Timeline for lower bounding first derivative of polynomial
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Feb 14, 2011 at 0:30 | comment | added | Yemon Choi | Ram, you misunderstood me. If you know the minimum value of p, call it $M$, and you know some other value of p, call it $K>M$, then the mean value theorem tells you that there exists u in your interval such that $K-M \leq |f'(u)|(b-a)$. I agree with all the other commenters that your original question may not admit an answer; just because you want such a bound to exist, that is not a reason to expect such a bound to exist. | |
| Feb 13, 2011 at 22:13 | answer | added | Helge | timeline score: 1 | |
| Feb 13, 2011 at 20:33 | answer | added | user9072 | timeline score: 1 | |
| Feb 13, 2011 at 10:32 | comment | added | Ram | that would go in recursion of finding min of till d-th derivative of f e.g. min|f'|= min{f'(a),f'(b),min{f'(c)}} where c is root of f" and f"(c)>=0 I need much cleaner approach like "Markoff Theorem" | |
| Feb 13, 2011 at 7:16 | comment | added | Yemon Choi | How about using the Mean Value Theorem, if you know the values of f at some points? | |
| Feb 13, 2011 at 4:01 | history | edited | Ram | CC BY-SA 2.5 |
added 186 characters in body; added 8 characters in body
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| Feb 12, 2011 at 13:57 | answer | added | Joe Silverman | timeline score: 4 | |
| Feb 12, 2011 at 11:18 | answer | added | Ilies Zidane | timeline score: 0 | |
| Feb 12, 2011 at 11:04 | answer | added | user5810 | timeline score: 0 | |
| Feb 12, 2011 at 10:44 | history | edited | Ram | CC BY-SA 2.5 |
added 317 characters in body; Post Made Community Wiki
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| Feb 12, 2011 at 8:39 | history | asked | Ram | CC BY-SA 2.5 |