomes the 0/0 always, which makes the limit is undefined. Marc van Leeuwen (talk) 07:02, 3 October 2011 (UTC)
thread_title:  α can be any complex number? 

message: As remarked by User:Rickhev1 in a recent edit, the proof of convergence was not complete in that it referred to the inequality \scriptstyle \left|{\alpha \choose k}\right| \le \frac {M} {k^{1+\operatorname{Re}\,\alpha}}, which gives sufficient conditions for simple/absolute convergence, but of course is not enough to show that those conditions are necessary. However, the asymptotics  \scriptstyle {\alpha \choose k} = \frac{(-1)^k} {\Gamma(-\alpha)k^ {1+\alpha} } \,(1+o(1)) as \scriptstyle k\to\infty , that was quoted above, does suffice. I made minor changes in the proof in order to make it complete. --pma 16:45, 20 December 2011 (UTC)
thread_title:  Proof 

message: Someone should write relation btwn "binomial Coef" , permutation nPr=n!/(n-r)! & combination nCr*r!=nPr  — Preceding unsigned comment added by 211.208.163.62 (talk) 11:17, 24 April 2014 (UTC)
replies: text: This is not the place for this, the properties of binomial coefficients are treated in that specific article.--LutzL (talk) 09:51, 25 August 2014 (UTC)
thread_title:  binomial coefficient & probability relation 

message: Why does Newton's binomial theorem redirect here and not to Binomial theorem? — MFH:Talk 04:37, 25 August 2014 (UTC)
replies: text: Versions of the binomial theorem date back to ancient times, Persians, Arabs and Fibonacci used them to approximate (square) roots, they are consistently documented since the 15th century. Newton's contribution is the binomial series. The "theorem" connecting the series to non-integer powers belongs to this article. The elementary binomial theorem article surely has a link here.--LutzL (talk) 09:58, 25 August 2014 (UTC)
thread_title:  redirect of  [[Newton's binomial theorem]] 

message: Isn't there an overlap between conditions ii and iv:
: ii. If |x| = 1, the series converges absolutely if and only if either Re(α) > 0 or α = 0. 
: iv. If x = −1, the series converges if and only if either Re(α) > 0 or α = 0. 
If x = −1 then |x| = 1 and one can always use the ii instead of iv.
Dalba 05:30, 2 November 2019 (UTC)
replies: text: The idea was to characterize absolute convergence first, then simple convergence. Of course the case x = −1 is included in |x| = 1, but what remains to say after items i, ii and iii, is about the simple convergence at x = -1. The content of iv is that for x=-1 there is simple convergence if and only if there is absolute convergence. One may state it like that, yet I found it more direct and clear to state each item independently from the others.pma 08:37, 2 November 2019 (UTC)
thread_title:  Overlap in conditions for convergence? 

message: It is surely germane to the topic that the negative binomial series generalizes the geometric series.  If nothing else, this article should at least link to geometric series.  But my edit was reverted by User:Quantling, after a self-revert (apparently realizing their mistake that indeed the geometric series is a special case), with the edit summary that this is "boring".  The object of an encyclopedia article is not to be interesting in the subjective opinion of editors, but to provide relevant information on a topic.  Tito Omburo (talk) 16:06, 9 November 2023 (UTC)
replies: text: I apologize for "but boring"; that can be interpreted as a personal attack and I don't mean or want that.  Please see the next section where (I hope) I have done a better job of explaining myself. —Quantling (talk | contribs) 16:53, 9 November 2023 (UTC)
thread_title:  But boring?