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The notation (x, y, z, ...) used for infinite sequences is ambiguous (at least wikipedia says so). So there is always more than one possible "rule", which means it is possible to continue the sequence in different ways? Is there a proof for that? Is there a proof that there are always infinite possible "solutions"/"rules"? I would be happy to know the proof or just the name of it. Thank you in advance

amWhy
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Tim Othy
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  • Suppose, for a sequence ${a_n}$, you are given the first 3 terms $a_1=x,\ a_2=y,\ a_3=z$ (as per your notation). Let the rule be "from the 4th term onwards, add j to the previous term", that is $a_n=a_{n-1}+j,\ n\ge 4$, where you can choose your $j$ as you like. Then you have as many rules as $j$. – HackR Oct 21 '22 at 21:40
  • sure, yes, that works. but thats not really what i was looking for. sorry for not asking preciously. i am not looking for a rule defined piecewise. – Tim Othy Oct 21 '22 at 22:32
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    A Lagrange interpolation polynomial can trivially pass through your points and continue in infinitely many ways. On a side note, you seem to think that "being defined piecewise" means something in math. It doesn't. – Ivan Neretin Oct 21 '22 at 22:57
  • i understand. thank you – Tim Othy Oct 26 '22 at 20:06

2 Answers2

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Consider the sequences $$a_n:=(2n+1)_{n=1}^\infty=1, 3, 5, 7, \dots $$ $$b_n:=\left (\frac {18111} {2}n^4 -90555n^3+\frac {633885} {2}n^2-452773n+217331 \right)_{n=1}^\infty=1, 3, 5, 7, \dots$$

You would believe that $a_n=b_n$ for all $n\in \Bbb{N}$ by just reading the first four terms. But you will soon find that $a_5\ne b_5$ as $$a_5=9$$ and $$b_5=217341$$

That is why the notation $a_1,a_2,a_3,\dots $ is ambiguous.

Edit: You can't prove that some notation is ambiguous. You just use examples to show it is not always the best to use this notation. And the notation is not always ambiguous. Most of the time it is quite clear what is meant by writing $a_1,a_2,\dots$ from the context in which it is being written. So the claim that the notation is always ambiguous is just wrong.

Seeker
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  • okay, yes, I see. it is obvious if is defined piecewise. i should have asked more precisely. it should be given that they are not defined piecewise. – Tim Othy Oct 21 '22 at 22:27
  • @TimOthy I have edited my answer to show why that notation is ambiguous. – Seeker Oct 21 '22 at 22:58
  • but this is just one example. i am looking for a proof that it is always ambiguous. – Tim Othy Oct 22 '22 at 11:29
  • @TimOthy It is not clear what you mean by a "proof" that this notation is always ambiguous. Given any finite sequence of terms, I can cook up an infinite number of ways of continuing that sequence. There are "dumb" continuations (e.g. let every subsequent term be equal to $47$), and less "dumb" continuations (e.g. assume that the sequence denotes a set of points $(n,a_n)$, and come up with a polynomial which passes through these points---take $a_n$ to be the value of that polynomial at $n$ for each $n$). – Xander Henderson Oct 28 '22 at 00:31
  • Of course, in most "real" situations, there is no ambiguity, because we are not relying on the notation alone. We can also look at the surround context and infer what an author meant. – Xander Henderson Oct 28 '22 at 00:31
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To show that there are infinitely many possible rules you can use to continue the sequence $$ 1,2,3,\ldots $$ (or any other sequence where you know the first few entries) requires a definition of "rule". If a rule is "describe what comes next, forever" then here are infinitely many rules:

  • continue with $1,1, 1, \ldots$ (1's forever)
  • continue with $2,2,2, \ldots$ (2's forever)
  • continue with $3,3,3, \ldots$ (3's forever)
  • and so on.

Of course in many particular situations the author writes the first few terms to suggest a particular pattern. On an exam you are supposed to guess the pattern - and that kind of question is always essentially ambiguous.

The pattern $$ 1,2,3, \ldots $$ usually suggests the positive integers, but in a context where you were studying recursion it might mean the start of a sequence you get by adding the three previous terms. Then it would continue $$ 1,2,3,6,11,20, \ldots \ . $$ Those are the tribonacci numbers.

Ethan Bolker
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