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Say I have a sequence that goes:

12, 16, 21, 27, 34, 42, 51 and so on in a manner that the increment between $a_1$ and $a_2$ is 4, between $a_2$ and $a_3$ is is 5 etc.

The formula which describes this series is $(n(n+7)/2)+12$.

My question is: would there be a way to find out the formula above given I have only the "description of its behavior" or a number of its terms, say $a_1$ to $a_7$?

Now, I have a hunch that this may not be possible, given meme formulae such as $f(x)=18111/2x^4-90555x^3+633885/2x^2-452773x+217331$, which produces the series 1, 3, 5, 7, 217341, meaning that even the most "normal looking" sequences can go wild, and also this website (oeis.org) which seems to archive lots of sequences.

Still, I'm guessing there might be a way that produces one of the possible formulae, even if probably no method can infer them all.

Is it possible?

BTW, I don't have any higher education on Mathematics, which means, in my country, that with what I was given at high school, I would be ready to start taking Calculus 1 in college. Nevertheless, I'm very interested and often watch niche videos on YouTube without much trouble keeping up.

paul garrett
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Fabio Freitas
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    I hope you'll pardon me for making a few small corrections... in particular, your example sequence is not "geometric" (not is it "arithmetic") in the usual senses of the word. – paul garrett Aug 30 '22 at 21:07
  • Hm, I actually thought that all sequences were one type or the other. Thank you for the corrections! – Fabio Freitas Aug 30 '22 at 21:11
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    Have a look at https://www2.math.upenn.edu/~wilf/AeqB.html – Rob Arthan Aug 30 '22 at 21:12
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    Well, for any set of $m<\infty$ real numbers $(y_1,...,y_m)$ and all set of $m$ distinct integers $k_1,...,k_m$, there exists a polynomial $f\colon \Bbb R \to \Bbb R$ such that $f(k_i)=y_i$ for $i=1,\ldots,m$. So I guess that the answer to your question is yes, and for a finite set of given terms, there even always infinitely many functions generating this "finite sequence"... – Surb Aug 30 '22 at 21:15
  • Interesting, I didn't even consider that this wasn't a given. The actual question, though, is about finding any of these functions besides guessing them. My bet is that either it's impossible, or there's a method that will output only a fraction of the many possible functions to produce the sequence. – Fabio Freitas Aug 30 '22 at 21:21
  • Surb's comment sums this up nicely. But even worse, if you were to give me infinitely many terms in the sequence - or even all of them - there are still infinitely many functions which give those values at precisely $n= 1, 2, \ldots$. – mathematics2x2life Aug 30 '22 at 21:21
  • I think that many of the commenters are focusing too much on the "finite number of terms" aspect of things, and not the "I have a description of how to generate the terms". The example in the question is a first order inhomogeneous linear difference equation. The answer to the first question is "there are infinitely many possible solutions", but the second question is very answerable (and has related answers on the site). – Xander Henderson Aug 30 '22 at 21:33
  • @FabioFreitas I have closed this question as a duplicate of other questions. None of these questions is a precise duplicate, but they all deal with similar kinds of recurrence relations. – Xander Henderson Aug 30 '22 at 21:34
  • This question might also be useful in learning some of the vocab: https://math.stackexchange.com/q/246221/ – Xander Henderson Aug 30 '22 at 21:36
  • For the other version of the question, there is some relevant meta discussion: https://math.meta.stackexchange.com/q/21093/ and https://math.meta.stackexchange.com/q/28409/ , for example. – Xander Henderson Aug 30 '22 at 21:42
  • @XanderHenderson Comparing to those answers you provided, I don't think I managed to express myself correctly. All of those answers have a starting algorithm, while in my case, with "I have a description of how to generate the terms" I didn't mean "(n(n+7)/2)+12", but "the first increment is 5, than 6, than 7 and so on". Perhaps another way I could have approached this would be asking how one would go about writing that phrase I quoted in mathematical notation. – Fabio Freitas Aug 30 '22 at 21:43
  • @FabioFreitas Yes, "the first difference is 4, then 5, then 6" gives rise to the inhomogeneous linear recurrence relation $a_{n+1} - a_{n} = n+3$. You then have the initial condition $a_1 = 12$. The dupe targets tackle this kind of problem. – Xander Henderson Aug 30 '22 at 21:44
  • @XanderHenderson on the meta discussion, this answer was actually what I was looking for, although I already had points 1 and 3 by myself. This second answer on the second question actually talks about polynomial interpolation, which seems to be what I was looking for: I now know that there is a method for getting one or *the simpler** function, even though I don't think I grasp it. – Fabio Freitas Aug 30 '22 at 21:52

1 Answers1

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At least as a place-holder response: given any finite sequence of numbers (real, complex, whatever), there is a definite algorithmic procedure to determine a polynomial $P$ whose values $P(1), ... , P(n)$ are those given entities. Likewise, if we want a linear combination of exponentials (instead of a linear combination of $x^n$'s), we can do that as well, though the determination of coefficients is far less elementary than in the polynomial case.

One form of the computations in the polynomial case use the finite-difference operator $\Delta(f)(x)=f(x+1)=f(x)$, and the sequence of polynomials that have the property that $\Delta f_{n+1}=f_n$. Namely, $x(x+1)(x+2)\ldots(x+n)$ or similar.

To be clear, though, given a finite sequence, there is truly no uniquely-determined "next" element in that sequence, unless there are restrictions. Sure, as in school exams and so on, sometimes there are "obvious", but, even then, given $1,2,4,8,16$, one exponential continuation is that the next thing is $32$, but the polynomial interpolation gives $31$.

EDIT: yes, there is also a notion of (Kolmogorov) complexity of a means/description of/for computing the sequence elements. So we might ask for "the/a simplest" formula/algorithm that reproduces the given finite sequence. And we might ask for a sort of meta-algorithm that produces that algorithm, given the finite sequence. Not trivial issues, although some versions are answerable.

paul garrett
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  • and once you have found THE polynomial of minimum degree (<= n-1) or any function, passing through the n points, you can always add a polynomial / function which has zeros at those points ( plus everywhere else) – G Cab Aug 30 '22 at 22:40
  • @GCab indeed! So, to be clear, "the next element in the sequence" really can be anything, just by using a polynomial of the next higher degree. :) – paul garrett Aug 30 '22 at 22:43
  • not just next higher, but even infinite, like $\sin x ,; 1/ \Gamma(-x)$ etc. – G Cab Aug 30 '22 at 23:05
  • pls. correct to $\sin(x/\ pi)$ – G Cab Aug 30 '22 at 23:12