I discovered a formula for the nth term of any sequence of numbers

[karpmage]

Hi there,

I recently discovered a formula for the nth term that works for any finite sequence of numbers. I was just wondering whether a formula has already been discovered, and if not, how and where I should publish it.

To give you an example of what i mean:

one formula for the nth term of any sequence with two numbers is:
(Un)=(u1)+((u2)-(u1))(n-1)

I've tried googling a formula but have come up with nothing. I asked my maths teacher and he didn't know either. He recommended that I consider publishing it.

(I'm worried that if I post the formula on this website, someone might steal it and I won't get credit for its discovery (if it is in fact my discovery). I'd also like your opinion on this.)

Thanks.

 

[jbunniii]

one formula for the nth term of any sequence with two numbers is:
(Un)=(u1)+((u2)-(u1))(n-1)

This isn't a formula for an arbitrary sequence, only an arithmetic progression, i.e. one which adds a fixed number at each step, such as 3, 8, 13, 18, 23, 28, ... Of course it's well known, almost trivial. Most high school algebra books mention it at some point.

There can't be any formula for predicting an element of a general sequence in terms of the other elements. Proof: I'll tell you the other elements, you tell me what number your formula predicts for the remaining one, and I'll spoil your day by picking a different number.

 

[karpmage]

This isn't a formula for an arbitrary sequence, only an arithmetic progression, i.e. one which adds a fixed number at each step, such as 3, 8, 13, 18, 23, 28, ... Of course it's well known, almost trivial. Most high school algebra books mention it at some point.

There can't be any formula for predicting an element of a general sequence in terms of the other elements. Proof: I'll tell you the other elements, you tell me what number your formula predicts for the remaining one, and I'll spoil your day by picking a different number.

I used this formula as an example as I knew it was very well known. You're right when you say that you can pick a different number. If you're given any sequence of numbers and asked to find the next number (and the sequence isn't defined as arithmetic, geometric, etc.), you can give any number and you will still be correct, provided that you have a formula to back it up.

 

[jbunniii]

I used this formula as an example as I knew it was very well known. You're right when you say that you can pick a different number. If you're given any sequence of numbers and asked to find the next number (and the sequence isn't defined as arithmetic, geometric, etc.), you can give any number and you will still be correct, provided that you have a formula to back it up.

My point is that your formula can't work for all possible sequences. So what class of sequences does it work for?

 

[pwsnafu]

Hi there,

I recently discovered a formula for the nth term that works for any finite sequence of numbers.

Sounds like polynomial interpolation.

If you're given any sequence of numbers and asked to find the next number (and the sequence isn't defined as arithmetic, geometric, etc.), you can give any number and you will still be correct, provided that you have a formula to back it up.

You can always find such a formula: see Newton polynomial.

 

[CompuChip]

Or I can write down a formula stating that the elements of the sequence are equal to the N numbers that I gave you for n <= N and to some arbitrary other sequence not matching your prediction for n > N.

Anyhow, I think we are now misinterpreting the question. For example, I can give you {1, 1} as the first two elements. How will you know if I'm talking about {1, 1, 1, 1, 1}, {1, 1, 2, 3, 5} or {1, 1, 2, 2, 3}?

 

[karpmage]

From what I can see, there's no real point publishing what I've found. So I'm going to post my formula here.

The formula takes the form of the binomial expansion, where:

uⁿ= (u-1)⁰/0! + (u-1)¹*(n-1)/1! + (u-1)²*(n-1)(n-2)/2! + (u-1)³*(n-1)(n-2)(n-3)/3! + ...

you then need to expand the sequence and replace uⁿ with u_n+1 (for example u³ gets replaced with u₄. Only "uⁿ" gets replaced by u_n)

This will give you:

u_n= u₁ + (u₂-u₁)(n-1)+(u₃-2u₂+u₁)(n-1)(n-2)/2 + (u₄-3u₃+3u₂-u₁)(n-1)(n-2)(n-3)/6 + ...

Which works for any given sequence of numbers.

I'm not sure whether this is the same as what you guys have posted.

 

[micromass]

I've tried googling a formula but have come up with nothing. I asked my maths teacher and he didn't know either. He recommended that I consider publishing it.

(I'm worried that if I post the formula on this website, someone might steal it and I won't get credit for its discovery (if it is in fact my discovery). I'd also like your opinion on this.)

Sorry man, you were beaten by the 6 year old Gauss: http://www.education2000.com/demo/demo/botchtml/arithser.htm

 

[karpmage]

My point is that your formula can't work for all possible sequences. So what class of sequences does it work for?

the issue isn't with the formula, it's with the number of numbers in the sequence. As I said, this gives a formula for the nth term for any sequence where only the first two terms are given. Of course, a sequence with only two terms could be arithmetic, or geometric, or something else altogether. With the information given, it's impossible to determine the next term, but a formula for the nth term can be found.

 

[karpmage]

Sorry man, you were beaten by the 6 year old Gauss: http://www.education2000.com/demo/demo/botchtml/arithser.htm

it doesn't have to be an arithmetic series. The formula that I displayed in my first post isn't the one that I'm talking about. The formula that you've shown isn't a formula for the nth term, at least not that I can see.

 

[johnqwertyful]

the issue isn't with the formula, it's with the number of numbers in the sequence. As I said, this gives a formula for the nth term for any sequence where only the first two terms are given. Of course, a sequence with only two terms could be arithmetic, or geometric, or something else altogether. With the information given, it's impossible to determine the next term, but a formula for the nth term can be found.

Sequences, by definition, are infinite.

Also "impossible to determine the next term" but somehow you can find the nth term? What about n=3?

 

[karpmage]

Sorry, my formula was wrong because I expanded (u-1)³ wrong. It's corrected now.

 

[karpmage]

Or I can write down a formula stating that the elements of the sequence are equal to the N numbers that I gave you for n <= N and to some arbitrary other sequence not matching your prediction for n > N.

Anyhow, I think we are now misinterpreting the question. For example, I can give you {1, 1} as the first two elements. How will you know if I'm talking about {1, 1, 1, 1, 1}, {1, 1, 2, 3, 5} or {1, 1, 2, 2, 3}?

That's more of a problem with sequences in general. You can say that about any sequence. For example, how would know, with absolute certainty, if {1,2,4,8} leads to {1,2,4,8,16,32} from the formula 2ⁿ or {1,2,4,8,15,26} as my formula states (by stopping to input values at a point such that it is a cubic. In actual fact, you could say that it is a quintic and put whatever values you want in.)

 

[pwsnafu]

u_n= u₁ + (u₂-u₁)(n-1)+(u₃-2u₂+u₁)(n-1)(n-2)/2 + (u₄-3u₃+3u₂-u₁)(n-1)(n-2)(n-3)/6 + ...

This is what I think you are claiming:

We have some finite list of numbers, for example {1,3,5,7,11,13}. We substitutes these numbers into the expression I quoted, u₁ = 1, u₂ = 3, etc. Then the right hand side is some expression in the variable n. If we set n = 3, for example, the right hand side evaluates to 5.

Is this correct? Are you claiming anything more? Why did your teacher suggest you should publish?

 

[karpmage]

Sequences, by definition, are infinite.

Also "impossible to determine the next term" but somehow you can find the nth term? What about n=3?

Let me rephrase, it's impossible to determine what is the intended next term, or intended formula for the nth term (except by luck) because there is an infinite number of terms that could take the place of n=3. Essentially, there is no need to do any calculations when determining the next term, since any answer is correct. This kind of makes questions like "Ex [2] The next term of 5, 11, 17, 23,... is ________." (http://www.math-magic.com/sequences/next_term.htm) pointless, unless it is explicitly stated to be an arithmetic or geometric, etc. sequence. One example where you could put in any answer is "The next term of 2, 5, 14, 41,... is _______." Does this mean that examiners can't mark these questions incorrect? I'd like to know.

I also said that you can find "A" formula, rather "the intended" formula

 

[pwsnafu]

Let me rephrase, it's impossible to determine what is the intended next term, or intended formula for the nth term (except by luck) because there is an infinite number of terms that could take the place of n=3. Essentially, there is no need to do any calculations when determining the next term, since any answer is correct. This kind of makes questions like "Ex [2] The next term of 5, 11, 17, 23,... is ________." (http://www.math-magic.com/sequences/next_term.htm) pointless, unless it is explicitly stated to be an arithmetic or geometric, etc. sequence. One example where you could put in any answer is "The next term of 2, 5, 14, 41,... is _______." Does this mean that examiners can't mark these questions incorrect? I'd like to know.

I also said that you can find "A" formula, rather "the intended" formula

In mathematics? There is no wrong answer.

In an IQ test? It's to test whether you can identify patterns so there is a right answer.

 

[karpmage]

This is what I think you are claiming:

We have some finite list of numbers, for example {1,3,5,7,11,13}. We substitutes these numbers into the expression I quoted, u₁ = 1, u₂ = 3, etc. Then the right hand side is some expression in the variable n. If we set n = 3, for example, the right hand side evaluates to 5.

Is this correct? Are you claiming anything more? Why did your teacher suggest you should publish?

This is all I'm claiming. When he said that he suggested that I publish, I think he was talking generally, as I spend a lot of time thinking about stuff like this. He didn't look into too much detail when I showed him either. He didn't actually check up on whether it was real or not either, he just didn't know off of the top of his head, so I probably shouldn't have put that in. Is this basically the same thing as polynomial interpolation/linear interpolation/Newton polynomial? Did they have a general formula?

 

[karpmage]

In mathematics? There is no wrong answer.

In an IQ test? It's to test whether you can identify patterns so there is a right answer.

These type of questions came up in my igcse exam. They were phrased similarly to the ones I posted. I remember one of them gave a sequence of 5 or 6 numbers, and then asked for the next term and a formula for the nth term. It was something like 3*((2^n)-1). However my sequence would also have given the correct number. As for the ones that didn't ask for the next number, would I have been incorrect if I didn't give the intended answer, but explained why it still works?

 

[karpmage]

http://en.wikipedia.org/wiki/Finite_difference

I just found this. The segment on Newton Series is pretty much exactly the formula that I put down, just phrased a little differently. I'm going to have to concede that Newton beat me to it.

 

[karpmage]

Sounds like polynomial interpolation.



You can always find such a formula: see Newton polynomial.

I didn't understand the formulas on the Wikipedia page that you posted here, but now I do. I see that this discussion could have ended a lot sooner.

 

[micromass]

These type of questions came up in my igcse exam. They were phrased similarly to the ones I posted. I remember one of them gave a sequence of 5 or 6 numbers, and then asked for the next term and a formula for the nth term. It was something like 3*((2^n)-1). However my sequence would also have given the correct number. As for the ones that didn't ask for the next number, would I have been incorrect if I didn't give the intended answer, but explained why it still works?

How would you use your sequence to get the right answer?

Let's say you are given 2,3,5,7,11,13,... (= the primes) How would you determine the next number using your method? If you can do this, then you have discovered a formula for the sequence of prime numbers.

 

[Simon Bridge]

In addition to what the others have already said...

I recently discovered a formula for the nth term that works for any finite sequence of numbers.

That is the claim under discussion.
When considering the replies, you should realize that this is a specific claim in mathematics ... a mathematician will read this to mean that the discovered formula can predict the next number in the series.

(I'm worried that if I post the formula on this website, someone might steal it and I won't get credit for its discovery (if it is in fact my discovery). I'd also like your opinion on this.)

This comes up a lot - one of the advantages of making a discovery public is that lots of people witness you as the originator of the discovery. In the event of a dispute, you'll have a record of exactly when and how you made it first. If you keep it secret, then someone else may beat you to the punch, and then you have no proof you got there first.

You certainly should not tell people you've made an important discovery until you are prepared to tell them what it is.

That's more of a problem with sequences in general. You can say that about any sequence. For example, how would know, with absolute certainty, if {1,2,4,8} leads to {1,2,4,8,16,32} from the formula 2ⁿ or {1,2,4,8,15,26} as my formula states (by stopping to input values at a point such that it is a cubic. In actual fact, you could say that it is a quintic and put whatever values you want in.)

Which is why your formula does not satisfy the claim you made in post #1 (quoted above). The general nature of sequences defeats you. It can, however, work for a particular class of sequences. That is, if it is supposed to be predictive... but I suspect the following quote shows that you don't mean to predict the actual next number ... i.e. if I fed you the sequence 3141 - the formula would give you "a" next number consistent with these, but probably won't continue 5926 (the sequence of digits in pi).

Let me rephrase, it's impossible to determine what is the intended next term, or intended formula for the nth term (except by luck) because there is an infinite number of terms that could take the place of n=3. Essentially, there is no need to do any calculations when determining the next term, since any answer is correct. This kind of makes questions like "Ex [2] The next term of 5, 11, 17, 23,... is ________." (http://www.math-magic.com/sequences/next_term.htm) pointless, unless it is explicitly stated to be an arithmetic or geometric, etc. sequence. One example where you could put in any answer is "The next term of 2, 5, 14, 41,... is _______." Does this mean that examiners can't mark these questions incorrect? I'd like to know.

Depends on what is being examined.

Exams are to be taken in the context of the material that is being examined and not just the literal content of the exam paper. The student is expected to make a judgement about what is intended - that is part of the test.

I suspect you mean that you have found a systematic way to generate a number that is consistent with the sequence of numbers that have come before. i.e. out of the infinite possible continuations of a sequence of numbers, this formula is a way of picking one.

That is not the same as a formula for finding the next number in a sequence ... at least not the way mathematicians understand those words. To illustrate this - consider the sequence (suggested by micromass) of primes.

The formula would generate "a" number for the next in the sequence, consistent with the previous ones (and the assumptions that went into the formula), but it probably won't be a prime number. The additional information that it is has to be a prime number makes all the difference.

Back to the exam situation - it is the context of the exam question that restricts the kinds of answers that will be considered correct.

eg. If it is a pattern-recognition test in math, then 31415 may be recognized as the sequence of digits in pi ... so the next three numbers would be 926, but a puzzle from MENSA would suggest that we should be suspicious of such an easy answer and look further, like the sequence of chimes on a clock that chimes once each half-hour... so the continuation would be 161... thogh that would be a fairer puzzle if an earlier sequence were chosen like 19110111112111213 ...

This help?

 

[micromass]

Depends on what is being examined.

Exams are to be taken in the context of the material that is being examined and not just the literal content of the exam paper. The student is expected to make a judgement about what is intended - that is part of the test.

I suspect you mean that you have found a systematic way to generate a number that is consistent with the sequence of numbers that have come before. i.e. out of the infinite possible continuations of a sequence of numbers, this formula is a way of picking one.

That is not the same as a formula for finding the next number in a sequence ... at least not the way mathematicians understand those words. To illustrate this - consider the sequence (suggested by micromass) of primes.

The formula would generate "a" number for the next in the sequence, consistent with the previous ones (and the assumptions that went into the formula), but it probably won't be a prime number. The additional information that it is has to be a prime number makes all the difference.

Back to the exam situation - it is the context of the exam question that restricts the kinds of answers that will be considered correct.

eg. If it is a pattern-recognition test in math, then 31415 may be recognized as the sequence of digits in pi ... so the next three numbers would be 926, but a puzzle from MENSA would suggest that we should be suspicious of such an easy answer and look further, like the sequence of chimes on a clock that chimes once each half-hour... so the continuation would be 161... thogh that would be a fairer puzzle if an earlier sequence were chosen like 19110111112111213 ...

This help?

I would like to add to this excellent post that IQ-tests and pattern-recognition tests are not mathematics. If I see on an IQ test the following sequence

1,~2,~4,~8,~16,~32,~64

then an IQ test will want you to say that 128 is the correct next value. But in mathematics, things don't work that way. In mathematics, there is no single correct next value in the sequence, unless they explicitly define what the correct next value is (but that usually defeats the entire concept of pattern recognition).

 

[Simon Bridge]

I would like to add to this excellent post

Aw shucks, twern't nothin.

that IQ-tests and pattern-recognition tests are not mathematics.

I will agree - provisionally. However: pattern recognition is often given in many countries as part of a math course (pre-secondary usually).

The course, so early on, usually has not covered many possible series so the test is just seeing if students can recognize them. When they are encountered, especially later, there is always some metadata to be taken into account. Otherwise it's meaningless.

After a while, students learn to state their assumptions explicitly - and it's a harder lesson that that sounds. We all make many assumptions without realizing them. I suspect OP needs to start doing that.

These exercises can be useful as a precursor to mathematical modelling as in fitting curves to data or projecting a trend from previous data. Empiricism yes? The impossibility of finding the next in sequence from the previous ones by themselves is pretty much a form of the philosophical "problem of induction" right? So we make a bunch of assumptions and hold the conclusions provisional.

I'm interested in how OP understands all this ...

 

[micromass]

Of course pattern recognition is an extremely important skill in mathematics. If you're not good in pattern recognition, then you won't be a very good mathematician, or scientist.

My statement only covered IQ tests and pattern recognition tests. They are not math in the sense that the answer isn't well-defined. There is no real "correct" answers. Or rather: there is no definition for what the correct answer should be. Or: there is no algorithm for checking whether your answer is the right one.

It is of course entirely possible to make pattern recognition tests which are mathematical (that is: for which there is a good answer). In fact, most of the exercises on mathematical induction tend to rely on pattern recognition. Of course, the crucial part is that we can also actually check our answer. If we can do that, then it's math.

 

[CompuChip]

In mathematics? There is no wrong answer.

In an IQ test? It's to test whether you can identify patterns so there is a right answer.

I was asked to take such an "IQ test" as part of a job application procedure recently. I more or less openly told them what I thought of these tests, both on how they claim to measure your intelligence as well as how they claim that 3 out of 4 answers are wrong.

Didn't get the job :)

 

[jbunniii]

I was asked to take such an "IQ test" as part of a job application procedure recently. I more or less openly told them what I thought of these tests, both on how they claim to measure your intelligence as well as how they claim that 3 out of 4 answers are wrong.

Didn't get the job :)

Next time I interview someone I should start by asking them to supply the next number in this pattern: $3, 1, 4, 1, 5, \ldots$. If they answer $1$ then they're robots and I don't want to work with them, so I can dismiss them immediately before we waste any more of each other's time. If they answer $9$ then they're thinking mathematically and we can move on to question 2. ("Prove the Banach-Tarski paradox.")

 

[micromass]

Next time I interview someone I should start by asking them to supply the next number in this pattern: $3, 1, 4, 1, 5, \ldots$. If they answer $1$ then they're robots and I don't want to work with them, so I can dismiss them immediately before we waste any more of each other's time. If they answer $9$ then they're thinking mathematically and we can move on to question 2. ("Prove the Banach-Tarski paradox.")

What if they answer 10 and say that they prefer to work in base 9?

 

[jbunniii]

What if they answer 10 and say that they prefer to work in base 9?

Then they can have my boss's job.

 

[CompuChip]

Next time I interview someone I should start by asking them to supply the next number in this pattern: $3, 1, 4, 1, 5, \ldots$. If they answer $1$ then they're robots and I don't want to work with them, so I can dismiss them immediately before we waste any more of each other's time. If they answer $9$ then they're thinking mathematically and we can move on to question 2. ("Prove the Banach-Tarski paradox.")

Haha, that one's really nice! You're probably right, it wouldn't even have occurred to me to continue with 1 (followed by 6, 1, 7, ...) before you mentioned that.

 

[Vargo]

If you carefully read the original post, you would see that there is no intrinsic flaw in his claim. He is not trying to create a nonsense formula for the nth term in a sequence given a list of the first k terms.

Instead he has a finite sequence u_1, ..., u_n. Given those numbers, he finds an explicit polynomial P(n) such that P(i)=u_i. There is no infinite sequence and it has nothing to do with pattern recognition. It is simply a polynomial interpolation and that is all he is claiming.

If the original post were not clear (which is understandable since the OP is not trained in maths), surely some of the follow up posts would have cleared this up??

 

[johnqwertyful]

finite sequence

No such thing exists.

 

[micromass]

No such thing exists.

While you're formally correct, it's a bit hairsplitting don't you think? We all know what he meant with it and many people do use this term.

 

[johnqwertyful]

While you're formally correct, it's a bit hairsplitting don't you think? We all know what he meant with it and many people do use this term.

Not at all. I think it's important to use correct terminology; the way things are formally defined in a formal setting.

 

[micromass]

Not at all. I think it's important to use correct terminology; the way things are formally defined in a formal setting.

It is formal. Many good math books use the terminology "finite sequence". For example: "Mathematical analysis" by Apostol, Definition 2.12.

 

[micromass]

Or "Real Analysis" by Royden, page 9.

I'm sure I can find many others.

 

[micromass]

Or basically just google "finite sequence".

Here is a blog by Terrence Tao who uses the term: http://terrytao.wordpress.com/2007/...nalysis-and-the-finite-convergence-principle/

 

[mathwonk]

i don't think so. i have one in mind whose nth term is ?

 

[Simon Bridge]

If you carefully read the original post, you would see that there is no intrinsic flaw in his claim. He is not trying to create a nonsense formula for the nth term in a sequence given a list of the first k terms.

Instead he has a finite sequence u_1, ..., u_n. Given those numbers, he finds an explicit polynomial P(n) such that P(i)=u_i. There is no infinite sequence and it has nothing to do with pattern recognition. It is simply a polynomial interpolation and that is all he is claiming.

The formula in the first post is not the one OP had in mind - that was posted later. That didn't help matters.

If the original post were not clear (which is understandable since the OP is not trained in maths), surely some of the follow up posts would have cleared this up??

Yet so many people are responding as if the claim is that the formula is predicting the next term in the intended sequence. Like:

i don't think so. i have one in mind whose nth term is ?

... I don't think OP is actually claiming to be able to predict the next term in mathwonk's sequence ... just to be able to come up with "a" next term given what has gone before and the assumptions (reasoning?) built-in to the formula. The formula would be predictive for a particular kind of sequence.

The wording in post #1 is unclear on this point, and muddied, rather than clarified, by crossed wires in later responses, but I think we are at the point where OP can be encouraged to confirm or deny the interpretation unambiguously.

I submit: further discussion is pointless until that happens.

 

[pwsnafu]

The wording in post #1 is unclear on this point, and muddied, rather than clarified, by crossed wires in later responses, but I think we are at the point where OP can be encouraged to confirm or deny the interpretation unambiguously.

I submit: further discussion is pointless until that happens.

He already did so, see the start of post #17. This thread is done, and the original poster said so himself in post #20.

 

[Simon Bridge]

From post #20:

I didn't understand the formulas on the Wikipedia page that you posted here, but now I do. I see that this discussion could have ended a lot sooner.

... not specific.
The confusion still continued though: witness the remaining posts.

I agree the thread is finished with no further input from OP.

 

[Borek]

Locked.