Talk:Binomial series

Missing cases

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)[reply]

Inconsistency

{\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)[reply]

In any case, it seems to me that the use of this Pchh symbol is not relevant here, so I suggest to leave it in the discussion of general Newton's series. PMajer 11:18, 28 April 2007 (UTC)[reply]

Complex values of α

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)[reply]

Easier

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)[reply]

You mean easier as compared with : ${\frac {\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -k+1)}{k!}}$ , do you? Yes, but the double parenthesis looks uglier :-) 79.38.22.37 (talk) 14:47, 14 August 2008 (UTC)[reply]

2-ray ground reflect model

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)[reply]

Asymptotics on the coefficients

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)[reply]

α can be any complex number?

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)[reply]

What you are missing is that the terms of the series ultimately become zero when

\alpha \in \mathbb {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)[reply]

This article does not help the novice understand the subject

Too many Wikipedia authors use articles they edit to display their own knowledge rather than help educate those who do not yet understand the subject. There's nothing like opening an article with a paragraph which could only be understood by someone who is already familiar with the subject. The Wikipedia articles on mathematics subjects are distinguished from similar articles which can be found elsewhere on the internet by the fact that they provide no help to those who are approaching the subject for the first time. I have no problem with what is already there. I just think the opening paragraph or two could be in plain English with minimal jargon and equations. Just my opinion. — Preceding unsigned comment added by 81.187.233.172 (talk) 16:16, 10 December 2011 (UTC)[reply]

I respect your opinion, but you should be aware that a mathematical article like this one requires a minimum of prerequisites and a minimum knowledge about the general subject. The opening paragraph has exactly the scope of giving you the coordinates of its subject. Here we are talking of a Taylor series for a complex function. If you do not know what is a Taylor series, please go there and start reading that article. That article starts telling you that we are talking about infinite series methods to represent a function: if you do not know what is a function, and you do not know what is a convergent series, please go to the linked articles, etc. To make an example for you: if you go to any article e.g. in the German wikipedia, and you do not read German, it makes little sense complaining about the difficulty of the subject. --pm a 09:54, 11 December 2011 (UTC)[reply]

"I respect your opinion" is perhaps an overly-generalist view of how mathematics develops and has developed. I see exactly where the first commentary is coming from. How do you penetrate the jargon and (particularly)the (very economical) notation to get to the ideas at their root? I have designed a talk that leads people through the arithmetic that is the Pascal Triangle to the algebra and analysis that is the binomial series for negative fractional index. I am of the priesthood that knows what is going on, but I am not afraid to start with the specific and go from there to the general. That, after all, is how mathematics has developed, and continues to develop. — Preceding unsigned comment added by 31.52.122.90 (talk) 08:36, 6 March 2017 (UTC)[reply]

@31.52.122.90: I suggest you tone down your language a bit rather than calling PMajer's "I respect your opinion" "ridiculous". Wikipedia's articles are written by volunteer editors on their spare time. It is also important to note that Wikipedia is not a textbook.

I see you mentioned Pascal's triangle. That, however is not directly applicable when

\alpha \notin \mathbb {N}

, in which case the reader needs the notion of an infinite series. So you can't avoid certain "jargon" like "infinite series".

With that said, I do think the lead can be revised to explain how one formally considers binomial series as a way to generalize a binomial expansion. Oftentimes as well, what I find missing in mathematics is motivation: one does not come up with definitions without a good reason to.--Jasper Deng (talk) 08:53, 6 March 2017 (UTC)[reply]

Thank you, Jasper Deng. I am new to Wikipedia commenting and editing and I had no intention other than to be arresting. Of course "infinite series" is an unavoidable technical term: it is also an analytically involved term. It is, however, not difficult to illustrate "divergence" arithmetically, so we have a way into motivating a discussion of "convergence". That is what I propose in my talk. — Preceding unsigned comment added by 31.52.122.90 (talk) 09:23, 6 March 2017 (UTC)[reply]

@31.52.122.90: We don't have to explain that in this article, though. The article on infinite series covers that. I should note though that it may be good to add an example of divergence of a binomial series in particular (versus convergence of a general series); one good example is trying to "evaluate" the binomial series where

\alpha =x=-1

, which is just alternating 1's and -1's and hence divergent. This might be suitable to add to the section on convergence.--Jasper Deng (talk) 09:31, 6 March 2017 (UTC)[reply]

Proof

As remarked by User:Rickhev1 in a recent edit, the proof of convergence was not complete in that it referred to the inequality $\scriptstyle \left|{\alpha \choose k}\right|\leq {\frac {M}{k^{1+\operatorname {Re} \,\alpha }}}$ , 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. --pm a 16:45, 20 December 2011 (UTC)[reply]

binomial coefficient & probability relation

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)[reply]

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)[reply]

redirect of Newton's binomial theorem

Why does Newton's binomial theorem redirect here and not to Binomial theorem? — MFH:Talk 04:37, 25 August 2014 (UTC)[reply]

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)[reply]

Overlap in conditions for convergence?

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)[reply]

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.pm a 08:37, 2 November 2019 (UTC)[reply]

But boring?

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)[reply]

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. —

Q

uantling (talk | contribs) 16:53, 9 November 2023 (UTC)[reply]

Relationship to the geometric series

@Tito Omburo: et al.: the article includes the recently added paragraph

The negative binomial series includes the case of the geometric series (when $\alpha =1$ ) and, more generally, series obtained by differentiation of the geometric series (for all positive integers $\alpha$ ).^[1]

However, I think this is wrong. The negative binomial sequence is the set of coefficients of the Maclaurin series expansion $(1 - x) - α$ . When $α = 1$ , this sequence is $1, 1, 1, ...$ , which is a geometric sequence, but only one of many. When $α$ is any other value, it is not a geometric sequence.

As to differentiation of the geometric sequence, what does that mean? A geometric series is something like $1 + 1/2 + 1/4 + 1/8 + ...$ ; what does it mean to differentiate that? A geometric sequence replaces those additions with commas, but I still don't know what it means to differentiate it. Yes, a generating function like $(1 - x) - α$ (for negative binomial) or $(1 - αx) -1$ (for geometric) can be differentiated with respect to its indeterminant $x$ , but that is a whole different ball of wax. Even if there were reason to believe that the coefficients in the derivative of the generating function were of interest, that makes the negative binomial sequence for integer $α$ be the $α$ -th "derivative of the geometric series" that started as $1, 1, 1, ...$ , but has no connection to any other geometric sequences of any sort.

The supplied reference does not include the word "negative", so I don't see how it backs up the new paragraph.

I propose we delete the recently added paragraph. — $Q$ uantling (talk | contribs) 16:45, 9 November 2023 (UTC)[reply]

Still not sure what the issue is with this addition. It is surely relevant that one gets the binomial series (as commonly understood to be a power series) with negative integral

\alpha

by differentiation of a geometric series (again a power series). Indeed, this is probably the most commonly used special case, apart from the standard binomial theorem itself, yet until now a mystifying omission from the article. (Although seeing your reaction to this rather uncontroversial fact, I find it less mystifying.) The supplied reference says that the geometric series is a special case of the binomial series in the case of

p=-1

. Tito Omburo (talk) 17:07, 9 November 2023 (UTC)[reply]

@Tito Omburo can you please try to stay polite here and in edit summaries? –jacobolus (t) 18:13, 9 November 2023 (UTC)[reply]

@Tito Omburo: Please help me to understand your view.

I think that you are starting with $(1 - x) - α$ the generating function for the negative binomial sequence. Is that right, or are you starting with $(1 - αx) -1$ , the generating function for the geometric sequence?
Whichever you start with, I think you are differentiation with respect to $x$ , not with respect to $α$ , right?
After differentiating, I think that you are immediately looking at the coefficients of $[x n]$ ; is that right or you evaluating the expression by setting $x$ to some value, such as 1?
With $α$ , you are just looking at its values at positive integers, right?

Thank you —

Q

uantling (talk | contribs) 18:20, 9 November 2023 (UTC)[reply]

Where did I say that you take a derivative of a power series with respect to

\alpha

? See the article. Tito Omburo (talk) 18:28, 9 November 2023 (UTC)[reply]

FWIW, the generating function for the negative binomial series at positive integer

\alpha

is

te^{td/dx}(1-x)^{-1}=t(1-x-t)^{-1}

. Tito Omburo (talk) 18:37, 9 November 2023 (UTC)[reply]

I think I've figured out how my perspective is different from that of Tito Omburo. I think of the generating function of the geometric sequence as $(1 - αx) -1$ , meaning that I can find the $n$ -th term by looking at the coefficient of $x n$ . That coefficient is $α n$ which is the $n$ -th term of a geometric sequence with parameter $α$ . However, by geometric series I believe that Tito Omburo means $(1 - x) -1$ , without the $α$ . In this case, the $n$ -th term is $x n$ , which is the $n$ -th term of a geometric sequence with parameter $x$ . If this was already obvious to you, I apologize for being slow to figure it out. (If I have it wrong, I apologize for that too.) — $Q$ uantling (talk | contribs) 19:24, 9 November 2023 (UTC)[reply]

Yes right. Tito Omburo (talk) 12:04, 10 November 2023 (UTC)[reply]

Good. I think the current article text explains things well. Thank you for that. —

Q

uantling (talk | contribs) 14:24, 10 November 2023 (UTC)[reply]

References

^ George Andrews (2018), "The geometric series in calculus" (PDF), The American Mathematical Monthly, 105 (1): 36–40, doi:10.1080/00029890.1998.12004846

[1] George Andrews (2018), "The geometric series in calculus" (PDF), The American Mathematical Monthly, 105 (1): 36–40, doi:10.1080/00029890.1998.12004846

[1]