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Is the sequence $0, 0, 0, 0 ...$ geometric? If so how would you define it? In order to define a geometric sequence you need the first term, and the ratio of terms. In this case you could have:

$a = 0$
$r = k$ for some $k \in \mathbb{R}$

Is this still geometric, even though a single unique definition doesn't exist (a non variable $r$)?

EDIT: This is an interesting debate. But you could also say that $0, 0, 0, 0 ...$ is an arithmetic sequence. So to all who are saying that it is geometric, can a sequence be both geometric and arithmetic?

talfred
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  • The problem is You can not have $\frac{a_{i+1}}{a_{i}}=r\neq \frac{0}{0}$ , where $i$ is any $ith $ term. – Someone May 14 '15 at 19:13
  • In response to the edit. Well since a sequence defined with an exponent $x$ will always change (increase or decrease depending on whether the $x$>1) at a faster rate than a sequence defined linearly, it appears that the only hope of having the sequences remain equal at all terms would be to have the sequence be defined in such a way that it is constant at all terms. Which explains why 0 is unique in this respect. – J.Gudal May 14 '15 at 19:36
  • @J.Gudal So you can have a sequence which is both geometric and arithmetic? – talfred May 14 '15 at 19:43
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    $1,1,1,1,1,...$ is a sequence geometric, arithmetic and harmonic. :) – Someone May 14 '15 at 19:44
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    @talfred Yes. Consider $U_{k} = 1^{k} $ $V_{k} =0k+1$ – J.Gudal May 14 '15 at 19:46

6 Answers6

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Ultimately it will depend on what you consider the definition of a geometric series, as you can see by the other answers here. If the definition relies on calculating a common ratio $r = a_{k+1}/a_k$, then you will run into division by $0$.

You can avoid that problem by saying that a geometric sequence is one whose terms obey the property $a_ka_{k+2} = a_{k+1}^2$. Another way is to define the sequence as $a_k = cr^k$.

Personally, I would say that the sequence $0,0,0,\ldots$ is a geometric sequence in the same way that a point is a circle of radius $0$.

Théophile
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    A very nice reference on your last sentence. :) – Atvin May 14 '15 at 19:21
  • @Theophile. Any interest in heading the International Committee for Standardizing Mathematical Definitions and Notation. :) – matt biesecker May 14 '15 at 19:22
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    @mattbiesecker Ha! Next thing you know, we'd be replacing $\pi$ in the literature with $\tau /2$ ... – Théophile May 14 '15 at 19:23
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    I like your last point! So you say it is geometric, but then I say given the sequence $0, 0, 0, 0...$ and given that it is geometric, what is $r$ for this sequence? – talfred May 14 '15 at 19:24
  • $r$ can be anything, since $c=0$? – Atvin May 14 '15 at 19:25
  • @Atvin but does this uniquely define the sequence? Seeing as you could take $r \in \mathbb{R}$ there exists infinitely many definitions for this sequence. So does a geometric sequence need a single unique definition, or can it be defined in more than one way? – talfred May 14 '15 at 19:27
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    Here's another metaphor similar to my last sentence: as a vector, $\langle 0,0 \rangle$ has no clearly defined direction. Do we still allow it as a vector? – Théophile May 14 '15 at 19:31
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    @Théophile You are the master of metaphors tonight. Could I hear your take on the edit in the original post? – talfred May 14 '15 at 19:36
  • I thought a point was a non-empty interval of length 0 :P – BCLC May 14 '15 at 19:37
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    @talfred Sure: I think the sequence $0,0,0, \ldots$ can be many things: geometric, arithmetic, exponential, and so on. It's not a particularly interesting sequence, but that doesn't mean we should shun it. :) The way I see it, it's much better to include the trivial sequence in the definition than to exclude it. Otherwise, we'd end up with such cumbersome statements as: "Any sequence of the form $a_k = cr^k$ is either a geometric sequence or the zero sequence." Similarly, "The solution to the ODE $y' = y$ is an exponential function or the zero function." It's exhausting even typing that out. – Théophile May 14 '15 at 19:54
  • @Théophile It looks like we need a rigorous definition of a geometric sequence! I believe a similar issue to the one we are having would occur for prime numbers, if "greater than 1" was not included in the definition. This debate about geometric sequences troubles me, as it is "wishy-washy" and Maths should be rigorous and definite. – talfred May 14 '15 at 20:02
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    @talfred The use of different definitions between different authors doesn't mean that there's lack of rigour. If the same author used two different definitions in the same work, then there would be a problem. It's more a question of which definition is most useful or appropriate, and this can change over time. The zero sequence is certainly not geometric according to Euclid, but we've moved on from there. – Théophile May 14 '15 at 20:27
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Just to play devil's advocate, here are some theorems we lose if we allow a sequence that ends in zeroes to be geometric:

  • A geometric sequence converges if and only if the common factor is in $(-1,1]$. (Counterexample: $0\cdot 2^{n-1}$ does converge).

  • A geometric sequence has a sum if and only if the common factor is numerically less than $1$. (Counterexample: $0\cdot 2^{n-1}$ does have a sum).

  • If $(a_n)$ and $(b_n)$ are geometric sequences and for some $i$ we have both $a_i=b_i$ and $a_{i+1}=b_{i+1}$, then $a_i=b_i$ for all $i$. (Counterexample: $0,0,0,\ldots$ versus $1,0,0,\ldots$).

Pragmatically it is probably most useful to exclude these sequences from being "geometric" such that the statements of the general properties can be kept simple. If we need to apply the theorems in a situation where the sequence might be degenerate, it is usually very simple to handle degenerate cases by ad-hoc methods.

hmakholm left over Monica
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  • Thank you for your contribution! I personally think that it is not geometric, as you can construct a counter-argument for it not being geometric, hence it is not rigorously accepted as being geometric. You could say the reverse: but it's more ideal in my eyes to not consider it geometric, until it has been rigorously proved (geometric or not) beyond all doubt. – talfred May 14 '15 at 20:07
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    @talfred: It is not a matter of proof! Which definition to use is a choice we make. – hmakholm left over Monica May 14 '15 at 20:08
  • So depending on the definition you choose, you end up with a different outcome! It should either be geometric, or it is not. Just like $1$ is not a prime number, because it is included in the definition of a prime number. – talfred May 14 '15 at 20:11
  • @talfred: Once you decide what you want the word "geometric sequence" to mean, the sequence is either geometric, or it is not. Just like once you decide what you want the word "prime" to mean, the meaning you decide on will either apply to $1$ or it won't. – hmakholm left over Monica May 14 '15 at 20:15
  • But $1$ is mathematically accepted as not being prime. Anyone saying that it is prime would be considered wrong. – talfred May 14 '15 at 20:18
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    @talfred: It is true that the particular definition of "prime" that most mathematicians have chosen to use leads to $1$ not being prime. That doesn't negate the fact that using this definition was a choice, rather than a necessary truth that is amenable to proof or disproof. – hmakholm left over Monica May 14 '15 at 20:20
  • @talfred: And I don't know why you bring up primes, but if you just take a look at http://en.wikipedia.org/wiki/Prime_number you will see that the definition of prime is different for different people in different contexts. Anyway the basic point which Henning has been trying to get across is that a definition is a choice of what is to be given what name, and a proof is a sequence of deductions from axioms that are allowed by inference rules. Definitions simply go into the list of axioms to fix the interpretation of certain predicate symbols or function symbols. – user21820 May 15 '15 at 01:15
  • Your last point is invalid as stated. $1,1,1,1,\ldots$ and $1,-1,1,-1,\ldots$ are bona fide distinct geometric sequences that coincide in more than one position. This could be repaired by requiring coincidence in two successive positions. But personally I think this is a rather unimportant property, and cannot imagine any real use being made of it. – Marc van Leeuwen May 15 '15 at 07:04
  • @Marc: Fixed. (And it might be that the devil has a poor case here, but I work with what I have to work with). – hmakholm left over Monica May 15 '15 at 14:38
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It depends on the exact definition, using @Atvin definition, it is not geometric. Using an alternate definition, e.g. "a geometric sequence has the form $a_n = c r^{n-1},$ where $c,r$ are constants." Taking $c=0$ generates your sequence.

matt biesecker
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  • Interesting take on that definition. Kind of interesting, that with the one I used, we got a non-existant thing, and with this, it seems ok. – Atvin May 14 '15 at 19:19
  • $0/0$ is not non-existent. Its indeterminate. There is a distinction. Indeterminate means that there exists a value, we just don't know what it is. – J.Gudal May 14 '15 at 19:54
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    @J.Gudal: $0/0$ is usually actually undefined. You may be thinking of $0^0$ which is an indeterminate form, but also defined as $1$. – hmakholm left over Monica May 14 '15 at 19:57
  • @Makholm http://mathworld.wolfram.com/Indeterminate.html – J.Gudal May 14 '15 at 19:58
  • @J.Gudal: $0/0$ happens to be both an indeterminate form and undefined. – hmakholm left over Monica May 14 '15 at 20:00
  • @Makholm But there is a subtle distinction.between the two. $0/0$ has no value that we can ascertain. But suppose if we decided that $2 \times 0=0 \rightarrow 2= \frac{0}{0} $ then we have defined $0/0=2$ right? But starting from $0/0$ we cannot determine what this value would be. I may be deeply mistaken though, so I'm happy to be corrected. – J.Gudal May 14 '15 at 20:05
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Using the definition: $\frac{ a_{n+1}}{a_n} = q$

It can't be a geometric series, since by definition it comes, that $\frac{ a_{n+1}}{a_n} = q$(for every $n \in \mathbb{N}$), where $q$ quotient is fixed, but in that case, you have $\frac00$, which leads to divising with $0$, and you can't do that. :)

Atvin
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  • This is my opinion. However you could still "generate" the series by taking $r = 1$ and $a = 0$, which would generate $0, 0, 0, 0 ...$. Because I've just generated a sequence using the two defining features $a$ and $r$ is this not geometric? – talfred May 14 '15 at 19:21
  • What is $a$ and what is $r$? :) – Atvin May 14 '15 at 19:23
  • Well @talfred , it depends on definition. You can use various definition to formulate a geometric progression. But one definition of common-ratio is the one stated in this answer. So you can't have that as $\frac{0}{0}$. However you can avoid this problem by defining your geometric series without the help of $r$. – Someone May 14 '15 at 19:23
  • @Mann This is my issue! There doesn't appear to be a strict mathematical argument as to whether it is geometric or not. I don't like this, maths should be rigorous. – talfred May 14 '15 at 19:32
  • Now that it seems, maybe it indeed can be considered as one. When properly defined. For e.g., the nice analogies by theophile :) @talfred – Someone May 14 '15 at 19:33
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    @talfred: Mathematics can be rigorous once we have decided which definitions and/or axioms to work from. However, the choice of which definitions it is worthwhile to use is a more pragmatic and fuzzy question. Ultimately it comes down to opinion about which definitions are so useful that they should get the short words. – hmakholm left over Monica May 14 '15 at 20:06
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We could also define a geometric sequence as any sequence such that each term ,after the first term, is the geometric mean of its successor and predecessor.

In which case, the sequence given would satisfy this definition.

I suppose it comes down to which definition we are using.

J.Gudal
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As others have noted, it all depends on how precisely one defines the term geometric sequence. One can define anything one likes; for instance Wikipedia gives the definition that the general term at position $n$ (starting from $0$) should be of the form $ar^n$ with the rather ridiculous condition that $r\neq0$ though $a$ is unconstrained; that would make $0,0,\ldots$ into a geometric series of undetermined ratio $r\neq0$ (taking $a=0$), but $1,0,0,\ldots$ would not be a geometric series (here the ratio would clearly have to be $0$, but that was forbidden).

Part of the problem here is that one tries to define the unqualified notion "geometric progression" (or sequence), while in practice one most often deals with geometric progressions of specified ratio $r$ (in which case the general term is simply $ar^n$ where $a$ is the initial term); this shifts the focus from generating a geometric progression (which really never poses any problem) to recognising a given sequence as being geometric (this leads to considering ratios of successive terms, which is problematic if they are $0$; it still does not explain why one should forbid $r$ to be $0$, but not $a$).

To take some distance from the rather uninspiring "I found a source that says this" approach, let me view geometric sequences as an instance of anther notion, that of eigenvector. In the vector space of infinite sequences, the geometric sequences with ratio $r$ are precisely the eigenvectors of the "guillotine operator", which removes the initial term of a sequence and shifts the other terms to fill its place, for the eigenvalue$~r$. From this perspective, one sees why the zero sequence should have a special status; normally the zero vector is explicitly forbidden as eigenvector (because it would be so for any value of$~\lambda$, whether or not it is an eigenvalue), although it is included in any eigenspace. For the exact same reason allowing the zero sequence as a geometric sequence is somewhat problematic (no unique ratio can be determined for it) but it is OK (in fact necessary) to include the zero sequence in the subspace of geometric sequences with given ratio$~r$.

Note that viewing geometric sequences as eigenvectors is quite natural; they are used in this role when solving linear recurrence relations with constant coefficients.

Marc van Leeuwen
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