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.

Why did you repost your message without replying to responses first?
Anway, here's my reply again.

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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I can be also contacted through http://www.physicsinsights.org

On 31 §°§Ü§ä, 12:29, sal wrote:
Why did you repost your message without replying to responses first?
Anway, here's my reply again.

On Tue, 30 Oct 2007 23:23:18 -0700, caoyanwh2003 wrote:
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

As written, that's wrong. The integral of an infinitesimal value is
zero. You've got the integral of dx^2:

¡Ò(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, ¡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

--
Nospam becomes physicsinsights to fix the email
I can be also contacted throughhttp://www.physicsinsights.org- §³§Ü§â§Ú§Ó§Ñ§ß§Ö §ß§Ñ §è§Ú§ä§Ú§â§Ñ§ß§Ú§ñ §ä§Ö§Ü§ã§ä -

- §±§à§Ü§Ñ§Ù§Ó§Ñ§ß§Ö §ß§Ñ §è§Ú§ä§Ú§â§Ñ§ß§Ú§ñ §ä§Ö§Ü§ã§ä -

What about case when dx and sindx like a function cannot expres
themselves in numerical expression at the same time? In other words
for instance if the dx inclenes to zero more quickly than the function
sindx itself, than what...?
The answer is the so call belated functions. I recommend You to see
USM www.kanevuniverse.com and especially partI, partII and pages 198
to 202. Thank You!

On Thu, 01 Nov 2007 04:59:00 -0700, Joro wrote:

On 31 Окт, 12:29, sal wrote:
Why did you repost your message without replying to responses first?
Anway, here's my reply again.

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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throughhttp://www.physicsinsights.org- Скриване на цитирани� тек�т -

- Показване на цитирани� тек�т -

What about case when dx and sindx like a function cannot expres
themselves in numerical expression at the same time?

I have no idea what you mean by this. In general they express themselves
just fine.

It might help if you said what you mean by "dx". The OP was apparently
using "physicist's sloppy infinitesimals" and treating them like regular
numbers which is rarely a good plan. What you're using dx for, though, I
can't guess.


In other words for
instance if the dx inclenes to zero more quickly than the function sindx
itself, than what...?

But x and sin(x) "incline to zero" at the same rate:

sin(x) = x - (1/6)x^3 + (1/120)x^5 - ...

Close to zero, all terms but the first are ignorable, so consequently the
ratios of the rate at which x and sin(x) "incline to zero" goes to 1.

sin(x)/x = 1 - (1/6)x^2 + (1/120)x^4 - ...

and

lim(x-0) [ sin(x)/x ] = 1


The answer is the so call belated functions. I recommend You to see USM
www.kanevuniverse.com and especially partI, partII and pages 198 to 202.
Thank You!


--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org