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



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