Timeline for Is $f(x)$ is more curvature than $g(x)$ then length of $f(x)$ seem longer than length of $g(x)$?
Current License: CC BY-SA 4.0
8 events
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| Jul 13, 2018 at 16:33 | comment | added | Iosif Pinelis | @AlexM. : Andreas Blass stated the point of the example exactly right. | |
| Jul 13, 2018 at 13:03 | comment | added | Andreas Blass | @AlexM. I think the point of the example is that we have equality of the two curvatures at points with the same $x$-coordinate, not at points the same distance along the curves. | |
| Jul 13, 2018 at 12:42 | comment | added | Alex M. | @IosifPinelis: "we would have that the length of the graph of a smooth enough function $f$ over $[a,b]$ would be determined by the curvature of the graph and the values of $f$ at $a,b$." But don't we have this? It seems to me that this is an immediate consequence of the fundamental theorem of curves: given the curvature and torsion (in this case $0$), there exists a unique curve possessing them, modulo isometries (theorem 2.1.7). In particular, its length will be unambiguously determined by the curvature function alone. | |
| Jul 12, 2018 at 23:23 | comment | added | Đào Thanh Oai | Your answer is very nice. | |
| Jul 12, 2018 at 23:02 | vote | accept | Đào Thanh Oai | ||
| Jul 12, 2018 at 22:21 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 12, 2018 at 22:04 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 4 characters in body
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| Jul 12, 2018 at 21:57 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |