Timeline for Polynomials and divided differences
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2013 at 20:04 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
corrected typo in title; clarified grammar in body; added tag
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| May 27, 2013 at 15:33 | comment | added | George | Actually this is the point of the question, that in the class of all continuous functions on [0, 1], only the polynomials of degree < or = N satisfy that condition. Intuitively, this is suggested by the fact that the set of knots 1/(N+m), ..., (N+m)/(N+m), when m takes all the natural numbers, is dense in [0, 1]. | |
| May 27, 2013 at 13:09 | comment | added | Todd Trimble | I don't know enough about the background of this question to determine whether it is "off-topic" -- but I ask people who vote to close on such or similar grounds to make sure they do know. | |
| May 27, 2013 at 12:28 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
formatting
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| May 27, 2013 at 12:14 | comment | added | JCM | Cheers for confirming. | |
| May 27, 2013 at 11:16 | comment | added | George | Thank you for your remark, in the question of course that f must be supposed at least continuous on [0, 1], but as I said in the Remark, it could be supposed to be differentiable too, if that hypothesis would be helpful for the proof. Irene | |
| May 27, 2013 at 11:12 | comment | added | JCM | In your question, it is not clear to me whether or not continuity of $f$ on $[0,1]$ is a requirement ... is it? Otherwise $f:[0,1]\to\mathbb{R}$ given by, $$ f(x) = \begin{cases} 0 &;x\in\mathbb{Q}\cap [0,1] \newline 1 &;x\in[0,1]\backslash \mathbb{Q} \end{cases}$$ may be relevant. | |
| May 27, 2013 at 7:13 | history | asked | George | CC BY-SA 3.0 |