Talk:Geometric progression

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[edit] Paragraph problem

I don't really agree with the paragaph:

One cannot see why the proportion called arithmetical is any more arithmetical than that which is called geometrical, nor why the latter is more geometrical than the former. On the contrary, the primitive idea of geometrical proportion is based on arithmetic, for the notion of ratios springs essentially from the consideration of numbers

That seems a rationalisation based on number, but it is not certain that mathematics started as number.

It is "not certain" that it did? Indeed, isn't it certain that it did not? The notion of real number grew out of geometry!

For example, take a square, double the length of its sides, double again, and again. Clearly the side lengths are in geometric progression; so too are the areas. Take a different square, add 2 to the length of the sides (2 what?), add 2 again, and again. This time the side lengths are in arithmetic progression, but the areas are not. It seems natural to call the former geometric, leaving arithmetic to the latter. In the medieval quadrivium, arithmetic was pure number, geometry was number in space, music number in time, and astronomy number in space and time; but I doubt that was the order in pre-history.--Henrygb 13:13, 21 Mar 2004 (UTC)

Both Geometric sequence and Geometric series deal with the other, so i"m putting them together under Geometric progression and redirecting.

[edit] progression, series or sequence?

Why isn't this article called "geometric sequence" but "geometric progression"? I don't find it clear what is meant with progression. --Abdull 13:43, 19 Jun 2005 (UTC)

a sequence means that the terms are just followed on after the other. a series means that every term is summed to the next. a prograssion just means that the numbers progress, keep going, withut specifyin if the terms are added or not
Cako 20:49, 2 November 2006 (UTC)goldencako

[edit] Interest Rate bad example

The example for return on capital interest is not correct and misleading. Interest rates are calculated using compound interest rates. If one could get $16000 out of $2000 in 6 years, that would be quite awesome. I am not familiar with wikipedia to fix this example, thanks for taking note of it. --Vastinnocentaims 15:20, 16 August 2005 (UTC)

[edit] Mistake in ratio?

I think the exponent in the ratio (second image under Formulae) should be 1 / (n - 1) instead of just n - 1. Anyone can verify this? Aggelos Orfanakos 00:09, August 30, 2005 (UTC)

[edit] missing formula

it appears that the current version of the page (on my display)is missing the first formula, namely the mathethematical definition of the sequence itself, before going into the definitions of the scale factor and common ratio

[edit] Clarification please?

It's been a few years since I studied this subject, so I came on here to help myself solve a little problem I was having, and found that this article was of little help. Can someone please clarify the geometric series part of the article? What is X? I can tell Xn is the Nth term of the geometric sequence, but what is X? Maybe put an example in?--149.135.21.11 14:33, 13 March 2006 (UTC)

[edit] Notation

a_n = a\,r^{n-1} \quad \mbox{where n is an integer such that }n \ge 1

In the case of the above, would it not be more elegant to say "n\in\mathbb{N}" in place of "\mbox{where n is an integer such that }n \ge 1"?

I'd change it myself, but I wouldn't want to mess around with something if there's a specific reason for that form. --MightyPenguin 11:53, 20 March 2006

\mathbb{N} can contain zero, depending on the circumstances (see natural number). As it's slightly ambiguous, it's probably better to leave it as is. siafu 21:44, 2 May 2006 (UTC)


[edit] Credit for Series Simplification

Who is to be given credit for discovering the geometric series simplification? Gauss? 131.120.10.130 19:34, 19 July 2006 (UTC)

[edit] Product

Can anyone add a subsection of the product of a geometric sequence?

[edit] should i add this:

Hey, I was wondering if it would be appropriate to add this to the article: Why do we call it a Geometric Series? If someone knows the answer, please add this here.

[edit] Product

The product of a geometric sequence can be expressed as

P = \prod_{i=0}^{n-1} ar^i

However, it can also be expressed using only the first and last termes and the total number of terms.

P=(a_1 \cdot a_2)^{n/2}

Which leads to the interesting conclusion, that

\sqrt[n]{P}=a_{\frac{(n+1)}{2}}.


If so, should i also include some proofs?
Cako 20:48, 2 November 2006 (UTC)goldencako
A danger for the first claim above is when n is odd, in which case the question arises as to what x1 / 2 is for negative or complex x. E.g. consider the case where n = 1 and a1 = − 1.
As for the second claim, it is evidently only defined for the case that n is odd. Even then, it doesn't hold if r is allowed to be non-real: suppose that n is 3, then multiplying a by cis(2π/3) doesn't change P but does change each term ai (including a_{\frac{(n+1)}{2}}). The claim might hold if n is odd and \sqrt[n]{P} is read as ‘one of the n'th roots of P’ instead of the usual ‘the principal n'th root of P’. (See Nth root.)
More generally: The principles behind [[WP::No original research]] may apply to such derivations: It's good to cite a reliable external source for such derivations, to give more confidence that the corresponding proof has been checked by an expert. (Purely-mathematical deductions are something of a boundary case for the no-original-research principle; e.g. [[WP::Attribution]] says that “straightforward mathematical calculations” don't count as original research.)
Pjrm 13:11, 1 April 2007 (UTC)

[edit] Interesting

I have someone following me around deleting my comments. Anyone want to comment on the following? --JohnLattier 23:06, 3 November 2006 (UTC)


- \sum_{k=0}^\infty ar^k = \lim_{n\to\infty}{\sum_{k=0}^{n} ar^k} - - - Remember, the infinite sum only converges towards the limit. The limit represents the number to which the sum is convergent. The equal sign is improper. --JohnLattier 05:52, 3 November 2006 (UTC) - - I suggest: - - \sum_{k=0}^\infty ar^k \approx \lim_{n\to\infty}{\sum_{k=0}^{n} ar^k} - --JohnLattier 05:55, 3 November 2006 (UTC)

Ugh. You're wrong, and this article is not going to incorporate your suggestion. End of story. Melchoir 23:58, 3 November 2006 (UTC)
I believe it's inappropriate to make such a response to a proposed change in Wikipedia without giving at least some evidence for your position.
(In this case, some evidence is that the Infinite series article claims that the \sum_{n=0}^\infty notation specifically means the limit of the sequence of partial sums. Granted, citing a Wikipedia article is not particularly strong evidence for what should go in another Wikipedia article, but it's considerably better than nothing.)
Even with strong evidence, it is not “end of story”, because opposing evidence can be given. (E.g. in the current case, the dispute is merely about the meaning of a notation, and one notation can sometimes have different meanings. Here, the meaning in the case of divergent series in particular has varied over time, and may well continue to evolve. Maybe the exact proposed change should not be adopted, but maybe some aspect of such an objection could reasonably be incorporated into the relevant article.)
If one disagrees with a proposed change, then one's own conviction does count as evidence, but is to be set against the opposing evidence that the proposer apparently believes that the proposed change is a good one. It suffices to give the evidence one has on hand (“I am confident that the proposed claim is false”, possibly giving some reason for the person to trust your judgement if you think that they too are confident of their position). I understand that one may strongly want to prevent certain changes, but I believe that the response above tends to have the undesirable side effect of dissuading future constructive changes.
Pjrm 14:50, 1 April 2007 (UTC)

[edit] Constant Factor

Why is there a constant factor "a" in all of these formulae? Isn't it more concise to just say \sum_{k=0}^\infty r^k = \frac{1}{1-r} instead of :\sum_{k=0}^\infty ar^k = \frac{a}{1-r} ? Dougthebug 23:36, 5 February 2007 (UTC)

Ans: Usually when deriving a formula you use a general form of the equation. You just can't assume things like a = 1. You need to derive equations using generic values. Sachhidh

[edit] Proof that 1 + 2 + 4 + 8 + ... = -1

n = 1 + 2 + 4 + 8 + 16 + ...

2n = 2 + 4 + 8 + 16 + 32 + ...

n = 1 + 2n

0 = 1 + n

n = -1

I talked to a mathematician about this proof and he didn't see any flaw in it. He said that the only thing that it proves is that if the series does converge, that it would converge to -1. An example of where it would converge to -1 would be in 2-adic. —The preceding unsigned comment was added by 61.108.11.194 (talk • contribs) 06:41, 8 February 2007 (UTC)

That sounds correct, and it could be made more precise. For example, you could say that any summation method for divergent series that is both stable and linear, and sums 1 + 2 + 4 + 8 + · · ·, sums it to −1. Melchoir 22:17, 8 February 2007 (UTC)
A work colleague gave me a variant of this proof, and referred to it as “proof that the universe is two's complement”. (The two's complement representation of −1 is all-bits-one, which in some sense ought to mean 1+2+4+⋯.) Pjrm 06:28, 19 April 2007 (UTC)

Using the formula for the limiting sum anyway: = 1/(1-2) = -1

Looking at the series to the left of the term.. 1/2, 1/4.. etc we get limiting sum 1.

It's analogous to "running the integral" backwards of this exponential function and finding it's limiting area.

(remark)

I'd say its not 'n = 1 + 2n' but rather 'n = 1 + 2n-inf' —The preceding unsigned comment was added by Morgenrodeo (talk • contribs) 08:14, 28 March 2007 (UTC).

There is an error with the proof. you cannot multiply infinity by two since by definition it is the largest number possible. —The preceding unsigned comment was added by 172.207.30.115 (talk • contribs) 06:55, April 1, 2007 (UTC).

[edit] Partial Sum

The line reading: Here n = 1+2+4+8+…….. 2n = 2+4+8+16+…………

Here we can easily say 2n>n

2n-n = (2+4+8+16+…………) – (1+2+4+8+……..) 2n-n = -1 n = 2n+1, which contradict 2n>n so n ≠ -1

[edit] Don't merge individual series into here

The beauty of summary style is that we can develop examples in separate articles, which include unique applications and background. All convergent geometric series are not identical, and they can't be covered in a single article. Melchoir 18:47, 16 March 2007 (UTC)

Most of the content of the individual articles is already here — all convergent geometric series are (pretty much) identical except for where they're referred to in the literature. 1/2 + 1/4 + ... -> Zeno and 0.999, 9/10 + 9/100 + ... to 0.999, 1/2 - 1/4 + 1/8 - ... to Numbers, 1/4 + 1/16 + ... to the previous series (if it is notable), etc. Perhaps there should be an article should be examples of geometric series, split out from the "geometric series" section, here, but the ones here don't deserve an article because no one could type it it exactly, and search probably wouldn't fit it either. (The spaced "dots" rather than unspaced dots or an ellipsis is another problem.... (!)). — Arthur Rubin | (talk) 22:40, 16 March 2007 (UTC)
Well, the individual articles are mostly substubs created without my knowledge; that's why they currently lack unique content. You could make them into redirects, but then I'd just have to recreate them when I get around to it.
Anyway, this article has become an unreadable, tangled list of boring quasi-information. It's a concrete argument against the incremental improvement model. As a reader, if I followed a link to some series and it took me here, I'd feel betrayed. Melchoir 23:21, 16 March 2007 (UTC)

I am for the merge unless the article can be expanded into something unique enough to merit its own article--Cronholm144 22:55, 12 May 2007 (UTC)

[edit] incorrect description of final outcome in one case

  1. Positive, the terms will all be the same sign as the initial term.
  2. Negative, the terms will alternate between positive and negative.
  3. Greater than 1, there will be exponential growth towards positive infinity.

I think that the third line should read:

Greater than 1, there will be exponential growth towards infinity (of the same sign as the first term).

eg, -1, -2, -4, -8 .... common ratio is 2, and is > 1

This is only the second 'post' I have made to Wikipedia. I would rather put this as a discussion item than wade in and edit the main page.

If someone with more experience agrees, would they please alter the main page.

Thank you Jsryork 11:21, 7 May 2007 (UTC)

[edit] What's the purpose of "a" in all these formulae?

Why is a needed in all these formulae? It's clear that it's a constant and therefore it can always come out of the summation... 83.67.217.254 20:41, 21 August 2007 (UTC)

It can come out of the summation when you're solving for the sum of the series, but it's needed when expressing the series as "a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number". If you didn't have a factor of a in each term, you could only express geometric series for which the first term is 1. —David Eppstein 05:09, 27 August 2007 (UTC)

[edit] Discrepancies in term numbering in formulae

1) This page has several discrepancies in numbering of terms:

"Elementary properties" says a_n = a *r^(n-1) i.e. terms start from n=1

however the summation formula and everything from the section "Geometric series" onwards uses: a_k = ar^k starting with k=0

2) The term-index used varies between n, k and even i

Please use one consistent notation and one term-index throughout; most people would favor the notation a_k = ar^k (i.e. numbering always starts from 0) and index k for simplicity. It also keeps the summation formula simpler.

Smcinerney 04:08, 3 September 2007 (UTC)

I have revised the Elementary Properties section to be consistent with the Series section. Journeyman (talk) 23:18, 26 March 2008 (UTC)

[edit] Geometric series article

Hi, I just added an article entirely on the subject of infinite geometric series. Geometric progressions and geometric series are important enough that they deserve separate articles. Jim 21:06, 22 September 2007 (UTC)

[edit] Formula for infinite series

\frac{1}{\frac{1}{a^1}+\frac{1}{a^2}+\frac{1}{a^3}+\frac{1}{a^4}+\frac{1}{a^5}\ldots}=\frac{1}{\sum_{n=1}^{\infty}\frac{1}{a^n}}=a-1;a>0

User:twentythreethousand 22:39, 27 September 2007

[edit] Infinite series for dummies

I had an assignment involving infinite geometric series and at the time I couldn't understand the justification for the formula presented here, as it looked like half of the formula just dissapeared. Later I found that it was because the r^infinity equalled zero. I tried to simplify this line of math so that others wouldn't be as confused as I was... is it mathematically correct tho, ne comments Laubpatr (talk) 04:34, 16 March 2008 (UTC)

The argument you give is not quite correct, since r∞ is not well-defined. However, it is true that if |r| < 1, then the value of rn can be made arbitrarily close to zero by selecting a sufficiently large value of n, that is, limn→∞ rn = 0. Michael Slone (talk) 08:23, 16 March 2008 (UTC)