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What is f'(0), the derivative of f at 0? It doesn't even exist, therefore it has no discontinuity at 0. Darboux's theeorem says that there is no way to create

What is f'(0), the derivative of f at 0? It doesn't even exist, therefore it has no discontinuity at 0. Darboux's theeorem says that there is no way to create a jump in the derivative, in part because a derivative at a point is defined in terms of limits from both sides, so the limits must be the same.

jerf 8y ago
Ah, I see. This is a definition other than the one I understood, which is most assuredly less precise than the one understood by people who have extensively studied analysis. (This being the internet, let me be very clear that I'm basically saying my definition was wrong.) (I'm pretty decent in mathematics in the general sense, reasoning from axioms, proofs, etc. But as I came up on the computer science side, I'm very lopsided into discrete mathematics, which is a bit unusual. Almost every other way to become a good mathematician makes you lopsided into real analysis and the fields that build on that.)
- jordigh 8y ago
  There's a certain cultural thing in mathematics that is kind of hard to convey unless you've been around other mathematicians about what is a definition and what constitutes proof. And these things definitely change culturally with time. Our current standards of analysis are quite modern. From a modern viewpoint, nobody really bothered to define continuity and limits until the 19th century, even though calculus is from the 17 century. The 19th century cultural fixes had a purpose: actual misunderstandings and/or errors crept up in previously published works. So when we get pedantic about what things and use early 20th century formalism, we do so kind of as a reaction to historical misunderstandings. When we ask, "what is the derivative of f at 0?", we're trying to show holes in understanding that have been patched by more modern frameworks.
thaumasiotes 8y ago
> What is f'(0), the derivative of f at 0? It doesn't even exist, therefore it has no discontinuity at 0. This is definitely wrong. The derivative of |x| is -1 where x < 0, and 1 where x > 0, and doesn't exist where x = 0. That is a perfect match to the definition of a jump discontinuity -- the limit from the left is not equal to the limit from the right. It's not at all necessary for the function to exist at x = 0 in order for it to have a discontinuity at x = 0. But hey, don't take my word for it; why not check the definition on Wolfram? http://mathworld.wolfram.com/JumpDiscontinuity.html http://mathworld.wolfram.com/JumpDiscontinuity.html The original claim was "the derivative, if it exists, cannot have a jump discontinuity." This is badly stated. You're defending the idea that if the derivative exists at a particular point, then there is no jump discontinuity in the derivative at that point. But there can be a function f which satisfies both of these properties: - f is the derivative of some other function F. ("The derivative of F exists.") - f has a jump discontinuity, somewhere. ("The derivative of F has a jump discontinuity.")
- disconcision 8y ago
  The definition you link states that a function has a discontinuity at a point /in its domain/ if yadda yadda. 0 is not in the domain of f'. See for example: https://math.stackexchange.com/questions/1431796/if-a-function-is-undefined-at-a-point-is-it-also-discontinuous-at-that-point https://math.stackexchange.com/questions/1431796/if-a-functi...
  - thaumasiotes 8y ago
    That is a question of your personal focus. For example, I'd expect a theorem that applied to "functions from ℝ to ℝ" to apply to f(x) = 1/x unless a specific qualifier was given.

jerf 8y ago

Ah, I see. This is a definition other than the one I understood, which is most assuredly less precise than the one understood by people who have extensively studied analysis. (This being the internet, let me be very clear that I'm basically saying my definition was wrong.) (I'm pretty decent in mathematics in the general sense, reasoning from axioms, proofs, etc. But as I came up on the computer science side, I'm very lopsided into discrete mathematics, which is a bit unusual. Almost every other way to become a good mathematician makes you lopsided into real analysis and the fields that build on that.)

jordigh 8y ago
There's a certain cultural thing in mathematics that is kind of hard to convey unless you've been around other mathematicians about what is a definition and what constitutes proof. And these things definitely change culturally with time. Our current standards of analysis are quite modern. From a modern viewpoint, nobody really bothered to define continuity and limits until the 19th century, even though calculus is from the 17 century. The 19th century cultural fixes had a purpose: actual misunderstandings and/or errors crept up in previously published works. So when we get pedantic about what things and use early 20th century formalism, we do so kind of as a reaction to historical misunderstandings. When we ask, "what is the derivative of f at 0?", we're trying to show holes in understanding that have been patched by more modern frameworks.

thaumasiotes 8y ago

> What is f'(0), the derivative of f at 0? It doesn't even exist, therefore it has no discontinuity at 0. This is definitely wrong. The derivative of |x| is -1 where x < 0, and 1 where x > 0, and doesn't exist where x = 0. That is a perfect match to the definition of a jump discontinuity -- the limit from the left is not equal to the limit from the right. It's not at all necessary for the function to exist at x = 0 in order for it to have a discontinuity at x = 0. But hey, don't take my word for it; why not check the definition on Wolfram? http://mathworld.wolfram.com/JumpDiscontinuity.html http://mathworld.wolfram.com/JumpDiscontinuity.html The original claim was "the derivative, if it exists, cannot have a jump discontinuity." This is badly stated. You're defending the idea that if the derivative exists at a particular point, then there is no jump discontinuity in the derivative at that point. But there can be a function f which satisfies both of these properties: - f is the derivative of some other function F. ("The derivative of F exists.") - f has a jump discontinuity, somewhere. ("The derivative of F has a jump discontinuity.")

disconcision 8y ago
The definition you link states that a function has a discontinuity at a point /in its domain/ if yadda yadda. 0 is not in the domain of f'. See for example: https://math.stackexchange.com/questions/1431796/if-a-function-is-undefined-at-a-point-is-it-also-discontinuous-at-that-point https://math.stackexchange.com/questions/1431796/if-a-functi...
- thaumasiotes 8y ago
  That is a question of your personal focus. For example, I'd expect a theorem that applied to "functions from ℝ to ℝ" to apply to f(x) = 1/x unless a specific qualifier was given.