In fact, Michiu Kaku was quoted as saying that string theories were an accidental discovery, and that they shouldn't have been discovered for another hundred years, because we don't have the mathematics yet. |
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In fact, Michiu Kaku was quoted as saying that string theories were an accidental discovery, and that they shouldn't have been discovered for another hundred years, because we don't have the mathematics yet. |
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No, I said I didn't understand the point of Riemann sums, not what they are. Why use an estimate when a definite integral gives you an exact answer, and does so much much quicker? Care to shed some light on that, because I still don't understand it. I understand that it's useful in teaching what an definite integral is, but I see zero obvious use cases for it. |
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Last edited by ninja9578; 08-14-2011 at 05:32 PM.
I don't really get how you can conceptualise integrals otherwise. An integral is fundamentally a sum. How do you know that doing an integral of a velocity vs. time curve gives you the distance travelled? How do you know that doing an integral of f'(t) gives you the difference in f(t)? Or, how would you work out the integral for something simple like the arc length of y = f(x)? |
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You can sidestep Riemann sums by taking the limit of approximating step functions. This is really all riemann sums are to begin with. The step functions are just a little cleaner and more abstract. You can treat all the exact same functions in this way. You can treat a wider range of functions using the Lebesgue integral and measure theory in general. But anybody that can do that crap understands Riemann sums and that either they or sequences of step functions are necessary for formally proving the existance of definite integrals. |
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Previously PhilosopherStoned
I'm just interested in how you could even deduce the form of various integrals without apparently understanding that you are summing little pieces in a limit. That's what I tried to communicate with the arclength thing. Another one would be, how do you know how to find the volume under a function of two variables (i.e. a double integral)? |
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Oh yeah, many (i.e. most) definite integrals can only be approximated. It's only the basic ones that one runs into in calculus classes that are easy enough to do without approximation. So that's another reason to use Riemann sums... |
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Previously PhilosopherStoned
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DV Buddy: BlueKat
You're implying there is more than one galaxy!? |
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