Proof of inconsistencies in BOGUS calculus

[John Gabriel]

In the Bogus mainstream calculus:

x^3 implies derivative at x=0 and has tangent line [A]
(x-1)^(1/3)+1 implies NO derivative at x=1 and has tangent line [B]

[A] and [B] are proof of inconsistency

So derivative does not depend on tangent line in the BOGUS calculus.

In the New calculus:

x^3 implies NO derivative at x=0 and has NO tangent line [C]
(x-1)^(1/3)+1 implies NO derivative at x=1 and has NO tangent line [D]

[C] and [D] are proof of consistency

[Dan Christensen]

On Monday, January 16, 2017 at 6:40:35 PM UTC-5, John Gabriel wrote:
> In the Bogus mainstream calculus:
>
> x^3 implies derivative at x=0 and has tangent line [A]
> (x-1)^(1/3)+1 implies NO derivative at x=1 and has tangent line [B]
>
> [A] and [B] are proof of inconsistency
>
> So derivative does not depend on tangent line in the BOGUS calculus.
>

Just admit it, Troll Boy -- your goofy system just doesn't work. It blows up for points of inflection (and linear functions, too). The derivative of y=x^3 is y'=3x^2 for ALL x in R. Until you fix this bug in your system, you will never get anywhere.


Dan

[John Gabriel]

On Monday, 16 January 2017 15:51:05 UTC-8, Dan Christensen wrote:
> On Monday, January 16, 2017 at 6:40:35 PM UTC-5, John Gabriel wrote:
> > In the Bogus mainstream calculus:
> >
> > x^3 implies derivative at x=0 and has tangent line [A]
> > (x-1)^(1/3)+1 implies NO derivative at x=1 and has tangent line [B]
> >
> > [A] and [B] are proof of inconsistency
> >
> > So derivative does not depend on tangent line in the BOGUS calculus.
> >
>

Shut up you dumb bastard. What happened to your policy? Yeah right, we know. You are full back in troll mode. What a moron.

[Harry Stoteles]

By dumb bastard you are talking about you and
your new "calculus", bird brain John Gabriel birdbrain.

[7777777]

tiistai 17. tammikuuta 2017 1.40.35 UTC+2 John Gabriel kirjoitti:
> In the Bogus mainstream calculus:
>

bogus in the sense that it does not know how to properly deal with
the infinitesimal ratio dy/dx, so who can?

and you are not prepared to face the challenge because you think that
this problem does not exist, right?

[John Gabriel]

On Tuesday, 17 January 2017 12:21:42 UTC-8, 7777777 wrote:
> tiistai 17. tammikuuta 2017 1.40.35 UTC+2 John Gabriel kirjoitti:
> > In the Bogus mainstream calculus:
> >
> bogus in the sense that it does not know how to properly deal with
> the infinitesimal ratio dy/dx, so who can?

For many reasons, but there is no infinitesimal ratio. As for who can - I, and I alone, have done it in the New Calculus.

>
> and you are not prepared to face the challenge because you think that
> this problem does not exist, right?

There is no challenge. The New Calculus is 100% rigorous.

[burs...@gmail.com]

By rigoros you dont mean you and your new
calculoose bird brain John Gabriel birdbrains?

[John Gabriel]

The baboons at Reddit are suddenly interested in the specifics of the New Calculus. They've realised that they stepped into shit big time and are now looking for ways to refute it. Just too hilarious!!!

[Dan Christensen]

Did you tell them how your wonky definition of a derivative blows up for functions as simple as y=x and y=x^3? Now, THAT'S hilarious!!!


Dan

[John Gabriel]

1/17/17

[burs...@gmail.com]

Am Mittwoch, 18. Januar 2017 05:05:46 UTC+1 schrieb John Gabriel:
> 1/17/17

Is this the proof of inconciousness in new

[7777777]

keskiviikko 18. tammikuuta 2017 1.02.47 UTC+2 John Gabriel kirjoitti:

> For many reasons, but there is no infinitesimal ratio. As for who can - I, and I alone, have done it in the New Calculus.
> >

and you think that calculus is possible without the infinitesimal ratio dy/dx ?

you think that the finite difference
df/n = (f(x+n) - f(x))/n is enough?

[John Gabriel]

No. But f'(x) = { f(x+n)-f(x-m) } / (m+n) is the solution.

[7777777]

torstai 19. tammikuuta 2017 2.16.57 UTC+2 John Gabriel kirjoitti:

>
> No. But f'(x) = { f(x+n)-f(x-m) } / (m+n) is the solution

it is the finite difference.

[John Gabriel]

It is the **solution**. The slope of a secant line is given by a finite difference.

[7777777]

and you think that the finite difference is enough?
that you can have calculus without the infinitesimal ratio?
I don't think that's possible. You made a mistake of rejecting the infinitesimals.

f'(x) = dy/dx <---------- infinitesimal ratio

You have

f'(c) = df/n

you have introduced a new unknown c

We want to know the derivative at point x, not a point c

[John Gabriel]

On Wednesday, 18 January 2017 21:56:46 UTC-8, 7777777 wrote:
> torstai 19. tammikuuta 2017 7.21.44 UTC+2 John Gabriel kirjoitti:
> > On Wednesday, 18 January 2017 20:41:36 UTC-8, 7777777 wrote:
> > > torstai 19. tammikuuta 2017 2.16.57 UTC+2 John Gabriel kirjoitti:
> > >
> > > >
> > > > No. But f'(x) = { f(x+n)-f(x-m) } / (m+n) is the solution
> > >
> > > it is the finite difference.
> >
> > It is the **solution**. The slope of a secant line is given by a finite difference.
>
> and you think that the finite difference is enough?

Yes. I have told you several times. What is the matter with you?

> that you can have calculus without the infinitesimal ratio?

Do you enjoy repeating your questions? Do you think that somehow by repeating yourself the answer will change?

> I don't think that's possible.

What you think is entirely irrelevant.

> You made a mistake of rejecting the infinitesimals.

Chuckle. This from one who believes in Ourobouros.

>
> f'(x) = dy/dx <---------- infinitesimal ratio

It's just a ratio. For the millionth time you moron, there is no such thing as an infinitesimal. You can pray as hard as you like, it won't change this FACT.

>
> You have
>
> f'(c) = df/n

That is complete shit. You have it! Not me.

>
> you have introduced a new unknown c

Bullshit.

>
> We want to know the derivative at point x, not a point c

I once thought you understood. Now I know you are just a troll.

[shio...@googlemail.com]

Cool story bro. Is that idea stolen from euclid, too?

[John Gabriel]

On Monday, 16 January 2017 15:40:35 UTC-8, John Gabriel wrote:

Grabiner writes:

"How then was the tangent to be defined at the point (0,O) for a curve like y =x^3 (FIGUREl), or to a point on a curve with many turning points (FIGURE 2)? "

Page 2 of
https://www.maa.org/sites/default/files/0025570x04690.di021131.02p02223.pdf

Grabiner is part of the BIG STUPID and in her September 1983 publication, she admitted that it was not possible to have a tangent line at x=0 when f(x)=x^3. If it were, then there wouldn't be a need to "define" a tangent.

See, there's the first problem with morons of the BIG STUPID: if the facts don't fit the theory, change the facts.

By definition, a tangent line intersects a curve in one point, extends finitely to both sides and crosses the curve nowhere.

Ironically, in their redefinition, the BIG STUPID used the difference quotient to define the tangent line slope, not noticing the circularity of their actions. Till that point, a derivative was derived from a tangent line, but now the tangent line was derived from the derivative. Circular? Of course.

Grabiner claims the Greeks didn't study many curves. Evidently she never read the Works of Archimedes - just like most of you morons haven't.

[burs...@gmail.com]

Am Freitag, 20. Januar 2017 12:15:20 UTC+1 schrieb John Gabriel:

By BIG STUPID you are talking about the proven
fact of your bird brain John Gariel birdbrains?

[John Gabriel]

The first and only rigorous formulation of calculus in human history is the New Calculus:

http://thenewcalculus.weebly.com

This is no longer open to debate. It is established fact.

[burs...@gmail.com]

Still trying a proof of the incontinence of the new
calculo bird brain John Gabriel birdbrains?


Am Freitag, 20. Januar 2017 17:38:47 UTC+1 schrieb John Gabriel:

[7777777]

you have only the mean value theorem, your

f'(x) = { f(x+n)-f(x-m) } / (m+n)

is the mean value theorem

The mean value theorem says that there is some unknown c in some interval.
It does not tell how to calculate the derivative at the point x.
To be able to calculate the derivative at the point x, we need to introduce
the infinitesimal dx, we need to know dy/dx

The finite difference, and that's what the mean value theorem is based on,
is not enough, you can't have calculus based on it alone.

You have ended up using results of the standard calculus, you ended up using the
infinitesimal ratio dy/dx, yet at the same time you are telling the opposite.

[John Gabriel]

On Friday, 20 January 2017 21:13:48 UTC-8, 7777777 wrote:
> torstai 19. tammikuuta 2017 13.58.21 UTC+2 John Gabriel kirjoitti:
> > On Wednesday, 18 January 2017 21:56:46 UTC-8, 7777777 wrote:
>
> > >
> > > f'(x) = dy/dx <---------- infinitesimal ratio
> >
> > It's just a ratio. For the millionth time you moron, there is no such thing as an infinitesimal. You can pray as hard as you like, it won't change this FACT.
> >
> > >
> > > You have
> > >
> > > f'(c) = df/n
> >
> > That is complete shit. You have it! Not me.
> >
> > >
> > > you have introduced a new unknown c
> >
> > Bullshit.
> >
> > >
> > > We want to know the derivative at point x, not a point c
> >
> > I once thought you understood. Now I know you are just a troll.
>
> you have only the mean value theorem, your
> f'(x) = { f(x+n)-f(x-m) } / (m+n)
> is the mean value theorem

Correct.

>
> The mean value theorem says that there is some unknown c in some interval.

No. That is the non-remarkable part of the MVT. It states that there is an ordinate which is the arithmetic mean of ALL the ordinates in a given interval.

> It does not tell how to calculate the derivative at the point x.

Excuse me, it does.

f'(x) = { f(x+n)-f(x-m) } / (m+n)

f ' (x) IS THE DERIVATIVE AT THE POINT x.

> To be able to calculate the derivative at the point x, we need to introduce
> the infinitesimal dx, we need to know dy/dx

BULLSHIT.

>
> The finite difference, and that's what the mean value theorem is based on,
> is not enough, you can't have calculus based on it alone.

Yes, I have proved it is so.

>
> You have ended up using results of the standard calculus, you ended up using the
> infinitesimal ratio dy/dx, yet at the same time you are telling the opposite.

Bwaaa haaaa haaaa. Nope. My derivative definition although based on the mean value theorem was NEVER realised by any one before me. You can't show me any source that supports your false claims.

The definition is based on a parallel secant line. Has NOTHING to do with ill-formed concepts such as infinitesimals or anything else.

The standard calculus DOES NOT work because of infinitesimals. There are NO infinitesimals in the mainstream calculus.

[John Gabriel]

Moreover, the relationship between x, m and n was not realised by anyone before me. This resulted in the first useful new knowledge - the auxiliary equation. Much more was realised but since most of you stupids are still stuck on the derivative, you won't know. Your loss.

[7777777]

lauantai 21. tammikuuta 2017 7.38.46 UTC+2 John Gabriel kirjoitti:
> On Friday, 20 January 2017 21:13:48 UTC-8, 7777777 wrote:
> > torstai 19. tammikuuta 2017 13.58.21 UTC+2 John Gabriel kirjoitti:
> > > On Wednesday, 18 January 2017 21:56:46 UTC-8, 7777777 wrote:
>
> >
> > The mean value theorem says that there is some unknown c in some interval.
>
> No. That is the non-remarkable part of the MVT. It states that there is an ordinate which is the arithmetic mean of ALL the ordinates in a given interval.
>
> > It does not tell how to calculate the derivative at the point x.
>
> Excuse me, it does.
>
> f'(x) = { f(x+n)-f(x-m) } / (m+n)
>
> f ' (x) IS THE DERIVATIVE AT THE POINT x

f'(c) is the derivative at the point c, and the point c is the unknown point
of the mean value theorem. Whereas f'(x) is the derivative at point x,
the derivative that we want to know, that's where f'(x) = dy/dx

You are implying that the point c is the same point as the point x.
True here:
If f(x) = x^2 and n=m we have the symmetric case
f'(c) = (f(c+n) - f(c-n))/(n+n)

as an example, select f(x)= x^2, c=133, n=14

f´(c) = (f(c+n) - f(c-n))/(n+n) = (f(133+14) - f(133-14))/(14+14) =
(21609-14161)/28 = 266


On the other hand, we know from the standard calculus that f'(x) = 2x so that f'(c)=2c
which gives f´(133) = 2*133 = 266

But generally the points x and c are not the same:
if you choose f(x)= x^3 then f'(c) is not equal to f'(x)

Take another example, i.e. f(x) = x^3, c=12, n=2.
The thing doesn't work anymore:

f'(c) = (f(c+n) - f(c-n))/(n+n) =
(2744-1000)/(2+2) = 436

f'(x) = f'(12) = 432

The point c is not c=x anymore.

[John Gabriel]

On Saturday, 21 January 2017 00:20:15 UTC-8, 7777777 wrote:
> lauantai 21. tammikuuta 2017 7.38.46 UTC+2 John Gabriel kirjoitti:
> > On Friday, 20 January 2017 21:13:48 UTC-8, 7777777 wrote:
> > > torstai 19. tammikuuta 2017 13.58.21 UTC+2 John Gabriel kirjoitti:
> > > > On Wednesday, 18 January 2017 21:56:46 UTC-8, 7777777 wrote:
> >
> > >
> > > The mean value theorem says that there is some unknown c in some interval.
> >
> > No. That is the non-remarkable part of the MVT. It states that there is an ordinate which is the arithmetic mean of ALL the ordinates in a given interval.
> >
> > > It does not tell how to calculate the derivative at the point x.
> >
> > Excuse me, it does.
> >
> > f'(x) = { f(x+n)-f(x-m) } / (m+n)
> >
> > f ' (x) IS THE DERIVATIVE AT THE POINT x
>
> f'(c) is the derivative at the point c, and the point c is the unknown point
> of the mean value theorem. Whereas f'(x) is the derivative at point x,
> the derivative that we want to know, that's where f'(x) = dy/dx
>
> You are implying that the point c is the same point as the point x.

WRONG. In my definition, point c=x.

> True here:
> If f(x) = x^2 and n=m we have the symmetric case
> f'(c) = (f(c+n) - f(c-n))/(n+n)
>
> as an example, select f(x)= x^2, c=133, n=14
>
> f´(c) = (f(c+n) - f(c-n))/(n+n) = (f(133+14) - f(133-14))/(14+14) =
> (21609-14161)/28 = 266
>
>

I have already told you that what you've written there is irrelevant nonsense. There is no "symmetric case". m, n and c are related through the auxiliary equation.

> On the other hand, we know from the standard calculus that f'(x) = 2x so that f'(c)=2c
> which gives f´(133) = 2*133 = 266
>
> But generally the points x and c are not the same:
> if you choose f(x)= x^3 then f'(c) is not equal to f'(x)

Of course you baboon!!! Because for x^3, m =/= n. There is no "symmetric case" there you idiot. Pay attention!!!!!!! For the function x^3, m is NOT EQUAL to n.

Read the following article because you are fucking clueless about the New Calculus. Are you the same 7s from space time and the universe or some other idiot?

https://drive.google.com/open?id=0B-mOEooW03iLVnVyMmJZcEZGdzg

The relationship is given by:

3x(m-n)+m^2-mn+n^2=0

x in the auxiliary equation is a *fixed* value, so you can solve for m in terms of n. There is an example in the link.

I gave you the auxiliary equation in the last comment. Stop writing the same shit over and over again please.

>
> Take another example, i.e. f(x) = x^3, c=12, n=2.
> The thing doesn't work anymore:
>
> f'(c) = (f(c+n) - f(c-n))/(n+n) =
> (2744-1000)/(2+2) = 436
>
> f'(x) = f'(12) = 432
>
> The point c is not c=x anymore.

See previous comment.

[burs...@gmail.com]

Still no clue how the MVT works, only knowing how you can
def an aux in Geogebra, bird brain John Gabriel birdbrain.

Pitty Geogebra doesn't spit out the c for the MVT. But
if you were a little bit more clever you could program

it I guess, with some root finder. And it would work
many times numerically...

[burs...@gmail.com]

Look bird brain John Gabriel birdbrain, the
difference between your aux and MVT is extremely trivial,
and to see that we can use Geogebra which has autodiff.

First in your videos everybody sees that you program
Geogebra as follows (the below is not exactly a
screenshot from your Geogebra, but the idea is clear):

aux(x,m,n) := (f(x+n)-f(x-m))/(m+n) - f'(x)
/* in Geogebra CAS you can automatically let do the
computer the job of calculating f'(x) symbolically
from f(x). */

Now since years you are bugging everybody with this
is the new calculoose and you are doing all tricks,
like solving aux from x,m for n etc..

Which has nothing to do with the MVT. In the MVT
we have two points x and c. And we have what we
might call the extended aux:

aux(x,m,n,c) = (f(x+n)-f(x-m))/(m+n) - f'(c)
/* in Geogebra CAS gain I guess you could automatically
let do the computer the job of calculating f'(c)
symbolically from f(x). You need an extra step
substituting c for x. */

Now the MVT states: There exists a c for a given x,m,n,
such that aux(x,m,n,c)=0. So we have totally different
solution input/output patterns:

JGs aux: Result is some interval bound,
Input x,m
Output n

MVT: Result is some point inbetween,
Input x,m,n
Output c

Of couse in the above MVT we could also go only with
two parameters b=x+n and a=x-m, since then b-a=n+m.
This is the normal casting of MVT.

But so far I don't see any article or video of JG
where you get the MVT correctly. If if you sometimes
cite the MVT from Wiki, in the end you always make

a great mess.

[burs...@gmail.com]

An many many people have already told you that

"JG aux" does not aways work. There was the penis
curve by john_c. And there is famous f(x)=x^2 with
his inflection point.

On the other MVT always works, provided the
function f satisifies the criteria that are
part of the MVT theorem.

Last but not least setting m=n=0 is still
dubious to find f'(x) algebraically, after
all you let Geogebra do the job.

But I guess functions such as f(x)=e^(-1/x^2)
are difficult by your method. Especially if
we would require hands off Geogebra.

Am Samstag, 21. Januar 2017 14:02:59 UTC+1 schrieb burs...@gmail.com:

> JGs aux: Result is some interval bound,
> Input x,m
> Output n
>
> MVT: Result is some point inbetween,
> Input x,m,n
> Output c

> a great mess.

[burs...@gmail.com]

Am Samstag, 21. Januar 2017 14:15:28 UTC+1 schrieb burs...@gmail.com:
> "JG aux" does not aways work. There was the penis
> curve by john_c. And there is famous f(x)=x^2 with
> his inflection point.

Corr.:

"JG aux" does not aways work. There was the penis

curve by john_c. And there is famous f(x)=x^3 with
his inflection point x=0.

[7777777]

lauantai 21. tammikuuta 2017 14.36.43 UTC+2 John Gabriel kirjoitti:
>
> https://drive.google.com/open?id=0B-mOEooW03iLVnVyMmJZcEZGdzg
>
> The relationship is given by:
>
> 3x(m-n)+m^2-mn+n^2=0
>
> x in the auxiliary equation
>

your so called auxiliary equation is just the infinitesimal part of the derivative, that's why I said that you are working with the infinitesimals
although you are telling the opposite

As an example, consider f(x) = x^3, now your auxiliary equation is 3x(m-n)+m^2-mn+n^2=0, whereas the infinitesimal part of the derivative is 3x*dx + (dx)^2 and it can be seen that your m-n is the same as the infinitesimal dx

Another example, consider f(x) = x^2, now your auxiliary equation is m-n=0
and the infinitesimal part of the derivative is dx, and again it can be seen
that your m-n is the same as the infinitesimal dx

[John Gabriel]

On Saturday, 21 January 2017 09:59:14 UTC-8, 7777777 wrote:
> lauantai 21. tammikuuta 2017 14.36.43 UTC+2 John Gabriel kirjoitti:
> >
> > https://drive.google.com/open?id=0B-mOEooW03iLVnVyMmJZcEZGdzg
> >
> > The relationship is given by:
> >
> > 3x(m-n)+m^2-mn+n^2=0
> >
> > x in the auxiliary equation
> >
> your so called auxiliary equation is just the infinitesimal part of the derivative, that's why I said that you are working with the infinitesimals
> although you are telling the opposite

NOOOOOOO!!!! That is entirely wrong.

>
> As an example, consider f(x) = x^3, now your auxiliary equation is 3x(m-n)+m^2-mn+n^2=0, whereas the infinitesimal part of the derivative is 3x*dx + (dx)^2 and it can be seen that your m-n is the same as the infinitesimal dx

Not true. 3x^2+3x(m-n)+m^2-mn+n^2 IS the derivative EXACTLY.

>
> Another example, consider f(x) = x^2, now your auxiliary equation is m-n=0

Correct.

> and the infinitesimal part of the derivative is dx, and again it can be seen
> that your m-n is the same as the infinitesimal dx

Not true. 2x+n-m IS the derivative EXACTLY.

[genm...@gmail.com]

On Saturday, 21 January 2017 09:59:14 UTC-8, 7777777 wrote:

I have told you thousands of times, you cannot talk about things that do not exist. The idea of infinitesimal is rubbish. There are no infinitesimals. Even in mainstream mathematics, there is NO use of infinitesimals. You don't seem to understand this. Why is that so? Ask anyone in the BIG STUPID if there are any infinitesimals in their bogus calculus and they will respond NO. dx is NOT an infinitesimal. It is a symbolic number.

[j4n bur53]

Of course there are infinitessimals, how otherwise could
it be 0.333... != 1/3, bird brain John Gabriel birdbrains.

genm...@gmail.com schrieb:

[Ross A. Finlayson]

Damnit that's a bot
with a sock-puppet
faking the Socratic
to straw-man itself.

The iota-values as of the raw differential of
the infinitesimal as between zero and one are
usual and familiar in the domain of real analysts,
with their understanding of the available operations,
and the maintenance of consistency throughout.

Or: thanks, Leibniz, your notation survives.

[burs...@gmail.com]

Its very easy: 0.333...3 = 1/3 - 10^(-n)/3. Hence
we have 0.333... = lim n->oo 1/3 - 10^(-n)/3

or if we replace 10^(-n) by dx, we get the following:
0.333... = lim dx->0 1/3 - dx/3 = 1/3.

[Ross A. Finlayson]

No, you haven't established a convention where you
could leave out the relevant summations, indices,
and limits, and if you had, it wouldn't be here.

(There are those, there.)

Calculus has Leibniz' notation, but not
Leibniz' (nilpotent) infinitesimals.

Here when I say calculus I mean
"Riemann-Lebesgue-Stieltjes".

[burs...@gmail.com]

Hi,

Sorry I wss using imprecise notation, lets use R_bar
and R_dot, so that it is 100% clear, the representation
limes, or infinite long string concatenation operator
is not "lim" but "conc", so we have for R_bar:

0.333 _ _ _ = conc n->oo 1/3 - 10^(-n)/3 != 1/3

But in R_dot, where the infinitessimal vanishes we
get the following result:

0.333 ... = lim n->oo 1/3 - 10^(-n)/3 = 1/3

The easieast way is to define conc n->oo via the
set builder, namely as:

conc n->oo s(n) = { s(n) | n in N }

We then have 0.333 _ _ _ = {0, 3/10, 33/100, ...}, this
definition only works when s(j) <= s(k) for j<=k. Because
then for base representations the set is isomorphic to
the omega word, its just all its prefixes.

Since we have now switched to sets, we need to indeed
embed 1/3 also in a set, this done by a singleton, we
have immediately the following inequality:

{0, 3/10, 33/100, ...} != {1/3}

But this is not the desired inequality, it is missing
the Muckefunk shift. So the desired inequality is a
little bit more complex, we have to find a FISON barrier,

So we say A == B iff there is a FISON such that:

A n [FISON,oo) = B n [FISON,oo)

Hope this helps!

[burs...@gmail.com]

Am Samstag, 21. Januar 2017 21:25:28 UTC+1 schrieb burs...@gmail.com:
> So we say A == B iff there is a FISON such that:
>
> A n [FISON,oo) = B n [FISON,oo)

Corr.:

A n [FISON,oo) = B n [FISON,oo) and A n [FISON,oo) != {}

Legend:
n = intersection,
[FISON,oo) the infinite unit interval starting at FISON.

[burs...@gmail.com]

The benefit of == is that its all compatible with Euclids
magnitude measurement of non-incommenurable quantities
since it doesnt use points only lengths, and FISON is
assumed to be a rational number from Q

Here is an example:

{1/4, 1/3, 1/2} == {1/8, 1/4, 1/2}

Proof:
Take FISON = 1/2, then
{1/4, 1/3, 1/2} n [1/2,oo) = {1/2} != {}
{1/8, 1/4, 1/2} n [1/2,oo) = {1/2}

We further immediately see:

{0, 3/10, 33/100, ...} ==/== {1/3}

A further benefit of == is, since Q is not enumerable,
it is not impredicative. It is even also not predicative
since the universe has only 10^100 atoms. None of the
elite so far had the same IQ,

so I guess I am the greatest mathematician. If you
multiply my IQ with the number of pages, I could
immediatly publish if this homo G.S. wouldn't call
my new calculus rubbish all the time.

[burs...@gmail.com]

Am Samstag, 21. Januar 2017 21:44:50 UTC+1 schrieb burs...@gmail.com:
> so I guess I am the greatest mathematician. If you
> multiply my IQ with the number of pages, I could
> immediatly publish if this homo G.S. wouldn't call
> my new calculus rubbish all the time.

Corr.:

so I guess I am the greatest mathematician. If you

multiply my IQ with 100 to get the number of pages, I

[abu.ku...@gmail.com]

I am going to study a translation of Arhcie's -- not a.p's --
but a slope-tangent need not be infinite, of course,
unnless you want to calculate the crossing,
a la "elliptic curve" in modern Diophatine analysis. in any case,
the infinite line wuld require t00 much ink

[7777777]

lauantai 21. tammikuuta 2017 21.25.58 UTC+2 genm...@gmail.com kirjoitti:
> 2x+n-m IS the derivative EXACTLY.

Do you know what exactly is n-m in this case?
You don't seem to generally accept the idea of the symmetric case m=n, but why are you in this case accepting that m-n=0 ?
Would you like to make an exception to the the "rules"? Is the function
f(x) = x^2 an exceptional function, and then what is f(x)= x^3? It is
a non-exceptional function? Or what it should be called?

>
> I have told you thousands of times, you cannot talk about things that do not exist. The idea of infinitesimal is rubbish. There are no infinitesimals. Even in mainstream mathematics, there is NO use of infinitesimals. You don't seem to understand this. Why is that so? Ask anyone in the BIG STUPID if there are any infinitesimals in their bogus calculus and they will respond NO. dx is NOT an infinitesimal. It is a symbolic number.

dx is an infinitesimal.

You have adopted the idea that there are no infinitesimals from your so called "BIG STUPID". It is quite strange why, because you seem to have rejected most of the ideas of the "BIG STUPID", doesn't it make you part of them, doesn't it make you at least a little stupid? Btw, do you still think that we got the infinitesimals from Cantor? Do you accuse me of trying infect mathematics with the Cholera-Bacillus of infinitesimals?

[j4n bur53]

BTW the equalivalence == was already invented, i.e.
the thingy with the [FISON,oo), read here:

"We will only be interested in the behaviour of cheap nonstandard
mathematical objects {x = x_{\bf n}} in the asymptotic limit {{\bf n}
\rightarrow \infty}. To this end, we will reserve the right to delete an
arbitrary finite number of values from the domain {\Sigma} of the
parameter {{\bf n}}, and work only with sufficiently large choices of
this parameter. In particular, two nonstandard objects {x = x_{\bf n}},
{y = y_{\bf n}} will be considered equal if one has {x_{\bf n} = y_{\bf
n}} for all sufficiently large {{\bf n}}. (This is where the Frechet
filter is implicitly coming into play.)"

https://terrytao.wordpress.com/2012/04/02/a-cheap-version-of-nonstandard-analysis/

burs...@gmail.com schrieb:

[burs...@gmail.com]

But I am currently stuck with 0.333... < 1/3 how this
ordering could imply a non-standard measure, just too stupid.

[John Gabriel]

On Sunday, 22 January 2017 00:58:52 UTC-8, 7777777 wrote:
> lauantai 21. tammikuuta 2017 21.25.58 UTC+2 genm...@gmail.com kirjoitti:
> > 2x+n-m IS the derivative EXACTLY.
>
> Do you know what exactly is n-m in this case?

YES.

> You don't seem to generally accept the idea of the symmetric case m=n, but why are you in this case accepting that m-n=0 ?

See the 8th grade proof on my website. I have proved that the sum of the terms in m and n are 0.

> Would you like to make an exception to the the "rules"?

No. There are no rules.

> Is the function
> f(x) = x^2 an exceptional function, and then what is f(x)= x^3?

All functions are different for crying out loud!

> It is
> a non-exceptional function? Or what it should be called?
>
> >
> > I have told you thousands of times, you cannot talk about things that do not exist. The idea of infinitesimal is rubbish. There are no infinitesimals. Even in mainstream mathematics, there is NO use of infinitesimals. You don't seem to understand this. Why is that so? Ask anyone in the BIG STUPID if there are any infinitesimals in their bogus calculus and they will respond NO. dx is NOT an infinitesimal. It is a symbolic number.
>
> dx is an infinitesimal.
>
> You have adopted the idea that there are no infinitesimals from your so called "BIG STUPID". It is quite strange why, because you seem to have rejected most of the ideas of the "BIG STUPID", doesn't it make you part of them, doesn't it make you at least a little stupid?

I have never adopted wrong ideas from the BIG STUPID. Your assertions make you look stupid.

> Btw, do you still think that we got the infinitesimals from Cantor?

I have never thought that. In fact, I was the one who told you that Cantor did not believe in infinitesimals.

> Do you accuse me of trying infect mathematics with the Cholera-Bacillus of infinitesimals?

Well, you can't infect mathematics, but you have infected mythmatics.

[7777777]

sunnuntai 22. tammikuuta 2017 16.30.09 UTC+2 John Gabriel kirjoitti:
> On Sunday, 22 January 2017 00:58:52 UTC-8, 7777777 wrote:
> > lauantai 21. tammikuuta 2017 21.25.58 UTC+2 genm...@gmail.com kirjoitti:
> > > 2x+n-m IS the derivative EXACTLY.
> >
> > Do you know what exactly is n-m in this case?
>
> YES.
>

but you don't think that n-m can be the same as the infinitesimal dx?
And you are using the infinitesimal dx when you set m=n=0 because, in fact,
that is the same as dx = n = m = 0

and what you have achieved so far is:

the mean value theorem is your

f'(c) = (f(c+m) - f(c-n))/(m+n)

but you need to calculate infinitesimal values in order to be able to always to use the symmetric case:

f'(x) = (f(x + dx) - f(x - dx))/2dx

only in this way can you calculate the derivative at a point x
otherwise you end up introducing a new unknown: the point c

In other words, it is confusing because your method seems to calculate
the derivative at the point c which lies on some interval, whereas the standard
method calculates the derivative at a specific point x which is the
starting point of the interval. Also you don't seem to be calculating
the derivative at the same point as the standard method does. Therefore
it is difficult to compare these two methods. Anyway I don't think you can
base the whole calculus on only finite differences, you need more, you
need more than the mean value theorem.

I did not examine this more than this, I don't know why the function
f(x) = x^2 behaves "in a symmetric way", only that the behavior can be
seen because the "auxliary equation" is m-n = 0, meaning the symmetric
case m=n

[burs...@gmail.com]

JG is not using the symmetric cass. He is just using
autodiff and computes aux(x,n,m) symbolically in
Geogebra CAS, and sometimes solves it to be zero.

Besides that he produces a lot of spam videos
and articles about his new calculoose.

But the symmetric case is well know, you can make the
following simple transformation:

f(x + dx) - f(x - dx))

---------------------- =
2*dx

f(x + dx) - f(x) + f(x) - f(x - dx)
----------------------------------- =
2*dx

f(x + dx) - f(x) f(x - dx) - f(x)
---------------- + ----------------
dx - dx
----------------------------------- =
2

f'(x+) + f'(x-)
---------------
2

So its just the arithmetic mean between the left and
the right derivative. If you want to know what the
left and the right derivative are, you can check

this video:
IIT JEE Video lecture Mathematics on Differentiability by IIT KOTA faculty
https://www.youtube.com/watch?v=JBd8risw5Ko

[John Gabriel]

On Sunday, 22 January 2017 07:47:16 UTC-8, 7777777 wrote:
> sunnuntai 22. tammikuuta 2017 16.30.09 UTC+2 John Gabriel kirjoitti:
> > On Sunday, 22 January 2017 00:58:52 UTC-8, 7777777 wrote:
> > > lauantai 21. tammikuuta 2017 21.25.58 UTC+2 genm...@gmail.com kirjoitti:
> > > > 2x+n-m IS the derivative EXACTLY.
> > >
> > > Do you know what exactly is n-m in this case?
> >
> > YES.
> >
>
> but you don't think that n-m can be the same as the infinitesimal dx?

NOOOOOOOOOOOOO! For fuck sakes you moron!!!!! NOOOOOOOOOOOO!!!!!!

Look stupid, choose any pair (k,k) where k is any number and it will work for n-m.

[John Gabriel]

On Sunday, 22 January 2017 07:47:16 UTC-8, 7777777 wrote:

> Anyway I don't think you can base the whole calculus on only finite differences, you need more, you need more than the mean value theorem.

I am laughing my arse off. Calculus is the MEAN VALUE THEOREM. It would not work at all without it.

[bassam king karzeddin]

Still trying to grasp why (1/3 =/= 0.333... ), despite tons of proofs and many explanations, poor minded guy or most likely a hired Joker defending the ruined myths behind any doubt,

That is why, no wonder brusegan appeared suddenly with some other accounts, and available 48 hours a day (with his allies), trying their best to divert people from the shining common sense facts for the favor of very well established ignorance by very well famous reputable Journals and universities

It is more than enough that those fiction makers had been exploiting and living on other innocent humans efforts in the societies they belong to for many centuries and for their alleged peculiar talents, that had been proved and shown not better than pure juggling, exactly similar to their daily production that can not be denied

It is the time for those professional cowards to admit the truth that can not be hidden by spiders threads any more, or else be painted with shame forever

This devilish attitude would never be able to prevent people with only little common sense to be cheated again and again by your silly tricks

People and World Governments must realize that is a big crime against the human minds and also a bigger crime against human resources

Regards
Bassam King Karzeddin
22ed, Jan., 2017

[genm...@gmail.com]

Let's summarise a few reasons why it is a bad idea to define S = Lim S:

[i] It leads to non-equations such as 1/3 = 0.333...

In algebra, we use the equality sign between numbers. One might say that 0.333... is the number 1/3, because it represents the limit of the series 0.3+0.03+0.003+..., but the problem with this approach, is that academics misguidedly try to perform "infinite" arithmetic using 0.333... and arrive at further absurd results such as 1 = 0.999... To say 0.333... is the limit, is like decreeing that 1/3 = 0.333... In mathematics there is no place for rules or decrees, only logic and common sense.

[ii] Many academics get wrong the idea that 0.333... is actually an infinite sum, which is obviously impossible. Then their colleagues will deride them by claiming that it's not an infinite sum, only a representation of 1/3. Well, one cannot arrive at this representation without the fallacy of infinite sum. The representation is not a result of long division because long division is a finite process. There is also confusion among academics about the Euclidean algorithm and the long division algorithm. These are not the same!

[iii] A number theorem in mathematics states that given any p/q and base b, it is not possible to represent p/q in that base b, unless b contains all the prime factors of q. In order to claim that 1/3 can be represented in base 10, you need to find an m and n, such that 1/3 = m / (10^n), where m and n are both integers. Good luck! One would think that on this theorem alone, academics would have been smart enough to realise that S = Lim S is a very bad idea indeed.

[iv] This leads to numerous other wrong ideas with respect to set theory and is a major time waster with no practical application in science, technology or engineering. The only numbers ever used by humans are the rational numbers. There is no such thing as a "real" number. It is an illusion and a myth.

[7777777]

sunnuntai 22. tammikuuta 2017 18.50.41 UTC+2 John Gabriel kirjoitti:
> On Sunday, 22 January 2017 07:47:16 UTC-8, 7777777 wrote:
>
> > Anyway I don't think you can base the whole calculus on only finite differences, you need more, you need more than the mean value theorem.
>
> I am laughing my arse off. Calculus is the MEAN VALUE THEOREM.

now I can see, you are only a troll, thinking that mean value theorem
is all you need, troll boy.

[j4n bur53]

Whats a non-equation? In many posts you say 1/3 != 0.333...

for example in this thread: "Learn why 1/3 is not equal to 0.333...."
https://groups.google.com/d/msg/sci.math/DuZI5vH5uVQ/fTPzQjeBEAAJ

And in this video: First slide "Why is 1/3 != 0.333...?"
https://www.youtube.com/watch?v=PL7LkWgZdCI

So what is inbetween 1/3 and 0.333... ? Whats the difference
between the two? Some infinitessimal e, which doesn't exist to

your own accord? In what respect are they different, except the
obvious that the partial sums are different from the limit case?

genm...@gmail.com schrieb:

[7777777]

sunnuntai 22. tammikuuta 2017 17.58.24 UTC+2 burs...@gmail.com kirjoitti:
> JG is not using the symmetric cass. He is just using
> autodiff and computes aux(x,n,m) symbolically in
> Geogebra CAS, and sometimes solves it to be zero.
>
> Besides that he produces a lot of spam videos
> and articles about his new calculoose.

the troll boy does not always seem to know what he is doing.

>
> But the symmetric case is well know, you can make the
> following simple transformation:
>
> f(x + dx) - f(x - dx))
> ---------------------- =
> 2*dx
>
> f(x + dx) - f(x) + f(x) - f(x - dx)
> ----------------------------------- =
> 2*dx
>
> f(x + dx) - f(x) f(x - dx) - f(x)
> ---------------- + ----------------
> dx - dx
> ----------------------------------- =
> 2
>
> f'(x+) + f'(x-)
> ---------------
> 2
>
> So its just the arithmetic mean between the left and
> the right derivative.

yes, the troll boy has no idea of what is the symmetric case, why every
function becomes such at an infinitesimal range.

[burs...@gmail.com]

Hi,

By L'Hôpital's rule we can also compute the limes
from JGs formula. Assume he has solved aux(x,n,m)=0
from x,n to m, i.e. assume he has found a function:

m = g(x,n)

Now his tangent slope is:

f(x+n)-f(x-g(x,n))
------------------
n+g(x,n)

If we have lim n->0 n+g(x,n)=0 and lim n->0
f(x+n)-f(x-g(x,n))=0, i.e. 0/0 we can apply L'Hopitals
rule to compute the overal quotient limes:

f(x+n)-f(x-g(x,n))
lim n->0 ------------------ = (1)
n+g(x,n)

f'(x+n) + f'(x-g(x,n)) * g_n(x,n)
lim n->0 --------------------------------- =
1 + g_n(x,n)


Which in many cases should give the following, and
which is nothing else than a weighted arithmetic mean
of the left and the right derivative:

f'(x+) + f'(x-) * g_n(x,n)
-------------------------- (2)
1 + g_n(x,n)

Compare this with the symmetric case where g(x,n)=n,
and hence g_n(x,n)=1, which gives the usual arithmetic
mean. Namely:

f'(x+) + f'(x-) * 1
-------------------
1 + 1

So trivially the JG quotient (1) will give f'(x) for example
if f'(x+)=f'(x-) and g_n(x,n) defined, as can be seen from
the L'Hoital's rule conversion of the JG quotient in (2).

Bye
P.S.: This is a different derivation of JG quotient
than found in the past on mathforum.org . But it has the
same problem, namely using Geogebra CAS autodiff isn't
really a new calculoose.

https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

[John Gabriel]

The MVT IS the reason you can find areas, volumes, etc. Now if that makes me a troll, well, I'd rather be a troll, than a deluded idiot like yourself who believes in the ourobouros. Chuckle.

[burs...@gmail.com]

Since your JG quotient by L'Hôpital's rule gives
the below result even if the curve has two tangets
at a point:

f'(x+) + f'(x-) * g_n(x,n)
-------------------------- (2)
1 + g_n(x,n)

It is not suitable for students. Since when f'(x+)
!= f'(x-) we say f'(x) undefined. See at the end of
this video a curver where f'(x) is undefined at some point:

At t=09:00 you find such a curve:

IIT JEE Video lecture Mathematics on
Differentiability by IIT KOTA faculty
https://www.youtube.com/watch?v=JBd8risw5Ko

But the main problem remains, namely using Geogebra CAS
autodiff isn't really a new calculoose. What we
smell here is some major bull shit:

Bullshit course introduction
https://www.youtube.com/watch?v=TzFjRWaME6o

Am Sonntag, 22. Januar 2017 20:15:29 UTC+1 schrieb John Gabriel: