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| ▲ | lucianbr 2 hours ago | parent | next [-] |
| Person who draws comparison from current situation to derivatives points out that derivatives rules don't apply to current situation. Awesome stuff. |
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| ▲ | ajkjk 32 minutes ago | parent [-] | | I 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. |
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| ▲ | Neywiny 4 hours ago | parent | prev | next [-] |
| Not 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 |
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| ▲ | ajkjk 4 hours ago | parent | next [-] | | You 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 3 hours ago | parent [-] | | I'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. |
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| ▲ | alwa 3 hours ago | parent | prev [-] | | Not 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… |
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| ▲ | alwa 3 hours ago | parent | prev | next [-] |
| As 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. |
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| ▲ | umanwizard 4 hours ago | parent | prev [-] |
| Derivatives 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. |
