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Is there a recurrence relation which has no closed formula?
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From what I know, it is unknown whether $x_n=x_{n-1}^2 + 1$ has a closed form. Is there a recurrence relation which is known to have no closed form with a proof of inexistence?
Assuming a closed form is a non recursive description using the elementary operations of addition multiplication and power, or, assuming any other good definition of "closed form".
Edit
There are similar questions out there, but the answers are a little bit going around the question.
So is there one with proof or is it unknown? And if it depends on the definition of a closed form, then what are the (or some) options? How strong can a closed form definition be to still have a recursive relation that can be proved to not being able to have its form?
number-theory elementary-number-theory
asked 8 hours ago
AladinAladin
493 bronze badges
New contributor
Aladin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment
|
$begingroup$
From what I know, it is unknown whether $x_n=x_{n-1}^2 + 1$ has a closed form. Is there a recurrence relation which is known to have no closed form with a proof of inexistence?
Assuming a closed form is a non recursive description using the elementary operations of addition multiplication and power, or, assuming any other good definition of "closed form".
Edit
There are similar questions out there, but the answers are a little bit going around the question.
So is there one with proof or is it unknown? And if it depends on the definition of a closed form, then what are the (or some) options? How strong can a closed form definition be to still have a recursive relation that can be proved to not being able to have its form?
number-theory elementary-number-theory
asked 8 hours ago
AladinAladin
493 bronze badges
New contributor
Aladin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
This example is a good candidate for an example without a closed form, but I am not aware of a proof either.
$endgroup$
– Peter
8 hours ago
3
$begingroup$
Take a look at cs.stackexchange.com/questions/27598/…
$endgroup$
– John Douma
8 hours ago
$begingroup$
Fix a language for closed forms, each a finite string in a countable alphabet. There are countably many forms, so enumerate them. In the spirit of Richard's paradox, define $a_n$ as the $n$th form's $n$th term $+1$. Clearly, $a_n$ is none of the original forms, but it looks like I've defined it as one. This can be resolved if you know about metalanguages, but the moral is we need to formalise what a closed form is, or at least which notation is legal in it, to properly consider some questions. I wonder whether your question is one of them.
$endgroup$
– J.G.
6 hours ago
$begingroup$
Much of the difficulty here depends on formalizing what a closed form is. I'm not aware of a generally accepted definition. In practice it turns out that closed forms aren't as big a deal as you might think.
$endgroup$
– Qiaochu Yuan
5 hours ago
add a comment
|
$begingroup$
From what I know, it is unknown whether $x_n=x_{n-1}^2 + 1$ has a closed form. Is there a recurrence relation which is known to have no closed form with a proof of inexistence?
Assuming a closed form is a non recursive description using the elementary operations of addition multiplication and power, or, assuming any other good definition of "closed form".
Edit
There are similar questions out there, but the answers are a little bit going around the question.
So is there one with proof or is it unknown? And if it depends on the definition of a closed form, then what are the (or some) options? How strong can a closed form definition be to still have a recursive relation that can be proved to not being able to have its form?
number-theory elementary-number-theory
asked 8 hours ago
AladinAladin
493 bronze badges
New contributor
Aladin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
From what I know, it is unknown whether $x_n=x_{n-1}^2 + 1$ has a closed form. Is there a recurrence relation which is known to have no closed form with a proof of inexistence?
Assuming a closed form is a non recursive description using the elementary operations of addition multiplication and power, or, assuming any other good definition of "closed form".
Edit
There are similar questions out there, but the answers are a little bit going around the question.
So is there one with proof or is it unknown? And if it depends on the definition of a closed form, then what are the (or some) options? How strong can a closed form definition be to still have a recursive relation that can be proved to not being able to have its form?
number-theory elementary-number-theory
number-theory elementary-number-theory
asked 8 hours ago
AladinAladin
493 bronze badges
New contributor
Aladin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
AladinAladin
493 bronze badges
New contributor
Aladin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
Aladin
asked 8 hours ago
AladinAladin
493 bronze badges
New contributor
Aladin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
AladinAladin
493 bronze badges
asked 8 hours ago
AladinAladin
493 bronze badges
493 bronze badges
New contributor
Aladin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aladin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
This example is a good candidate for an example without a closed form, but I am not aware of a proof either.
$endgroup$
– Peter
8 hours ago
3
$begingroup$
Take a look at cs.stackexchange.com/questions/27598/…
$endgroup$
– John Douma
8 hours ago
$begingroup$
Fix a language for closed forms, each a finite string in a countable alphabet. There are countably many forms, so enumerate them. In the spirit of Richard's paradox, define $a_n$ as the $n$th form's $n$th term $+1$. Clearly, $a_n$ is none of the original forms, but it looks like I've defined it as one. This can be resolved if you know about metalanguages, but the moral is we need to formalise what a closed form is, or at least which notation is legal in it, to properly consider some questions. I wonder whether your question is one of them.
$endgroup$
– J.G.
6 hours ago
$begingroup$
Much of the difficulty here depends on formalizing what a closed form is. I'm not aware of a generally accepted definition. In practice it turns out that closed forms aren't as big a deal as you might think.
$endgroup$
– Qiaochu Yuan
5 hours ago
add a comment
|
2
$begingroup$
This example is a good candidate for an example without a closed form, but I am not aware of a proof either.
$endgroup$
– Peter
8 hours ago
3
$begingroup$
Take a look at cs.stackexchange.com/questions/27598/…
$endgroup$
– John Douma
8 hours ago
$begingroup$
Fix a language for closed forms, each a finite string in a countable alphabet. There are countably many forms, so enumerate them. In the spirit of Richard's paradox, define $a_n$ as the $n$th form's $n$th term $+1$. Clearly, $a_n$ is none of the original forms, but it looks like I've defined it as one. This can be resolved if you know about metalanguages, but the moral is we need to formalise what a closed form is, or at least which notation is legal in it, to properly consider some questions. I wonder whether your question is one of them.
$endgroup$
– J.G.
6 hours ago
$begingroup$
Much of the difficulty here depends on formalizing what a closed form is. I'm not aware of a generally accepted definition. In practice it turns out that closed forms aren't as big a deal as you might think.
$endgroup$
– Qiaochu Yuan
5 hours ago
2
2
$begingroup$
This example is a good candidate for an example without a closed form, but I am not aware of a proof either.
$endgroup$
– Peter
8 hours ago
$begingroup$
This example is a good candidate for an example without a closed form, but I am not aware of a proof either.
$endgroup$
– Peter
8 hours ago
3
3
$begingroup$
Take a look at cs.stackexchange.com/questions/27598/…
$endgroup$
– John Douma
8 hours ago
$begingroup$
Take a look at cs.stackexchange.com/questions/27598/…
$endgroup$
– John Douma
8 hours ago
$begingroup$
Fix a language for closed forms, each a finite string in a countable alphabet. There are countably many forms, so enumerate them. In the spirit of Richard's paradox, define $a_n$ as the $n$th form's $n$th term $+1$. Clearly, $a_n$ is none of the original forms, but it looks like I've defined it as one. This can be resolved if you know about metalanguages, but the moral is we need to formalise what a closed form is, or at least which notation is legal in it, to properly consider some questions. I wonder whether your question is one of them.
$endgroup$
– J.G.
6 hours ago
$begingroup$
Fix a language for closed forms, each a finite string in a countable alphabet. There are countably many forms, so enumerate them. In the spirit of Richard's paradox, define $a_n$ as the $n$th form's $n$th term $+1$. Clearly, $a_n$ is none of the original forms, but it looks like I've defined it as one. This can be resolved if you know about metalanguages, but the moral is we need to formalise what a closed form is, or at least which notation is legal in it, to properly consider some questions. I wonder whether your question is one of them.
$endgroup$
– J.G.
6 hours ago
$begingroup$
Much of the difficulty here depends on formalizing what a closed form is. I'm not aware of a generally accepted definition. In practice it turns out that closed forms aren't as big a deal as you might think.
$endgroup$
– Qiaochu Yuan
5 hours ago
$begingroup$
Much of the difficulty here depends on formalizing what a closed form is. I'm not aware of a generally accepted definition. In practice it turns out that closed forms aren't as big a deal as you might think.
$endgroup$
– Qiaochu Yuan
5 hours ago
add a comment
|
1 Answer
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$begingroup$
This may be kind of weak, but in a certain sense, Liouville's theorem guarantees there is no closed formula for $a_0 =0$ and $a_n = a_{n-1} + int_{n-1}^n e^{-x^2} text{d}x$.
answered 8 hours ago
Cade ReinbergerCade Reinberger
1,1273 silver badges11 bronze badges
$endgroup$
$begingroup$
No closed formula with only elementary functions. Certainly a closed formula using erf.
$endgroup$
– marty cohen
6 hours ago
$begingroup$
And that's an important point in many ways. Closed is something like ill definited
$endgroup$
– Cade Reinberger
6 hours ago
add a comment
|
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$begingroup$
This may be kind of weak, but in a certain sense, Liouville's theorem guarantees there is no closed formula for $a_0 =0$ and $a_n = a_{n-1} + int_{n-1}^n e^{-x^2} text{d}x$.
answered 8 hours ago
Cade ReinbergerCade Reinberger
1,1273 silver badges11 bronze badges
$endgroup$
$begingroup$
No closed formula with only elementary functions. Certainly a closed formula using erf.
$endgroup$
– marty cohen
6 hours ago
$begingroup$
And that's an important point in many ways. Closed is something like ill definited
$endgroup$
– Cade Reinberger
6 hours ago
add a comment
|
$begingroup$
This may be kind of weak, but in a certain sense, Liouville's theorem guarantees there is no closed formula for $a_0 =0$ and $a_n = a_{n-1} + int_{n-1}^n e^{-x^2} text{d}x$.
answered 8 hours ago
Cade ReinbergerCade Reinberger
1,1273 silver badges11 bronze badges
$endgroup$
$begingroup$
No closed formula with only elementary functions. Certainly a closed formula using erf.
$endgroup$
– marty cohen
6 hours ago
$begingroup$
And that's an important point in many ways. Closed is something like ill definited
$endgroup$
– Cade Reinberger
6 hours ago
add a comment
|
$begingroup$
This may be kind of weak, but in a certain sense, Liouville's theorem guarantees there is no closed formula for $a_0 =0$ and $a_n = a_{n-1} + int_{n-1}^n e^{-x^2} text{d}x$.
answered 8 hours ago
Cade ReinbergerCade Reinberger
1,1273 silver badges11 bronze badges
$endgroup$
This may be kind of weak, but in a certain sense, Liouville's theorem guarantees there is no closed formula for $a_0 =0$ and $a_n = a_{n-1} + int_{n-1}^n e^{-x^2} text{d}x$.
answered 8 hours ago
Cade ReinbergerCade Reinberger
1,1273 silver badges11 bronze badges
answered 8 hours ago
Cade ReinbergerCade Reinberger
1,1273 silver badges11 bronze badges
answered 8 hours ago
Cade ReinbergerCade Reinberger
1,1273 silver badges11 bronze badges
answered 8 hours ago
Cade ReinbergerCade Reinberger
1,1273 silver badges11 bronze badges
1,1273 silver badges11 bronze badges
$begingroup$
No closed formula with only elementary functions. Certainly a closed formula using erf.
$endgroup$
– marty cohen
6 hours ago
$begingroup$
And that's an important point in many ways. Closed is something like ill definited
$endgroup$
– Cade Reinberger
6 hours ago
add a comment
|
$begingroup$
No closed formula with only elementary functions. Certainly a closed formula using erf.
$endgroup$
– marty cohen
6 hours ago
$begingroup$
And that's an important point in many ways. Closed is something like ill definited
$endgroup$
– Cade Reinberger
6 hours ago
$begingroup$
No closed formula with only elementary functions. Certainly a closed formula using erf.
$endgroup$
– marty cohen
6 hours ago
$begingroup$
No closed formula with only elementary functions. Certainly a closed formula using erf.
$endgroup$
– marty cohen
6 hours ago
$begingroup$
And that's an important point in many ways. Closed is something like ill definited
$endgroup$
– Cade Reinberger
6 hours ago
$begingroup$
And that's an important point in many ways. Closed is something like ill definited
$endgroup$
– Cade Reinberger
6 hours ago
add a comment
|
Aladin is a new contributor. Be nice, and check out our Code of Conduct.
Aladin is a new contributor. Be nice, and check out our Code of Conduct.
Aladin is a new contributor. Be nice, and check out our Code of Conduct.
Aladin is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
This example is a good candidate for an example without a closed form, but I am not aware of a proof either.
$endgroup$
– Peter
8 hours ago
3
$begingroup$
Take a look at cs.stackexchange.com/questions/27598/…
$endgroup$
– John Douma
8 hours ago
$begingroup$
Fix a language for closed forms, each a finite string in a countable alphabet. There are countably many forms, so enumerate them. In the spirit of Richard's paradox, define $a_n$ as the $n$th form's $n$th term $+1$. Clearly, $a_n$ is none of the original forms, but it looks like I've defined it as one. This can be resolved if you know about metalanguages, but the moral is we need to formalise what a closed form is, or at least which notation is legal in it, to properly consider some questions. I wonder whether your question is one of them.
$endgroup$
– J.G.
6 hours ago
$begingroup$
Much of the difficulty here depends on formalizing what a closed form is. I'm not aware of a generally accepted definition. In practice it turns out that closed forms aren't as big a deal as you might think.
$endgroup$
– Qiaochu Yuan
5 hours ago