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Sep 9, 2017 at 2:07 comment added Jahan Claes @Pere Edited to include asymmetric graphs.
Sep 9, 2017 at 2:07 history edited Jahan Claes CC BY-SA 3.0
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Sep 9, 2017 at 2:03 comment added Jahan Claes @Pere It's true that it might be better to draw something obviously non-symmetric on the board, though. The graphs I showed looked a little too symmetric.
Sep 9, 2017 at 2:02 comment added Jahan Claes @Pere You don't need to assume the force is symmetrical, just that it is smooth. Non-symmetrical functions still have Taylor expansions!
Sep 8, 2017 at 22:54 comment added Pere Your argument is that if the force of the spring is symmetrical and smooth, it can be linearized. However, the question is how to justify that the force is symmetrical.
Sep 8, 2017 at 16:06 history answered Jahan Claes CC BY-SA 3.0