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One announcement from a company or government can suddenly change the derivative discontinuously. Derivatives irl do not follow the rules of calculus that you
by ajkjk 11mo ago
One announcement from a company or government can suddenly change the derivative discontinuously.
Derivatives irl do not follow the rules of calculus that you learn in class because they don't have to be continuous. (you could quibble that if you zoom in enough it can be regarded as continuous.. But you don't gain anything from doing that, it really does behave discontinuous)
- Neywiny 11mo agoNot sure what kinda calculus you took at least here in the states it's very standard to learn about such functions in class, and yes there is a difference between discontinuous and the slope being really large (though finite) for a brief period of time
- ajkjk 11mo agoYou rarely study delta and step functions in an introductory calculus class. In this case the first derivative would be a step function, in the sense that over any finite interval it appears to be discontinuous. Since you can only sample a function in reality there's no distinguishing the discontinuous version from its smooth approximation. (I suppose a rudimentary version of this is taught in intro calc. It's been a long time so I don't really remember.)
- Neywiny 11mo agoI'm sure it depends on who's teaching the class and what curriculum they follow, but we were doing piecewise linear functions well before differentiation so I think I do actually disagree as per your caveat. It's also possible that the courses triaged different material. As a calc for engineers not calc for math majors taker, my experience may have been heavier on deltas and steps.
- alwa 11mo agoNot to be all “do you know who X is,” but I did have to chuckle a little when I saw who it is that you’re teaching differentiation to here…
- School-Cotton 11mo agoDerivatives in actual calculus don’t have to be continuous either. Consider the function defined by f(x) = x^2 sin(1/x) for x != 0; f(0) = 0. The derivative at 0 exists and is 0, because lim h-> 0 (h^2 sin(1/h))/h = lim h-> 0 (h sin(1/h)), which equals 0 because the sin function is bounded. When x !=0, the derivative is given by the product and chain rules as 2x sin(1/x) - cos(1/x), which obviously approaches no limit as x-> 0, and so the derivative exists but is discontinuous.
- alwa 11mo agoAs seems to have sort of happened between March and April of this year, at least from the Ramp chart in TFA. I wonder what that was about.
- lucianbr 11mo agoPerson who draws comparison from current situation to derivatives points out that derivatives rules don't apply to current situation. Awesome stuff.
- ajkjk 11mo agoI don't understand your point. It seemed like the person I was replying to didn't understand how both claims could be simultaneously true so I was elaborating.