Timeline for What conditions imply that a function over $\mathbb{Z}$ is a polynomial? [closed]
Current License: CC BY-SA 3.0
30 events
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| Jul 2, 2015 at 15:01 | vote | accept | Halbort | ||
| Jul 2, 2015 at 15:01 | vote | accept | Halbort | ||
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| Jul 2, 2015 at 0:40 | vote | accept | Halbort | ||
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| Jul 1, 2015 at 19:39 | vote | accept | Halbort | ||
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| Jul 1, 2015 at 13:50 | comment | added | Halbort | Why is this off topic? | |
| Jul 1, 2015 at 5:37 | history | closed |
Bill Johnson Felipe Voloch Qiaochu Yuan Neil Strickland Alex Degtyarev |
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| Jul 1, 2015 at 4:37 | answer | added | Aaron Meyerowitz | timeline score: 2 | |
| Jul 1, 2015 at 3:58 | comment | added | Duchamp Gérard H. E. | @GeraldEdgar I think you gave the answer and this iterated difference method was used in the old times (Galileo ...). The first time $\Delta^n$ annihilates $F$ is the degree of $F$ plus one. | |
| Jul 1, 2015 at 3:07 | history | edited | Halbort | CC BY-SA 3.0 |
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| Jul 1, 2015 at 3:04 | comment | added | David Roberts♦ | I think perhaps by the way it was phrased. If it were something like: "What conditions on a function imply that it is a polynomial, if I am only allowed to use its values on the integers?" Otherwise it looks like a total rookie question. It is best to specify what sort of function, and what the domain is. Is it defined on $\mathbb{R}$? $\mathbb{C}$? Is it smooth? Analytic? Merely continuous? | |
| Jul 1, 2015 at 1:55 | comment | added | Halbort | Why was my question downvoted. | |
| Jul 1, 2015 at 1:53 | answer | added | Per Alexandersson | timeline score: 3 | |
| Jul 1, 2015 at 1:05 | comment | added | Halbort | Are there any other ways besides finite differences? | |
| Jul 1, 2015 at 1:01 | comment | added | Gerald Edgar | Only defined on the positive integers? Take Anthony's suggestion, and convert to this: take the difference finitely many times, and get identically zero. The difference $\Delta F$ of $F$ is: $\Delta F(n) = F(n+1)-F(n)$. | |
| Jul 1, 2015 at 0:51 | review | Close votes | |||
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| Jul 1, 2015 at 0:32 | history | edited | Halbort | CC BY-SA 3.0 |
added 122 characters in body; edited tags
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| Jul 1, 2015 at 0:16 | comment | added | Halbort | @Robert Israel Sorry I feel stupid. | |
| Jul 1, 2015 at 0:11 | comment | added | Halbort | @Robert Israel I may be missing something but doesn't $e^x$ satisfy your criteria | |
| Jun 30, 2015 at 23:39 | comment | added | Robert Israel | If $f$ is entire and there are at least two values that $f$ takes only finitely many times, then $f$ is a polynomial. | |
| Jun 30, 2015 at 21:38 | comment | added | Halbort | OK I see what you mean Geoff Robinson. Thank you for your help! | |
| Jun 30, 2015 at 21:37 | comment | added | Geoff Robinson | If you have the function value at enough points, and it really is polynomial, then the polynomial given by Lagrange interpolation will eventually stabilize to the right polynomial, and adding extra points and function values will not change it. If that doesn't happen, the function isn't polynomial. | |
| Jun 30, 2015 at 21:20 | comment | added | Otis Chodosh | A classic strengtening of @AnthonyQuas's answer: suppose that for any $x$ there exists $n=n(x)$ so that $f^{(n)}(x)=0$. Then $f$ is a polynomial. mathoverflow.net/questions/34059/… | |
| Jun 30, 2015 at 21:16 | comment | added | Halbort | Yes but how would I prove that something is a polynomial as opposed to some other type of function that passes through those values. | |
| Jun 30, 2015 at 21:14 | comment | added | Geoff Robinson | Lagrange interpolation if you know enough values of your function. You implicitly did this sort of thing with your sin example. A polynomial function which takes the value 0 infinitely often is identically 0, which sin(x) is not. | |
| Jun 30, 2015 at 21:11 | comment | added | Halbort | @Geoff Robinson How would I use that without constructing the polynomial. | |
| Jun 30, 2015 at 21:10 | comment | added | Geoff Robinson | A polynomial of degree $n$ is uniquely determined from its values at $n+1$ distinct points. | |
| Jun 30, 2015 at 21:10 | comment | added | Anthony Quas | If you differentiate it finitely many times and get 0 everywhere, maybe? | |
| Jun 30, 2015 at 21:09 | comment | added | Halbort | Yes exactly David Speyer thank you: +1 | |
| Jun 30, 2015 at 21:08 | comment | added | David E Speyer | A minor extension of Liouville's theorem: If $f: \mathbb{C} \to \mathbb{C}$ is holomorphic, and $|f(z)| = O(|z|^N)$ as $|z|\to \infty$, then $f$ is a polynomial of degree $\leq N$. Is this the sort of thing you are looking for? | |
| Jun 30, 2015 at 21:01 | history | asked | Halbort | CC BY-SA 3.0 |