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| Tags: calculous, curve, space, theorem |
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#1
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Cao's theorem 3
From when x¡ú0 there are sin x=x, ex-1=x, ln(1+x)=x, (1+x)^§Ñ-1=§Ñx, we can conclude follow theorem 1, ¡ßsin dx=dx ¡à ¡Òsin dx dx=¡Òdxdx=1 2, ¡ß edx-1=dx ¡à ¡Ò(edx-1)dx=¡Òdxdx=1 3, ¡ß ln(1+dx)=dx ¡à ¡Òln(1+dx)dx=¡Òdxdx=1 4, ¡ß (1+dx)^§Ñ-1=§Ñdx ¡à ¡Ò[(1+dx)^§Ñ-1]dx=¡Ò§Ñdxdx=§Ñ¡Òdxdx=§Ñ These all can show even if a very tiny digital such as dx in the integral formula, we cann't deal it with 0 and then calculate them again, that is incorrect. Because even if a very tiny digital such as dx¡ú0 , as after we calculate the integral formula , it is a number that cann't be ignored. The 4 can explain it throughly. 2007-10-31 |
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#2
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On Tue, 30 Oct 2007 23:23:18 -0700, caoyanwh2003 wrote:
Cao's theorem 3 From when x→0 there are sin x=x, ex-1=x, ln(1+x)=x, (1+x)^а-1=аx, we can conclude follow theorem 1, ∵sin dx=dx ∴ ∫sin dx dx=∫dxdx=1 As written, that's wrong. The integral of an infinitesimal value is zero. You've got the integral of dx^2: ∫(dx)^2 and that's zero, not 1, over any finite interval of integration. To see this more clearly, look at the limit which defines the (Riemann) integral, taking the integral of (dx)^2 from "a" to "b": lim_{n-infty} [ sum_0^{n-1} (((b-a)/n)^2) ] The summands are all the same value, so we can replace the sum with multiplication by the number of terms in the sum: lim_{n-infty} [ n * (((b-a)/n)^2) ] Multiplying out, that's: lim_{n-infty} [ (b-a)^2/n ] and that's certainly zero. Of course your whole notion of infinitesimals seems to be very half- baked, as well. It's never the case that "sin dx = dx" for nonzero dx. Rather, if dx is infinitesimal, then sin(dx) - dx ~ dx^2. That is, in simple terms, the difference between sin(dx) and dx is doubly infinitesimal, but it's never zero, save when dx == 0. 2, ∵ edx-1=dx ∴ ∫(edx-1)dx=∫dxdx=1 3, ∵ ln(1+dx)=dx ∴ ∫ln(1+dx)dx=∫dxdx=1 4, ∵ (1+dx)^а-1=аdx ∴ ∫[(1+dx)^а-1]dx=∫аdxdx=а∫dxdx=а These all can show even if a very tiny digital such as dx in the integral formula, we cann't deal it with 0 and then calculate them again, that is incorrect. Because even if a very tiny digital such as dx→0 , as after we calculate the integral formula , it is a number that cann't be ignored. The 4 can explain it throughly. 2007-10-31 -- Nospam becomes physicsinsights to fix the email I can be also contacted through http://www.physicsinsights.org |
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