Pathetic crank "Me" (real name Franz) tries to usurp my formula.

[Eram semper recta]

So, after I revealed my formula:

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

a local troll and crank going using the handle "Me" tried to usurp my formula as follows:

Q(x,h)= df [f(x+h)-f(x)]/h - f'(x)

Placing a "df" in front of my formula doesn't mean shit. There is no record or publication anywhere before I revealed the theorem.

Do not fall for this plagiarism and libel of my honourable and historic work.

The bastards on this forum and their fellow scum in the mainstream will do everything they can to minimise and belittle my work.


Early in January 2020, I revealed the historic geometric theorem which provides the general derivative for any smooth function:

https://drive.google.com/open?id=1RDulODvgncItTe7qNI1d8KTN5bl0aTXj

I also showed how it fixes the broken mainstream formulation of integral:

https://drive.google.com/open?id=1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y

I'll post a reminder and refresh this information so that new visitors don't get to miss it.

Also, download and study the most important mathematics book ever written:

https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view

It's free because it's priceless!

[Eram semper recta]

For the innocent:

My formula is complete and accurate and does not require any drivel such as "df" anywhere in it.

[Eram semper recta]

On Monday, 11 May 2020 08:02:39 UTC-4, Eram semper recta wrote:

Moreover, that one little theorem such as this changes ALL you ever knew about calculus is truly remarkable. More astounding is the fact that NO ONE before me even imagined it could be done.

Cauchy mused that algebra was insufficient, but with only the knowledge of the Ancient Greeks, it is possible to formulate ALL of calculus (differentiation and integration) from this incredibly simple little geometric theorem.

The morons of mainstream academia could not see the forest through the trees.

But this is always the case when prolific bullshit is presented by syphilitic brains that never understood the Elements of Euclid or mathematics in general.

[Me]

On Monday, May 11, 2020 at 2:02:39 PM UTC+2, Eram semper recta wrote:

> So, after I revealed my formula:
>

> [f(x+h)-f(x)]/h = f'(x) + Q(x,h) (*)

>
> "Me" tried to usurp my formula as follows:
>

> Q(x,h) =df [f(x+h)-f(x)]/h - f'(x)

>
> Placing a "df" in front of my formula doesn't mean shit.

Huh?! Low on medication again, dumbo?

Hint: In calculus your statement (or "formula")

[f(x+h) - f(x)]/h = f'(x) + Q(x,h)

follows TRIVIALLY from the definition

Q(x,h) =df [f(x+h)-f(x)]/h - f'(x) .

So no one placed a "df" in front of your trivial formula (*).

Hint: Actually, "=df" (or rather "=_df") is part of the metalanguage. Hence it in not part of ANY formula.

[Dave Parker]

Shut up idiot.

[Eram semper recta]

On Monday, 11 May 2020 10:36:33 UTC-4, Me wrote:
> On Monday, May 11, 2020 at 2:02:39 PM UTC+2, Eram semper recta wrote:
>
> > So, after I revealed my formula:
> >
> > [f(x+h)-f(x)]/h = f'(x) + Q(x,h) (*)
> >
> > "Me" tried to usurp my formula as follows:
> >
> > Q(x,h) =df [f(x+h)-f(x)]/h - f'(x)
> >
> > Placing a "df" in front of my formula doesn't mean shit.
>

>

> Hint: In calculus your statement (or "formula")

Which calculus you stupid fuck?! Show me one reference where you see this anywhere in any of your calculus books? You pathetic, lying, stupid crank!

>
> [f(x+h) - f(x)]/h = f'(x) + Q(x,h)
>
> follows TRIVIALLY from the definition
>
> Q(x,h) =df [f(x+h)-f(x)]/h - f'(x) .

There is NO definition necessary you fucking moron! [f(x+h) - f(x)]/h = f'(x) + Q(x,h) is an IDENTITY and Q(x,h) = [f(x+h)-f(x)]/h - f'(x) (not your drivel Q(x,h) =df [f(x+h)-f(x)]/h - f'(x)) follows from the IDENTITY.

>
> So no one placed a "df" in front of your trivial formula (*).
>
> Hint: Actually, "=df" (or rather "=_df") is part of the metalanguage.

What metalanguage, you stupid, lying, ignorant fuck???? The identity is plain algebra, you moron!

<drivel>

[Mostowski Collapse]

LoL

[Me]

On Monday, May 11, 2020 at 6:21:08 PM UTC+2, Eram semper recta wrote:

> The identity is plain algebra

Sure.

But since it seems that you are too dumb to understand what I've said, here's a slight variant:

Let Q(x,h) = [f(x+h) - f(x)]/h - f'(x). Then [f(x+h) - f(x)]/h = f'(x) + Q(x,h).

Again:

f and f' are known, x and h are variables. Hence we can define

Q(x,h) =df [f(x+h) - f(x)]/h - f'(x) .

From this we get (the trivial truth):

[f(x+h) - f(x)]/h = f'(x) + Q(x,h) ,
since
[f(x+h) - f(x)]/h = f'(x) + [f(x+h) - f(x)]/h - f'(x) .

It's really THAT SIMPLE, dumbo.

[Me]

On Monday, May 11, 2020 at 6:30:10 PM UTC+2, Me wrote:

> It's really THAT SIMPLE, dumbo.

Q(x,h) is just the difference between the slope of the secant line and the slope of the tangent line. Nothing "earth-shattering". :-)

[Dan Christensen]

On Monday, May 11, 2020 at 8:02:39 AM UTC-4, Eram semper RECTUM (formerly "John Gabriel" and "Jew Lover") wrote:

> So, after I revealed my formula:
>
> [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
>

Is that the formula that blows up for functions as simple as f(x)=|x|? Thought so.

Fix it, or scrap, John. It really is time to cut your losses and move on. You aren't getting any younger, John.


Even at his advanced age (60+?), John Gabriel is STILL struggling with basic, elementary-school arithmetic. As he has repeatedly posted here:

"1/2 not equal to 2/4"
--October 22, 2017

“1/3 does NOT mean 1 divided by 3 and never has meant that”
-- February 8, 2015

"3 =< 4 is nonsense.”
--October 28, 2017

"Zero is not a number."
-- Dec. 2, 2019

"0 is not required at all in mathematics, just like negative numbers."
-- Jan. 4, 2017

“There is no such thing as an empty set.”
--Oct. 4, 2019

“3 <=> 2 + 1 or 3 <=> 8 - 5, etc, are all propositions” (actually all are meaningless gibberish)
--Oct. 22, 2019

No math genius our JG, though he actually lists his job title as “mathematician” at Linkedin.com. Apparently, they do not verify your credentials.


Though really quite disturbing, interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated March 10, 2020) at https://groups.google.com/forum/#!msg/sci.math/PcpAzX5pDeY/1PDiSlK_BwAJ


Dan

Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com

[Ross A. Finlayson]

On Monday, May 11, 2020 at 5:39:01 AM UTC-7, Eram semper recta wrote:
- Moreover, that one little theorem such as this

changes ALL you ever knew about calculus is truly remarkable.

Sounds almost like something I wrote 20 years ago....

... and won't shut up.

But, it is just a factorial basis.

Also it's "justly" a factorial basis.

Here the point for making a well-defined factorial
basis is for roots also boundaries. (Out through the
roots.)

But, isn't JG a Chinese Burse bot?

This is the theory that the JG is just a bot,
being some "Chinese teacher in Africa that
is some Greek Jew", "best" something, idiot.

You can notice its response is all monotone.

Wouldn't "A bot", know?

Also I'm totally polite and correct where it's rude and wrong.

It's really pretty simple and not that deep.



Is it not a simple rip-off of the medium?


After studying calculus for three or four years,
then later with the advanced calculus and the
ordinary differential equations, these days I
frame those all up way high above in algebra.

Enumerating ordinary systems is like enumerating
probability distributions for their shape terms,
here ordinary systems as periodic and reciprocal.

I.e., it's illustrative and gives the brain a giant tool.

(Six years of calculus and advanced abstract algebra,
that people with math degrees have.)

[Eram semper recta]

On Monday, 11 May 2020 12:30:10 UTC-4, Me wrote:
> On Monday, May 11, 2020 at 6:21:08 PM UTC+2, Eram semper recta wrote:
>
> > The identity is plain algebra
>
> Sure.


>
> But since it seems that you are too dumb to understand what I've said, here's a slight variant:
>
> Let Q(x,h) = [f(x+h) - f(x)]/h - f'(x).

Idiot. There is nothing to "Let ...". The identity is fixed, you moron!

> Then [f(x+h) - f(x)]/h = f'(x) + Q(x,h).
>
> Again:
>
> f and f' are known, x and h are variables. Hence we can define

No. Only f is known. In geometry, both f'(x) and Q(x,h) are well defined as slopes.

> Q(x,h) is just the difference between the slope of the secant line and the slope of the tangent line. Nothing "earth-shattering". :-)

It was pretty earth-shattering before I revealed it to you, wasn't it moron?

What's even more remarkable is that no one before me had the intelligence to realise it.

Still waiting for you to show me in which "calculus" book it was published before I revealed it, you stupid crank! You should be ashamed of yourself. But trolls know no shame eh?

<drivel>

[Eram semper recta]

On Monday, 11 May 2020 12:30:10 UTC-4, Me wrote:

> On Monday, May 11, 2020 at 6:21:08 PM UTC+2, Eram semper recta wrote:

> Let Q(x,h) = <bla>

There is no "letting anything" or variables. The identity is fixed just as Pythagoras' identity is fixed: a^2 = b^2 + c^2.

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

is the Pythagoras of calculus. It is the Holy Grail which thousands of morons before me were unable to realise.

I'd say the Holy Grail of calculus is pretty earth-shattering. Chuckle.

[Me]

On Tuesday, May 12, 2020 at 3:51:35 PM UTC+2, Eram semper recta wrote:
> On Monday, 11 May 2020 12:30:10 UTC-4, Me wrote:
> > On Monday, May 11, 2020 at 6:21:08 PM UTC+2, Eram semper recta wrote:
> >

> > The identity is fixed [...]

Of course!

There's NO QUESTION that we have the "fixed" identity:

[f(x+h) - f(x)]/h = f'(x) + [f(x+h) - f(x)]/h - f'(x)

(for x e IR, assuming that f is defined on IR, f' exists "everywhere", and h e IR\{0}).

Now let Q(x,h) = [f(x+h) - f(x)]/h - f'(x). Then we get

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

(for x e IR and h e IR\{0}).

A rather TRIVIAL fact. :-)

> It is

really THAT SIMPLE, dumbo.

That's why your equation "is true but of no significance...".

[Me]

On Tuesday, May 12, 2020 at 1:39:44 PM UTC+2, Eram semper recta wrote:
> On Monday, 11 May 2020 12:30:10 UTC-4, Me wrote:

> > f and f' are known, x and h are variables.
> >

> No.

Yes. You know there's a METHOD in calculus which allows to DETERMINE f'(x) if (a) f is known and (b) f' exists at x. In many cases we can even state an explicit formula for f' (if there's a formula for f).

See: https://en.wikipedia.org/wiki/Derivative

> What's even more remarkable is that no one before me <bla>

Actually, a variant of your "formula" comes up in the context of "linear approximation" of a function.

Hint:

"If a real-valued function f is differentiable at the point /a/ then it has a linear approximation at the point /a/. This means that there exists a function h_1 such that

f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a), lim_(x->a)h_1(x) = 0 ."

Proof. If f is differentiable at the point /a/, f'(a) exists, then we can define a function h_1 with h_1(x) = (f(x) - f(a))/(x-a) - f'(a) for all x e IR\{a} and h_1(x) = 0 for x = a. Hence f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a) for all x e IR. Moreover, then lim_(x->a)h_1(x) = lim_(x->a)((f(x) - f(a))/(x-a) - f'(a)) = lim_(x->a)(f(x) - f(a))/(x-a)) - lim_(x->a)(f'(a)) = f'(a) - f'(a) = 0. qed

So instead of your trivial truth (some sort of tautology)

[f(x+h) - f(x)]/h = f'(x) + Q(x,h)

we consider the meaningful fact that for any point /a/ where f is differentiable there exists a function h_1 such that

f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a)
and
lim_(x->a)h_1(x) = 0 .

This means that for sufficiently small |x-a| we get

f(x) ~ f(a) + f'(a)(x-a).

See: https://en.wikipedia.org/wiki/Linear_approximation#Definition
and https://en.wikipedia.org/wiki/Taylor%27s_theorem#Motivation

[Sergio]

JG's "formula" instantly proves he is a QuACk and troll.


JG's "formula" is "infected", self-usurping, and self-vomiting.

[Me]

On Tuesday, May 12, 2020 at 5:33:21 PM UTC+2, Sergio wrote:

> JG's "formula" instantly proves he is a QuACk and troll.
>
> JG's "formula" is "infected", self-usurping, and self-vomiting.

C'mon. At least it's a string of symbols! Of course, without a proper "interpretation" it doesn't mean anything.

JG doesn't know how to interpret mathematical formulas correctly.

[Mostowski Collapse]

self-usurping, and self-vomiting, the cause
of self-evidence in new calculoose.

LoL

[Me]

On Tuesday, May 12, 2020 at 5:42:15 PM UTC+2, Me wrote:
> On Tuesday, May 12, 2020 at 5:33:21 PM UTC+2, Sergio wrote:
> >
> > JG's "formula" instantly proves he is a QuACk and troll.
> >
> > JG's "formula" is "infected", self-usurping, and self-vomiting.
> >
> C'mon. At least it's a string of symbols! Of course, without a proper "interpretation" it doesn't mean anything.

For example, just writing/stating

x * y = 1

doesn't "mean" that much. JG generally ignores any context and skips all quantifiers, when stating his "formulas".

[Dan Christensen]

On Monday, May 11, 2020 at 8:02:39 AM UTC-4, Eram semper recta wrote:
> So, after I revealed my formula:
>
> [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
>

Were you ever able to determine that, for f(x)=|x|, you have f'(x) = -1 for x<0, +1 for x>0 and undefined for x=0? NO??? That's like not being able to prove 2+2=4 in arithmetic. Oooops... you can't do that either, can you, John. Oh, well...

Really, don't you think it is time to cut your losses and move on? You aren't getting any younger.


Dan

[Me]

On Tuesday, May 12, 2020 at 1:39:44 PM UTC+2, Eram semper recta wrote:

> It was pretty earth-shattering

Well, here's another earth-shattering formula!

x = y + Q(x,y)

Proof: x = y + (x - y) is an identity. Now let Q(x,y) = x - y. Then x = y + Q(x,y) follows. qed

[Eram semper recta]

On Tuesday, 12 May 2020 11:10:34 UTC-4, Me wrote:
> On Tuesday, May 12, 2020 at 3:51:35 PM UTC+2, Eram semper recta wrote:
> > On Monday, 11 May 2020 12:30:10 UTC-4, Me wrote:
> > > On Monday, May 11, 2020 at 6:21:08 PM UTC+2, Eram semper recta wrote:
> > >
> > > The identity is fixed [...]
>
> Of course!
>
> There's NO QUESTION that we have the "fixed" identity:

Who told you about this identity crank? Was it not I?


> Now let Q(x,h) <drivel>

Like I said imbecile, there is no "letting anything".

The identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) is to calculus AS Pythagoras is to geometry. This is pretty earth-shattering.

You can whine all you like, but your bullshit and drivel don't convince anyone. Even high school students can see the truth of my historic theorem.

So yes, it is not only TRUE (like ALL theorems!) but also very significant because no one before me was able to realise it. And indeed, it is a rigorous formulation of calculus based ONLY on sound geometry. It solves both the problem of differentiation AND integration.

Truly significant and earth-shattering. Chuckle.

Because you are a crank, I refresh the links for others:

The historic theorem:

https://drive.google.com/open?id=1RDulODvgncItTe7qNI1d8KTN5bl0aTXj

How it fixes the bogus mainstream formulation:

https://drive.google.com/open?id=1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y

[Eduardo Fahqtardo]

Shut up idiot.

[Sergio]

totally agree,

JG's desperate use of *an error term to cure his infected math*, the
*QUacK function* Q(x,h), is the *Red Warning Sign* his math is
terminally wrong...

JG's equation is only usable at some secant somewhere, but NEVER correct
at the tangent point.


you can see how the infection has deeply spread into his sick math,
creats nauseam stomach upheaval and velocity fluid projection thusly to
keep the sick infected math out of the body, a pure and proper physical
rejection. Call the Rabbi.

[Me]

On Wednesday, May 13, 2020 at 12:54:24 AM UTC+2, Sergio wrote:

> Call the Rabbi.

Or the exorcist! :-P

:-)

[Eram semper recta]

No response from the moron "Me" (known as Franz) who made all the false claims. Hmm. Chuckle.

[Me]

On Tuesday, May 12, 2020 at 8:55:15 PM UTC+2, Eram semper recta wrote:

> the truth of my [...] theorem. [...] it is not only TRUE (like ALL

> theorems!) but also very significant because no one before me was
> able to realise it.

Actually, a variant of your "theorem" comes up in the context of "linear approximation" of a function.

[Eram semper recta]

On Tuesday, 12 May 2020 20:10:17 UTC-4, Me wrote:
> On Tuesday, May 12, 2020 at 8:55:15 PM UTC+2, Eram semper recta wrote:
>
> > the truth of my [...] theorem. [...] it is not only TRUE (like ALL
> > theorems!) but also very significant because no one before me was
> > able to realise it.
>
> Actually, a variant of your "theorem" comes up in the context of "linear approximation" of a function.

Hey crank. My theorem doesn't come up anywhere in Taylor series. Taylor series is based on Newton's interpolation polynomial which has ZERO to do with geometry. It is a theory based on FINITE DIFFERENCES.


Nowhere in the history of humanity has anyone ever known the identity of my theorem stated as:

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

It is to calculus what Pythagoras was to geometry. Truly significant and what you might call earth-shattering.

Get an education crank!

[Daniel Grace]

On Wednesday, May 13, 2020 at 1:36:52 PM UTC+1, Eram semper recta wrote:
>
> Get an education crank!
>

That's psychological projection.

[Me]

On Wednesday, May 13, 2020 at 2:36:52 PM UTC+2, Eram semper recta wrote:
> On Tuesday, 12 May 2020 20:10:17 UTC-4, Me wrote:
>
> Nowhere in the history of humanity has anyone ever known the identity of my

> theorem stated as: <bla>

And know one will EVER be interested in your trivial formula. *lol*

It seems that you do not know the meaning of the word "theorem" in mathematics, dumbo.

> earth-shattering

Sure. :-)

[red...@siu.edu]

In 1968 (a few years before JG announced his unique discovery) in the textbook Intermediate Mathematical Analysis Labarre on page 118 showed that

[f(x+h)-f(x)]/h = f'(x) + a(x, h),

for some function a, which is determined in the proof.

Labarre indicates that one can give a geometric proof if so desired.

Don

[Me]

Thanks for the reference (again), Don!

[Eram semper recta]

On Wednesday, 13 May 2020 12:08:45 UTC-4, red...@siu.edu wrote:
> In 1968 (a few years before JG announced his unique discovery) in the textbook Intermediate Mathematical Analysis Labarre on page 118 showed that
>
> [f(x+h)-f(x)]/h = f'(x) + a(x, h),

Liar!

There is no such thing on page 118:

https://ia801905.us.archive.org/27/items/in.ernet.dli.2015.134298/2015.134298.Intermediate-Mathematical-Analysis.pdf

Labarre was talking about the Lipschitz condition which has ZERO to do with my identity.

It was NOT stated as

[f(x+h)-f(x)]/h = f'(x) + Q(x, h).

Page 118 has:

f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ)

=> [f(x) - f(ⲝ)]/(x-ⲝ) = f'(ⲝ) - a(x,ⲝ)


Now that might look similar but it's not similar at all! For starters, the derivative is in terms of some number ⲝ and NOT x because Labarre was not talking about my theorem but about the Lipschitz Condition which is entirely different from what I am talking about.

First try to understand what it is you are reading before you jump to the conclusion that it is the same thing. Just because it looks similar, does not mean it is. In fact, my identity has NOTHING to do with a number ⲝ. Rather it is based in x which is the point of TANGENCY. Get it, you stupid, ignorant imbecile?

ⲝ is NOT the point of TANGENCY.

>
> for some function a, which is determined in the proof.

Oh yeah, and it is left to the reader to provide a geometric interpretation of the quantity a(x,ⲝ)(x - ⲝ)! LMAO. I don't know what that geometric interpretation might even be. Do you? But I DO KNOW what Q(x,h) is in terms of geometry. Lipschitz and Labarre had no fucking clue!

Please do tell us what the geometric interpretation is for a(x,ⲝ)(x - ⲝ) because there is NO such thing in my historic identity:

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

And be careful how you misquote my identity, you dumb cunt!!!!

It is NOT [f(x+h)-f(x)]/h = f'(x) + a(x, h)

OR even

[f(x)-f(ⲝ)]/h = f'(ⲝ) + a(ⲝ,x)

>
> Labarre indicates that one can give a geometric proof if so desired.

Pffft! What is such a proof? Do it and show us moron!!


>
> Don

[Eram semper recta]

Chuckle.

So finally, we get a page number (118) and a link:

https://ia801905.us.archive.org/27/items/in.ernet.dli.2015.134298/2015.134298.Intermediate-Mathematical-Analysis.pdf


And as expected, the ignoramus Don Redmond has no clue what is the Lipschitz condition and jumbles up everything on the basis that it looks similar.

Let me tell you a little anecdote:

A waiter brought wine to a table when asked for wine from a certain region. The wine was not from the requested region but a region nearby.

The patron who was a wine connoisseur spat it out and chided the waiter.

"Why did you not bring the wine I requested?" he asked.

The waiter replied that it was from a region nearby. At that moment a lady waiter walked by and he grabbed her by the arm at which point he instructed the devious waiter to place his forefinger in her vagina and his middle finger in her anus. He then asked the waiter to smell each finger.

"Does your forefinger smell like your middle finger?", he asked.

Those regions are very close to each other but there is no comparison!

Moral of the story is this: Look carefully at things that appear to be similar but are in fact very different!!!

[Mostowski Collapse]

Hey sign legasthenic. Do you mean:

=> [f(x) - f(ⲝ)]/(x-ⲝ) = f'(ⲝ) + a(x,ⲝ)

Substitute x=x+h and ⲝ=x, we get almost potato formula:

[f(x+h) - f(x)]/h = f'(x) + a(x+h,x)

Now set Q(x,h):=a(x+h,x), then you get:

[f(x+h) - f(x)]/h = f'(x) + Q(x,h)

The author then proofs lim x->ⲝ a(x,ⲝ)=0. With the
afforementioned substitution x=x+h and ⲝ=x it would say
lim h->0 a(x+h,x). And with Q(x,h):=a(x+h,x) it says:

lim h->0 Q(x,h) = 0

Who wouldn't have thought so? But since he
uses lim h->0, this is not the same as bingo bongo
new calculoose, which is based on division by zero.

[Eram semper recta]

On Wednesday, 13 May 2020 14:03:34 UTC-4, Mostowski Collapse wrote:
> Hey sign legasthenic. Do you mean:
>
> => [f(x) - f(ⲝ)]/(x-ⲝ) = f'(ⲝ) + a(x,ⲝ)
>
> Substitute x=x+h and ⲝ=x,

x = x + h birdbrain? LMAO. Nice try. What a moron!

> we get almost potato formula:
>
> [f(x+h) - f(x)]/h = f'(x) + a(x+h,x)

x is the coordinate at the point of TANGENCY birdbrain! LMAO!!! Too funny. ⲝ is some random number, you moron!!!

<drivel>

[Mostowski Collapse]

Yes, you first substitute x=x+h, you get:

[f(x+h) - f(ⲝ)]/(x+h-ⲝ) = f'(ⲝ) + a(x+h,ⲝ)

Then you substitute ⲝ=x, you get:

[f(x+h) - f(x)]/(x+h-x) = f'(x) + a(x+h,x)

Which is the same as:

[f(x+h) - f(x)]/h = f'(x) + a(x+h,x)

Whats your prolololoproblem?

[Eram semper recta]

This is what I have never understood about illiterates like Jan Burse. I mean the idiot doesn't even have a fucking clue what we are talking about, but he still believes that he can actually contribute to a topic way beyond his capabilities. Tsk, tsk.

[Me]

On Wednesday, May 13, 2020 at 8:03:34 PM UTC+2, Mostowski Collapse wrote:

> Do you mean:
>
> => [f(x) - f(ⲝ)]/(x-ⲝ) = f'(ⲝ) + a(x,ⲝ)
>
> Substitute x=x+h and ⲝ=x, we get almost potato formula:
>
> [f(x+h) - f(x)]/h = f'(x) + a(x+h,x)
>

> Now set Q(x,h) := a(x+h,x), then you get:

>
> [f(x+h) - f(x)]/h = f'(x) + Q(x,h)
>
> The author then proofs lim x->ⲝ a(x,ⲝ)=0. With the
> afforementioned substitution x=x+h and ⲝ=x it would say

> lim h->0 a(x+h,x). And with Q(x,h) := a(x+h,x) it says:
>
> lim h->0 Q(x,h) = 0
>
> Who wouldn't have thought so?

Hint: Me, May 12

<quote>

Actually, a variant of your "formula" comes up in the context of "linear approximation" of a function.

Hint:

"If a real-valued function f is differentiable at the point /a/ then it has a linear approximation at the point /a/. This means that there exists a function h_1 such that

f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a), lim_(x->a)h_1(x) = 0 ."

Proof. If f is differentiable at the point /a/, f'(a) exists, then we can define a function h_1 with h_1(x) = (f(x) - f(a))/(x-a) - f'(a) for all x e IR\{a} and h_1(x) = 0 for x = a. Hence f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a) for all x e IR. Moreover, then lim_(x->a)h_1(x) = lim_(x->a)((f(x) - f(a))/(x-a) - f'(a)) = lim_(x->a)(f(x) - f(a))/(x-a)) - lim_(x->a)(f'(a)) = f'(a) - f'(a) = 0. qed

So instead of your trivial truth (some sort of tautology)

[f(x+h) - f(x)]/h = f'(x) + Q(x,h)

we consider the meaningful fact that for any point /a/ where f is differentiable there exists a function h_1 such that

f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a)
and
lim_(x->a)h_1(x) = 0 .

This means that for sufficiently small |x-a| we get

f(x) ~ f(a) + f'(a)(x-a).

See: https://en.wikipedia.org/wiki/Linear_approximation#Definition
and https://en.wikipedia.org/wiki/Taylor%27s_theorem#Motivation

</quote>

Nothing new under the sun.

[Eram semper recta]

On Wednesday, 13 May 2020 14:11:16 UTC-4, Swiss retard Jan Burse aka Mostowski Collapse wrote:
> Yes, you first substitute x=x+h, you get:

No idiot You cannot substitute x = x+h because x is never equal to x+h. Get it retard? LMAO.

>
> [f(x+h) - f(ⲝ)]/(x+h-ⲝ) = f'(ⲝ) + a(x+h,ⲝ)
>
> Then you substitute ⲝ=x, you get:

NO. Because x is the POINT of TANGENCY and ⲝ is some random number, you birdbrain!

<drivel>

[Eram semper recta]

> Nothing new under the sun.

Look, I can ignore a lot of things, but I cannot ignore stupidity, ignorance and dishonesty.

Your attempts to minimise my work are futile. Chuckle. If you are so confident, then do explain what Labarre meant by "geometric explanation of a(x,ⲝ)(x-ⲝ) and show how one can switch ⲝ with x which is the POINT of TANGENCY.

[j4n bur53]

Labarre uses ⲝ where the tangent point is,
and not x. You can check yourself. He has
a picture on page 130, I made a screenshot:

What are the variables in Labarre doing:

https://gist.github.com/jburse/e1a79d40a1ec45f1d96074623beb6a02#gistcomment-3302964

Eram semper recta schrieb:

[Eram semper recta]

Do you know why Prof. Donnie Redmond hasn't even tried to explain these things? Because he has no clue what they mean. Chuckle. Redmond is about as clueless as you are - perhaps even more.

[Me]

On Wednesday, May 13, 2020 at 8:14:44 PM UTC+2, Eram semper recta wrote:

> No idiot. You cannot substitute x = x+h because <bla>

Add "substitution" to the list of concepts you don't understand.

[j4n bur53]

Conclusion, John Gabbermonkeys brain cancer
has progressed so far, he can even no
read Labarre? LoL

j4n bur53 schrieb:

[Eram semper recta]

Seriously?! LMAO.

Yeah right. x = x + h is nonsense unless h can be 0. I use substitution often when it makes sense.

[j4n bur53]

LoL, what a damsel.

Eram semper recta schrieb:

[Eram semper recta]

On Wednesday, 13 May 2020 14:21:21 UTC-4, Mostowski Collapse wrote:
> Conclusion, John Gabbermonkeys brain cancer
> has progressed so far, he can even no
> read Labarre? LoL
>
> j4n bur53 schrieb:
> > Labarre uses ⲝ where the tangent point is,
> > and not x. You can check yourself. He has
> > a picture on page 130, I made a screenshot:
> >
> >     What are the variables in Labarre doing:
> >
> > https://gist.github.com/jburse/e1a79d40a1ec45f1d96074623beb6a02#gistcomment-3302964

That is something entirely unrelated Birdbrain!

[j4n bur53]

Nope, its the same x,ⲝ as on page 118.

If you substitute:

ⲝ ---> x

x ---> x+h

We get your potato formula.

We also get your potato diagram on page 130,
after the substitution.

[Daniel Grace]

On Wednesday, May 13, 2020 at 7:21:23 PM UTC+1, Eram semper recta wrote:
> On Wednesday, 13 May 2020 14:19:46 UTC-4, Me wrote:
> > On Wednesday, May 13, 2020 at 8:14:44 PM UTC+2, Eram semper recta wrote:
> >
> > > No idiot. You cannot substitute x = x+h because <bla>
> >
> > Add "substitution" to the list of concepts you don't understand.
>

Don't forget to add "long division" to your list while you're at it:
https://groups.google.com/forum/?hl=en#!topic/sci.math/Pm-qKi-UMwg%5B226-250%5D

> Seriously?! LMAO.
>
> Yeah right. x = x + h is nonsense unless h can be 0. I use substitution often when it makes sense.

He meant substitute "x + h" wherever "x" is. That's not the same thing as setting h to 0.

[Eram semper recta]

On Wednesday, 13 May 2020 14:23:11 UTC-4, Mostowski Collapse wrote:
> LoL, what a damsel.

Idiot Jan Burse. You thought you had me on this one eh? What a moron.

LMAO.

[Me]

On Wednesday, May 13, 2020 at 8:23:11 PM UTC+2, Eram semper recta wrote:
> On Wednesday, 13 May 2020 14:21:21 UTC-4, Mostowski Collapse wrote:
> >
> > Labarre uses ⲝ where the tangent point is,
> > and not x. You can check yourself. He has
> > a picture on page 130, I made a screenshot:

https://gist.github.com/jburse/e1a79d40a1ec45f1d96074623beb6a02#gistcomment-3302964

> That is something entirely unrelated

No, it isn't, dumbo.

[Eram semper recta]

On Wednesday, 13 May 2020 14:27:09 UTC-4, Daniel Grace wrote:
> On Wednesday, May 13, 2020 at 7:21:23 PM UTC+1, Eram semper recta wrote:
> > On Wednesday, 13 May 2020 14:19:46 UTC-4, Me wrote:
> > > On Wednesday, May 13, 2020 at 8:14:44 PM UTC+2, Eram semper recta wrote:
> > >
> > > > No idiot. You cannot substitute x = x+h because <bla>
> > >
> > > Add "substitution" to the list of concepts you don't understand.
> >
>
> Don't forget to add "long division" to your list while you're at it:
> https://groups.google.com/forum/?hl=en#!topic/sci.math/Pm-qKi-UMwg%5B226-250%5D

Idiot. NO one understands long division as well as I do. I am sorry for you that you learned nothing from the lesson I gave you at that link.

>
> > Seriously?! LMAO.
> >
> > Yeah right. x = x + h is nonsense unless h can be 0. I use substitution often when it makes sense.
>
> He meant substitute "x + h" wherever "x" is. That's not the same thing as setting h to 0.

Of course it is in this context.

[Eram semper recta]

On Wednesday, 13 May 2020 14:24:44 UTC-4, Mostowski Collapse wrote:
> Nope, its the same x,ⲝ as on page 118.
>
> If you substitute:
>
> ⲝ ---> x
>
> x ---> x+h
>

<preceding drivel ignored>

> We get your potato formula.

It's one hell of a "potato" formula because you are so jealous of it eh? LMAO.

[Eram semper recta]

You can't just shift the point of tangency when you are talking about a conclusion regarding that point.

[Eram semper recta]

On Monday, 11 May 2020 08:02:39 UTC-4, Eram semper recta wrote:
> So, after I revealed my formula:

>
> [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
>

> a local troll and crank going using the handle "Me" tried to usurp my formula as follows:
>
> Q(x,h)= df [f(x+h)-f(x)]/h - f'(x)
>
> Placing a "df" in front of my formula doesn't mean shit. There is no record or publication anywhere before I revealed the theorem.
>
> Do not fall for this plagiarism and libel of my honourable and historic work.
>
> The bastards on this forum and their fellow scum in the mainstream will do everything they can to minimise and belittle my work.
>
>
> Early in January 2020, I revealed the historic geometric theorem which provides the general derivative for any smooth function:
>
> https://drive.google.com/open?id=1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
>
> I also showed how it fixes the broken mainstream formulation of integral:
>
> https://drive.google.com/open?id=1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y
>
> I'll post a reminder and refresh this information so that new visitors don't get to miss it.
>
> Also, download and study the most important mathematics book ever written:
>
> https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
>
> It's free because it's priceless!

So Redmond tried once again unsuccessfully to diminish my historic theorem and as before, he failed again. What is it with these jealous mainstream academics?

[Eram semper recta]

For the record, I studied analysis from Rudin's book. Before Redmond mentioned Labarre, I had never heard of his book. In any case, it is not relevant and does not affect the genuine statement of my theorem which has NEVER been published anywhere before I revealed it to the world.

I have read about the Lipschitz condition, but not in Labarre's book. Lipschitz was an idiot in my opinion and his condition has ZERO to do with my theorem.

[gabriel...@gmail.com]

On Monday, May 11, 2020 at 8:02:39 AM UTC-4, Eram semper recta wrote:
> So, after I revealed my formula:
>
> [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
>
> a local troll and crank going using the handle "Me" tried to usurp my formula as follows:
>
> Q(x,h)= df [f(x+h)-f(x)]/h - f'(x)
>
> Placing a "df" in front of my formula doesn't mean shit. There is no record or publication anywhere before I revealed the theorem.
>
> Do not fall for this plagiarism and libel of my honourable and historic work.
>
> The bastards on this forum and their fellow scum in the mainstream will do everything they can to minimise and belittle my work.
>
>
> Early in January 2020, I revealed the historic geometric theorem which provides the general derivative for any smooth function:
>
> https://drive.google.com/open?id=1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
>
> I also showed how it fixes the broken mainstream formulation of integral:
>
> https://drive.google.com/open?id=1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y
>
> I'll post a reminder and refresh this information so that new visitors don't get to miss it.
>
> Also, download and study the most important mathematics book ever written:
>
> https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
>
> It's free because it's priceless!

On page 118 LeBarre says:

Note that a(ⲝ, x) is simply the difference between the slope of the secant line through (ⲝ, f(ⲝ)) and (x, f(x)) and the slope of the tangent line at (ⲝ, f(ⲝ)). Thus the condition lim (x -> ⲝ) a(ⲝ, x) = 0 requires that the secant line slope approach the slope of the tangent line.

But NOWHERE does LeBarre state this as an identity. In fact, he derives

f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ)

from the epsilonics inequality on page 117 by multiplying the inequality by (x-ⲝ).

Also, on page 118, LeBarre says "Suppose we determine the number a(x,ⲝ) so that f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ), which is misleading, because a(x,ⲝ) if indeed the difference in slopes can ALWAYS be determined from my identity. At best, LeBarre/Lipschitz had a very superficial understanding and nothing close to my theorem.

LeBarre mentions nothing about similar triangles and nothing about the detailed geometry in my theorem.

In any case, there is no question that my theorem is far easier to understand and stated more clearly. To deny this, is again intellectual dishonesty.

[Daniel Grace]

On Wednesday, May 13, 2020 at 7:30:03 PM UTC+1, Eram semper recta wrote:
> On Wednesday, 13 May 2020 14:27:09 UTC-4, Daniel Grace wrote:
> > On Wednesday, May 13, 2020 at 7:21:23 PM UTC+1, Eram semper recta wrote:
> > > On Wednesday, 13 May 2020 14:19:46 UTC-4, Me wrote:
> > > > On Wednesday, May 13, 2020 at 8:14:44 PM UTC+2, Eram semper recta wrote:
> > > >
> > > > > No idiot. You cannot substitute x = x+h because <bla>
> > > >
> > > > Add "substitution" to the list of concepts you don't understand.
> > >
> >
> > Don't forget to add "long division" to your list while you're at it:
> > https://groups.google.com/forum/?hl=en#!topic/sci.math/Pm-qKi-UMwg%5B226-250%5D
>
>
> Idiot. NO one understands long division as well as I do. I am sorry for you that you learned nothing from the lesson I gave you at that link.
>

We know about GCDs already.

> >
> > > Seriously?! LMAO.
> > >
> > > Yeah right. x = x + h is nonsense unless h can be 0. I use substitution often when it makes sense.
> >
> > He meant substitute "x + h" wherever "x" is. That's not the same thing as setting h to 0.
>
> Of course it is in this context.

If you set h to 0, then you won't be making a substitution with "x + h", because you'll just be replacing "x" with another "x", i.e. doing nothing to change the expression / formula / equation etc.

[Me]

On Wednesday, May 13, 2020 at 9:12:01 PM UTC+2, gabriel...@gmail.com wrote:

> On page 118 LeBarre says:
>
> Note that a(ⲝ, x) is simply the difference between the slope of the secant
> line through (ⲝ, f(ⲝ)) and (x, f(x)) and the slope of the tangent line at
> (ⲝ, f(ⲝ)).

*lol* And *I* stated WEEKS AGO:

"Q(x,h) is just the difference between the slope of the secant line and the slope of the tangent line. Nothing "earth-shattering". :-)"

Can't help, but you are just a SILLY BIG CLOWN, dumb like shit!

[Mostowski Collapse]

Its easy to convert:

f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ)

by substituting x->x+h, ⲝ->x, Q(x,h):=a(x+h,x)
into this one:

(f(x+h)-f(x))/h = f'(x) + Q(x,h)

Every monkey on this planet can do that.
LeBarre also shows lim h->0 Q(x,h), without

using bingo bongo division by zero.

[Sergio]

On 5/13/2020 2:11 PM, gabriel...@gmail.com wrote:
> On Monday, May 11, 2020 at 8:02:39 AM UTC-4, Eram semper recta wrote:
>> So, after I revealed my formula:
>>
>> [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
>>
>> a local troll and crank going using the handle "Me" tried to usurp my formula as follows:
>>
>> Q(x,h)= df [f(x+h)-f(x)]/h - f'(x)
>>
>> Placing a "df" in front of my formula doesn't mean shit. There is no record or publication anywhere before I revealed the theorem.
>>
>> Do not fall for this plagiarism and libel of my honourable and historic work.
>>
>> The bastards on this forum and their fellow scum in the mainstream will do everything they can to minimise and belittle my work.
>>
>>
>> Early in January 2020, I revealed the historic geometric theorem which provides the general derivative for any smooth function:
>>
>> https://drive.google.com/open?id=1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
>>
>> I also showed how it fixes the broken mainstream formulation of integral:
>>
>> https://drive.google.com/open?id=1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y
>>
>> I'll post a reminder and refresh this information so that new visitors don't get to miss it.
>>
>> Also, download and study the most important mathematics book ever written:
>>
>> https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
>>
>> It's free because it's priceless!

no, it's free because it is WORTHLESS !!

>
> On page 118 LeBarre says:
>
> Note that a(ⲝ, x) is simply the difference between the slope of the secant line through (ⲝ, f(ⲝ)) and (x, f(x)) and the slope of the tangent line at (ⲝ, f(ⲝ)). Thus the condition lim (x -> ⲝ) a(ⲝ, x) = 0 requires that the secant line slope approach the slope of the tangent line.

... an admission that JG copied "his equation" from LeBarre,

>
> But NOWHERE does LeBarre state this as an identity. In fact, he derives
>
> f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ)
>
> from the epsilonics inequality on page 117 by multiplying the inequality by (x-ⲝ).
>

your equation is ALWAYS WRONG as it is *secant*, *not tangent*.

>
> LeBarre mentions nothing about similar triangles and nothing about the detailed geometry in my theorem.

Because LeBarre used real math, not some scribbling in the cow dirt

>
> In any case, there is no question that my theorem is bad math. To deny this, is again intellectual dishonesty.
>



To deny your JG equation is ALWAYS WRONG, is total dishonesty.

[Eram semper recta]

On Wednesday, 13 May 2020 16:13:52 UTC-4, Me wrote:
> On Wednesday, May 13, 2020 at 9:12:01 PM UTC+2, gabriel...@gmail.com wrote:
>
> > On page 118 LeBarre says:
> >
> > Note that a(ⲝ, x) is simply the difference between the slope of the secant
> > line through (ⲝ, f(ⲝ)) and (x, f(x)) and the slope of the tangent line at
> > (ⲝ, f(ⲝ)).
>
> *lol* And *I* stated WEEKS AGO:
>
> "Q(x,h) is just the difference between the slope of the secant line and the slope of the tangent line. Nothing "earth-shattering". :-)"

It is indeed very earth-shattering. If Lebarre actually understood what my theorem reveals, then he would also have shown how such an identity (which he NEVER realised!) would lead to integration. He DID NO such thing. Lebarre didn't even realise the theorem. In fact, he didn't state it at all in the way my theorem is stated completely in terms of geometry. There is no real analysis in my theorem.

In fact Lebarre failed to realise many of the facts which I revealed:

i. A general formula for derivatives
ii. Seamless integration

> Can't help, but you are just a SILLY BIG CLOWN, dumb like shit!

Speaking about yourself again? You know what's really funny you clown? I'll tell you: you jumping up and down claiming that the result is trivial when in fact it is very significant.

I mean if it were so trivial, how is it that no one knew about it?

It would also be interesting to note that in order for a student to learn all the shit in Lebarre's book (and I have spotted several errors in those pages mentioned), it would take many months. On the other hand, my theorem is simple and easy to learn - even a high school student can learn it.

Also, how is it that you claimed it was wrong - because you are a dipstick who knows shit about mathematics.

[Eram semper recta]

On Wednesday, 13 May 2020 16:30:37 UTC-4, Mostowski Collapse wrote:
> Its easy to convert:
>
> f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ)
>
> by substituting x->x+h, ⲝ->x, Q(x,h):=a(x+h,x)
> into this one:
>
> (f(x+h)-f(x))/h = f'(x) + Q(x,h)
>
> Every monkey on this planet can do that.

Except monkeys like you who knew shit about this until I revealed it to you eh? Chuckle. No one knew about my theorem before I revealed it.

(f(x+h)-f(x))/h = f'(x) + Q(x,h)

is to calculus as Pythagoras is to mathematics.

Lebarre was clueless about how the same formula is used to perform integration.

All that Redmond has done is give you a false hope. Chuckle.

> LeBarre also shows lim h->0 Q(x,h), without using bingo bongo division by zero.

He does no such thing. In order for lim h->0 Q(x,h) to be 0, h=0 which is impossible, hence definitely division by 0 in your bogus calculus. LeBarre had no clue about my theorem. His drivel is based on what he memorised in real analysis.

[Eram semper recta]

On Wednesday, 13 May 2020 15:58:55 UTC-4, Daniel Grace wrote:
> On Wednesday, May 13, 2020 at 7:30:03 PM UTC+1, Eram semper recta wrote:
> > On Wednesday, 13 May 2020 14:27:09 UTC-4, Daniel Grace wrote:
> > > On Wednesday, May 13, 2020 at 7:21:23 PM UTC+1, Eram semper recta wrote:
> > > > On Wednesday, 13 May 2020 14:19:46 UTC-4, Me wrote:
> > > > > On Wednesday, May 13, 2020 at 8:14:44 PM UTC+2, Eram semper recta wrote:
> > > > >
> > > > > > No idiot. You cannot substitute x = x+h because <bla>
> > > > >
> > > > > Add "substitution" to the list of concepts you don't understand.
> > > >
> > >
> > > Don't forget to add "long division" to your list while you're at it:
> > > https://groups.google.com/forum/?hl=en#!topic/sci.math/Pm-qKi-UMwg%5B226-250%5D
> >
> >
> > Idiot. NO one understands long division as well as I do. I am sorry for you that you learned nothing from the lesson I gave you at that link.
> >
>
> We know about GCDs already.

The point is not about GCDs, you stupid fuck. In fact, it has NOTHING to do with GCDs.

>
> > >
> > > > Seriously?! LMAO.
> > > >
> > > > Yeah right. x = x + h is nonsense unless h can be 0. I use substitution often when it makes sense.
> > >
> > > He meant substitute "x + h" wherever "x" is. That's not the same thing as setting h to 0.
> >
> > Of course it is in this context.
>
> If you set h to 0, then you won't be making a substitution with "x + h", because you'll just be replacing "x" with another "x", i.e. doing nothing to change the expression / formula / equation etc.

It doesn't matter what you set h equal to idiot.

x is NEVER equal to x + h.

Now while one can make certain substitutions and I often have, they have to be done in the correct context.

x = x + h is just plain drivel and all you'll get by falling for Jan Burse's garbage.

[Me]

On Wednesday, May 13, 2020 at 10:30:37 PM UTC+2, Mostowski Collapse wrote:

> It's easy to convert

>
> f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ)
>
> by substituting x->x+h, ⲝ->x, Q(x,h):=a(x+h,x) into this one:
>

> (f(x+h)-f(x))/h = f'(x) + Q(x,h) .

>
> Every monkey on this planet can do that.

Indeed! :-) Weeks ago I did something similar (not based on LeBarre book):

"Now it's rather straightforward to get "your" formula from

f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a) . (*)

In (*) we substitute (simultaneously) "x+h" for "x" and "x" for "a". This way we get

f(x+h) = f(x) + f'(x)h + h_1(x+h)h .

From this equation we get (for h =/= 0):

(f(x+h) - f(x))/h = f'(x) + h_1(x+h) .

So your Q(x,h) is just h_1(x+h). In other words, with the definition Q(x,h) =df h_1(x+h), we finally get:

(f(x+h) - f(x))/h = f'(x) + Q(x,h) ."

[Eram semper recta]

On Monday, 11 May 2020 08:02:39 UTC-4, Eram semper recta wrote:
> So, after I revealed my formula:
>
> [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
>
> a local troll and crank going using the handle "Me" tried to usurp my formula as follows:
>
> Q(x,h)= df [f(x+h)-f(x)]/h - f'(x)
>
> Placing a "df" in front of my formula doesn't mean shit. There is no record or publication anywhere before I revealed the theorem.
>
> Do not fall for this plagiarism and libel of my honourable and historic work.
>
> The bastards on this forum and their fellow scum in the mainstream will do everything they can to minimise and belittle my work.
>
>
> Early in January 2020, I revealed the historic geometric theorem which provides the general derivative for any smooth function:
>
> https://drive.google.com/open?id=1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
>
> I also showed how it fixes the broken mainstream formulation of integral:
>
> https://drive.google.com/open?id=1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y
>
> I'll post a reminder and refresh this information so that new visitors don't get to miss it.
>
> Also, download and study the most important mathematics book ever written:
>
> https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
>
> It's free because it's priceless!

So Redmond has encouraged the local morons by giving them a false hope that my theorem was known before I published it. The truth is that no such thing ever happened. Nowhere in LeBarre's book does he state that the identity is the result of a theorem. In fact, it is not even stated in the context of my theorem.

What is more telling is that LeBarre had only a superficial understanding because he did not define the integral using my theorem but instead used the bullshit of Riemann sums and limits, whereas I have proved in my articles that neither limit theory nor infinity or infinitesimals are required for either the derivative or integral.

The local idiots feel empowered by Redmond's drivel and all of a sudden, they are all experts on the Lipschitz condition (which has NOTHING to do with my theorem at all!) which just some time ago they never even heard about.

Redmond is a jealous and dishonest bastard who will do anything to minimise my great work. What a shame. What a vile academic. What a loser is Redmond. Tsk, tsk.

[Eram semper recta]

On Wednesday, 13 May 2020 19:28:24 UTC-4, Me wrote:
> On Wednesday, May 13, 2020 at 10:30:37 PM UTC+2, Mostowski Collapse wrote:
>
> > It's easy to convert
> >
> > f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ)
> >
> > by substituting x->x+h, ⲝ->x, Q(x,h):=a(x+h,x) into this one:
> >
> > (f(x+h)-f(x))/h = f'(x) + Q(x,h) .
> >
> > Every monkey on this planet can do that.
>
> Indeed! :-) Weeks ago I did something similar (not based on LeBarre book):

You did nothing but drivel. You don't even understand the Lipschitz condition, silly dipstick!

>
> "Now it's rather straightforward to get "your" formula from
>
> f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a) . (*)

Nope. Because what LeBarre was talking about has NOTHING to do with my theorem, you crank!

>
> In (*) we substitute (simultaneously) "x+h" for "x" and "x" for "a". This way we get

If you substitute "x" for "a", you get ZERO, you fucking MORON!!!!

h_1(x)(x-a)

Let's see:

h_1(x+h)(x+h-x) = h_1(x+h)h

What the fuck is h_1(x+h)h ???!!!!

THINK, you big ape. THINK.

Redmond tosses a fart bomb and you say it smells good! LMAO.

>
> f(x+h) = f(x) + f'(x)h + h_1(x+h)h .
>
> From this equation we get (for h =/= 0):
>
> (f(x+h) - f(x))/h = f'(x) + h_1(x+h) .
>
> So your Q(x,h) is just h_1(x+h).

Not at all. Chuckle.

<drivel>

[Me]

On Thursday, May 14, 2020 at 1:23:21 AM UTC+2, Eram semper recta wrote:

> x is NEVER equal to x + h. [...]
> x = x + h is just plain drivel and <bla>

Everybody -except you- knew what JB wanted to epxress. He actually mentioned "substitution" EXPLICITLY, dumbo.

So if you prefer, you may read Jan's "x = x + h" as "x->x+h" (meaning: "replace 'x' by 'x+h' in the expression...)

To avoid "problems" like this I'd prefer to write

...we substitute (simultaneously) "x+h" for "x" and "x" for "ⲝ" ,

here. This should be clear enough. :-)

And yes: Your famous "theorem" was already stated by LeBarre

IN A SLIGHTLY DIFFERENT FORM,

(and actually as a theorem) dumbo.

EVERYBODY -except you, that is- can SEE that. :-)

[Eram semper recta]

If Redmond even had the slightest confidence in his knowledge, then he would be on here trying to defend it. Truth is Redmond knows exactly what trollish game he is playing and all you morons fall for it!

[Eram semper recta]

Jan Burse is a proven idiot and a troll. Most of the time he has no clue what is being discussed, never mind the drivel he spews out like sewerage.

[Me]

On Thursday, May 14, 2020 at 1:38:24 AM UTC+2, Eram semper recta wrote:

Finally SOME MATH! (What a relieve!)

> > "Now it's rather straightforward to get "your" formula from
> >
> > f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a) . (*)
> >
> Nope.

Yes. Just READ THE FUCKING POST, you dumb asshole!!! (Actually, it's hard to decide who's the bigger asshole: you or JB.)

> > In (*) we substitute (simultaneously) "x+h" for "x" and "x" for "a".
> > This way we get
> >

> > f(x+h) = f(x) + f'(x)h + h_1(x+h)h .
> >

> If you substitute "x" for "a", you get ZERO, you fucking MORON!

*sigh* I said:

we substitute ***simultaneously*** "x+h" for "x" and "x" for "a"

So let's focs on the term:

> h_1(x)(x-a) (**)
>
> Let's see:

Yes. :-)

After the ***simultaneous*** substitution of "x+h" for "x" and "x" for "a" we get:

> h_1(x+h)(x+h-x) = h_1(x+h)h

Right.

> What the fuck is h_1(x+h)h ?!

Huh?!

A formula?

> THINK, you big ape. THINK.

I'll try. So what? Huh?!

What's the problem with the formula

h_1(x+h)h = h_1(x+h) * h ?

Note that "h_1" is just a function symbol for a function taking exactly one parameter. In this case x+h.

> Redmond <bla>

My derivation is in NO WAY (directly) related to anything Mr. Redmond said.

Well, you see: This way we get:

> > f(x+h) = f(x) + f'(x)h + h_1(x+h)h .
> >
> > From this equation we get (for h =/= 0):
> >
> > (f(x+h) - f(x))/h = f'(x) + h_1(x+h) .
> >
> > So your Q(x,h) is just h_1(x+h)

which a simple comparison with your formula

(f(x+h)-f(x))/h = f'(x) + Q(x,h) .

reveals.

In other words, with the definition Q(x,h) =df h_1(x+h), we finally get your "theorem":

(f(x+h) - f(x))/h = f'(x) + Q(x,h) ."

Note that we only applied extremely simple "transformations" to (*) to get "your theorem".

In other words, (*) is just "your theorem" _in a slightly different form_.

Holy shit! You really are a pain in the ass.

[Me]

On Thursday, May 14, 2020 at 1:58:38 AM UTC+2, Me wrote:
> On Thursday, May 14, 2020 at 1:38:24 AM UTC+2, Eram semper recta wrote:
> >
> > What the fuck is h_1(x+h)h ?!
> >
> Huh?!
>
> A formula?

Corr.: Should read: A term?

[Me]

On Thursday, May 14, 2020 at 1:58:38 AM UTC+2, Me wrote:

> In other words, (*) is just "your theorem" _in a slightly different form_.

So there's compelling evidence that "your" famous "theorem" is already well established in math (read: real analysis).

It seems that we own you, dumbo.

[Eram semper recta]

On Wednesday, 13 May 2020 19:58:38 UTC-4, Me wrote:
> On Thursday, May 14, 2020 at 1:38:24 AM UTC+2, Eram semper recta wrote:
>

> Yes. Just READ THE FUCKING POST, you dumb asshole!!! (Actually, it's hard to decide who's the bigger asshole: you or JB.)
>
> > > In (*) we substitute (simultaneously) "x+h" for "x" and "x" for "a".

No man. No man. I use substitutions too. Learn how to use substitutions correctly here:

https://drive.google.com/drive/folders/0B-mOEooW03iLUUlFR0ZwMjNNVjg

You can't make the substitutions that birdbrain Burse suggests - they are illogical and anti-mathematical nonsense.

x CANNOT be substituted by x+h because of a fundamental truth:

Things that are equal to the same thing are equal to each other.

Now ask yourself: what are x and x+h both equal to? THINK!!!!

If you for one moment allow yourself to think like Burse, you are lost.

> A formula?

>> Corr.: Should read: A term?

It doesn't matter what you call the "thing". What Redmond has given you has NOTHING to do with my theorem.

Nothing in LeBarre's book demonstrates that he understood anything outside of the inequalities he memorised from real analysis he was taught.

The Lipschitz Condition has ZERO to do with my theorem which is entirely geometric.

You are trying to dismantle my credibility by hook or crook - this shows you are intellectually dishonest.

Try to be honest with yourself, even though you don't like me, focus on the topic. I have taught you a lot. Maybe I should just stop wasting my time with you.

[Me]

On Thursday, May 14, 2020 at 2:13:56 AM UTC+2, Me wrote:

> So there's compelling evidence that "your" famous "theorem" is already well
> established in math (read: real analysis).
>
> It seems that we own you, dumbo.

Why, oh why did you open that can of worms?

[Eram semper recta]

On Wednesday, 13 May 2020 20:13:56 UTC-4, Me wrote:
> On Thursday, May 14, 2020 at 1:58:38 AM UTC+2, Me wrote:
>
> > In other words, (*) is just "your theorem" _in a slightly different form_.
>
> So there's compelling evidence that "your" famous "theorem" is already well established in math (read: real analysis).

Facepalm. No! Because that observation in Lebarre's book is NOT a theorem and he did not understand it in any context outside of the inequality. If indeed he had, he would have no doubt defined the definite integral in the way I do in my article. We know he did not!

>
> It seems that we own you, dumbo.

Chuckle. I am always several steps ahead of you. :-)

[Me]

On Thursday, May 14, 2020 at 2:15:57 AM UTC+2, Eram semper recta wrote:
> On Wednesday, 13 May 2020 19:58:38 UTC-4, Me wrote:
> >
> > In (*) we substitute (simultaneously) "x+h" for "x" and "x" for "a".
> >
> No man. No man.

Huh?! Nothing forbids this trivial substitution.

Well, right, you don't know the concept of substitution. (*sigh*)

> You can't make the substitutions <bla>

Of course we can. :-)

Hint (you silly dumbfuck):

It's just a simple "renaming".

Didn't they teach you *ANY* math in the jungle, apeman?

[Me]

On Thursday, May 14, 2020 at 2:19:12 AM UTC+2, Eram semper recta wrote:

> Chuckle. I am always several steps ahead of you. :-)

Yeah, well, that's just, like, your opinion, man. :-)

See: https://www.youtube.com/watch?v=pWdd6_ZxX8c

[Mostowski Collapse]

Hey damsel. What about 3=<4 invalid
and cant substitute x+h for x.
Will you write a book about it? Maybe

Crank "Me" could join you, and write
a chapter about the sum of a series
is not always the limit of the sequence

of partial sums.

LMAO!

[666]

torstai 14. toukokuuta 2020 2.16.22 UTC+3 Eram semper recta kirjoitti:
> On Wednesday, 13 May 2020 16:30:37 UTC-4, Mostowski Collapse wrote:
> > Its easy to convert:
> >
> > f(x) = f(ⲝ) + f'(ⲝ) (x-ⲝ) + a(x,ⲝ)(x-ⲝ)
> >
> > by substituting x->x+h, ⲝ->x, Q(x,h):=a(x+h,x)
> > into this one:
> >
> > (f(x+h)-f(x))/h = f'(x) + Q(x,h)
> >
> > Every monkey on this planet can do that.
>
> Except monkeys like you who knew shit about this until I revealed it to you eh? Chuckle. No one knew about my theorem before I revealed it.
>
>
> (f(x+h)-f(x))/h = f'(x) + Q(x,h)
>
> is to calculus as Pythagoras is to mathematics.

the same forumla can be found, for example:

https://fi.wikipedia.org/wiki/Funktion_differentiaali#Differentiaali_ja_korjaustermi

Q(x,h) is the error function ε(x,h)

[Eram semper recta]

Except that Q(x,h) is NOT an error. It is a *difference*.

Still, nowhere in any published work has my theorem been stated. Labarre's observation of the difference is NOT a theorem.

[nee...@gmail.com]

เมื่อ วันพฤหัสบดีที่ 14 พฤษภาคม ค.ศ. 2020 18 นาฬิกา 16 นาที 51 วินาที UTC+7, Eram semper recta เขียนว่า:

https://www.facebook.com/photo.php?fbid=1603253749829610&set=a.397252653763065&type=3&theater

[Eram semper recta]

On Wednesday, 13 May 2020 20:25:53 UTC-4, Me wrote:
> On Thursday, May 14, 2020 at 2:15:57 AM UTC+2, Eram semper recta wrote:
> > On Wednesday, 13 May 2020 19:58:38 UTC-4, Me wrote:
> > >
> > > In (*) we substitute (simultaneously) "x+h" for "x" and "x" for "a".
> > >
> > No man. No man.
>
> Huh?! Nothing forbids this trivial substitution.

Of course it is forbidden. It results in the following nonsense:

If you substitute "x" for "a", you get ZERO.

h_1(x)(x-a)

Hint: x-x=0 or have you changed your mind about this now? :-)

Let's see:

h_1(x+h)(x+h-x) = h_1(x+h)h

Hint: h=0

Again, what the hell is h_1(x+h)h ?

So the substitution in order to arrive at my identity is illegal.

Read my article (https://drive.google.com/drive/folders/0B-mOEooW03iLUUlFR0ZwMjNNVjg) to see how VALID substitutions work. You can't just substitute like an imbecile!

x = x + h ??! Are you insane? Since when does an object x equal to itself and still be equal to x+h? This is only possible if h=0. But we know that h CANNOT be equal to 0.

[Eram semper recta]

On Thursday, 14 May 2020 03:27:51 UTC-4, Moronic troll Jan Burse aka Mostowski Collapse driveled:

> Hey damsel. What about 3=<4 invalid
> and cant substitute x+h for x.

How can x+h be equal to x? Only if h=0, yes? Oh right ... this happens a lot in your bogus calculus. LMAO.

3 =< 4 is INVALID but you are intellectually too stupid to understand why.

[Mostowski Collapse]

Well an error is a difference.

Like the error between pi and 3.14 is:

0.001592...

[Eram semper recta]

On Monday, 11 May 2020 08:02:39 UTC-4, Eram semper recta wrote:
> So, after I revealed my formula:
>
> [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
>
> a local troll and crank going using the handle "Me" tried to usurp my formula as follows:
>
> Q(x,h)= df [f(x+h)-f(x)]/h - f'(x)
>
> Placing a "df" in front of my formula doesn't mean shit. There is no record or publication anywhere before I revealed the theorem.
>
> Do not fall for this plagiarism and libel of my honourable and historic work.
>
> The bastards on this forum and their fellow scum in the mainstream will do everything they can to minimise and belittle my work.
>
>
> Early in January 2020, I revealed the historic geometric theorem which provides the general derivative for any smooth function:
>
> https://drive.google.com/open?id=1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
>
> I also showed how it fixes the broken mainstream formulation of integral:
>
> https://drive.google.com/open?id=1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y
>
> I'll post a reminder and refresh this information so that new visitors don't get to miss it.
>
> Also, download and study the most important mathematics book ever written:
>
> https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
>
> It's free because it's priceless!

As I said, Redmond tossed you an old bone which you thought was going to disprove my claims. You all pounced on it but the fact is that my theorem has never been published or revealed by anyone before me.

Labarre's book is a bore and he doesn't understand calculus at all.

I was the FIRST to reveal the general formula for area and volume. If Labarre knew this, he might have made a connection between his *passing* observation and the fact that the definite integral can be expressed using my identity. He did not realise this connection.

Well, back to the drawing board for you. Rather than spending time trying to diminish the knowledge I reveal to you, it is wiser to try and understand how my brilliant mind works. Of course you'll never know precisely because you are not capable. However, you might learn to think properly and isn't this a good thing?

[Mostowski Collapse]

This is the absolute error,
the relative error would be:

(pi-3.14)/pi ~ 0.05%

[Eram semper recta]

On Thursday, 14 May 2020 07:30:07 UTC-4, Moronic troll aka Mostowski Collapse driveled:

> Well an error is a difference.

There is NO error in [f(x+h)-f(x)]/h = f'(x) + Q(x,h), you stupid fuck. This is an identity exactly like Pythagoras's a^2 = b^2 + c^2. Do you call c^2 the error term, you silly dachshund?

>
> Like the error between pi and 3.14 is:
>
> 0.001592...

LMAO. You are what happens when a stupid boy goes to college and his teacher makes him believe he is smart. That you passed your computer science exams doesn't mean you are anything special. Delusion can be a terrible thing ... tsk, tsk.

[Eram semper recta]

On Thursday, 14 May 2020 03:56:49 UTC-4, 666 wrote:

A differential is merely a finite difference.

In my New Calculus, differentials are well defined:

f'(x) = dy/dx = [ f(x+n)-f(x-m) ]/ (m+n)

dy = k { f(x+n)-f(x-m) } and dx = k { m+n } where k is any factor not equal to 0.

[Me]

On Thursday, May 14, 2020 at 9:56:49 AM UTC+2, 666 wrote:

> the same forumla can be found, for example:
> https://fi.wikipedia.org/wiki/Funktion_differentiaali#Differentiaali_ja_korjaustermi
>
> Q(x,h) is the error function ε(x,h)

Right. Month ago I already wrote:

"Let Q(x,h) = [f(x+h) - f(x)]/h - f'(x). Then [f(x+h) - f(x)]/h = f'(x) + Q(x,h)."

In the wikipedia article you mentiond they do exactly the same. :-)

First they define:

"e(x,h) = [f(x+h) - f(x)]/h - f'(x)" (*)

then they state:

"Multyplying this equation with h gives:

f(x+h) - f(x) = f'(x)h + he(x,h)."

Without multipliction, we just just get

[f(x+h) - f(x)]/h = f'(x) + e(x,h)

from (*).

So there's indeed compelling evidence that JG's "theorem" is already well established in "math".

[Mostowski Collapse]

-Q(x,h) is the error of [f(x+h)-f(x)]/h towards f'(x).

Whats wrong with you?

Here is have a banana John Gabbermonkey:

LITTLE BIG - GO BANANAS
https://www.youtube.com/watch?v=ADlGkXAz1D0

[Me]

On Thursday, May 14, 2020 at 1:16:51 PM UTC+2, Eram semper recta wrote:
> >
> > Q(x,h) is the error function ε(x,h)
> >
> Except that Q(x,h) is NOT an error.

*sigh* ε(x,h) is CALLED "error function", dumbo, since it DEPENDS on the parameters x and h. What's the matter with you? Dumb like shit?

Since Q(x,h), just *is* ε(x,h), we MAY call Q(x,h) "error function" as well.

> It is a *difference*.

OF COURSE it is a difference, actually it is DEFINED that way:

Q(x,h) =df [f(x+h) - f(x)]/h - f'(x) .

See?!

> Still, nowhere in any published work has my theorem been stated.

Huh?! A slight variant of your formula (or "theorem") can clearly be seen in that wikipedia artcle, dumbo!

Hint: They state:

"Multyplying this equation with h gives:

f(x+h) - f(x) = f'(x)h + he(x,h)."

Without multipliction, we just just have

[f(x+h) - f(x)]/h = f'(x) + e(x,h) .

I'd say that's EXACTLY your "formula" (if we denote the error function with "Q" instead of "e").

Quote JG: "I revealed my formula:

[f(x+h) - f(x)]/h = f'(x) + Q(x,h)

<bla bla>."

[Eram semper recta]

On Thursday, 14 May 2020 09:26:19 UTC-4, Me wrote:
> On Thursday, May 14, 2020 at 9:56:49 AM UTC+2, 666 wrote:
>
> > the same forumla can be found, for example:
> > https://fi.wikipedia.org/wiki/Funktion_differentiaali#Differentiaali_ja_korjaustermi
> >
> > Q(x,h) is the error function ε(x,h)
>
> Right. Month ago I already wrote:

You didn't write ANYTHING that *I* didn't reveal to you crank!

I revealed to you that:

[f(x+h) - f(x)]/h = f'(x) + Q(x,h)

That is an ORIGINAL identity NEVER before published by anyone.

And no crank! Q(x,h) is no more an "error-term" than c^2 is an error term in the Pythagorean identity a^2 = b^2 + c^2.

Get it stupid? LMAO.

[f(x+h) - f(x)]/h = f'(x) + Q(x,h) is to calculus AS Pythagoras is to ALL of mathematics.

On second thoughts, I want that engraved on a monument in my honour. Chuckle.

Were it not for my brilliant New Calculus, it would NEVER have been realised.

[Eram semper recta]

On Thursday, 14 May 2020 09:26:19 UTC-4, Me wrote:

<bullshit>

> So there's indeed compelling evidence that JG's "theorem" is already well established in "math".

Aaaah??? So now my theorem is valid eh? You stupid crank fuck.

[Me]

On Thursday, May 14, 2020 at 1:22:54 PM UTC+2, Eram semper recta wrote:
> On Wednesday, 13 May 2020 20:25:53 UTC-4, Me wrote:
> > On Thursday, May 14, 2020 at 2:15:57 AM UTC+2, Eram semper recta wrote:
> > > On Wednesday, 13 May 2020 19:58:38 UTC-4, Me wrote:
> > > >
> > > > In (*) we substitute (simultaneously) "x+h" for "x" and "x" for "a".
> > > >
> > > No man. No man.
> >
> > Huh?! Nothing forbids this trivial substitution.
> >
> Of course it is forbidden.

Nope.

Well, since you are too dumb to do it simultaneously, we can do it STEPWISE as well.

We start with

f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a) . (*)

In (*) we substitute FIRST "x+h" for "x".

This way we get

f(x+h) = f(a) + f'(a)(x+h-a) + h_1(x+h)(x+h-a) . (**)

NOW we substitute "x" for "a" in (**)

This way we get

f(x+h) = f(x) + f'(x+h-a)h + h_1(x+h)(x+h-x)

and hence (as before)

f(x+h) = f(x) + f'(x)h + h_1(x+h)h . (***)

Hint: You know, it's just a simple "renaming" (sort of).

The "reference point" was originally called /a/. Now it is called "x". In the same way the "variant point" (reatively to x) is no called "x+h" instead of "x". (the new variable) h is just the difference between these two points (orignally, instead of h the difference x-a was used in the formula).

[Eram semper recta]

On Thursday, 14 May 2020 09:50:51 UTC-4, Me wrote:
> On Thursday, May 14, 2020 at 1:22:54 PM UTC+2, Eram semper recta wrote:
> > On Wednesday, 13 May 2020 20:25:53 UTC-4, Me wrote:
> > > On Thursday, May 14, 2020 at 2:15:57 AM UTC+2, Eram semper recta wrote:
> > > > On Wednesday, 13 May 2020 19:58:38 UTC-4, Me wrote:
> > > > >
> > > > > In (*) we substitute (simultaneously) "x+h" for "x" and "x" for "a".
> > > > >
> > > > No man. No man.
> > >
> > > Huh?! Nothing forbids this trivial substitution.
> > >
> > Of course it is forbidden.
>
> Nope.
>
> Well, since you are too dumb to do it simultaneously, we can do it STEPWISE as well.
>
> We start with
>
> f(x) = f(a) + f'(a)(x-a) + h_1(x)(x-a) . (*)
>
> In (*) we substitute FIRST "x+h" for "x".

The above is your first error stupid. I stop here.

> "...is simply the difference between the slope of the secant line and the slope of the tangent line ..."

And it is also important to note that Labarre didn't realise this fact is true for ALL functions. Therefore, Labarre had NO idea that such a theorem exists!

Ha, ha! Take that you idiot Redmond!

[Eram semper recta]

On page 119, the moron Labarre claims in his unremarkable theorem that the finite difference (slope of secant line) is bounded by some number M. LMAO.

Pages 118-120 are just a lot of hot air and if Labarre were smart enough, he would have realised that

[f(x+h)-f(x)]/h = f'(x) + Q(x,h)

is TRUE for ALL functions. Not just "polynomials" as the crank "Me" claimed so many times. Chuckle.

[Eram semper recta]

On Monday, 11 May 2020 08:02:39 UTC-4, Eram semper recta wrote:
> So, after I revealed my formula:
>

> [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
>

> a local troll and crank going using the handle "Me" tried to usurp my formula as follows:
>
> Q(x,h)= df [f(x+h)-f(x)]/h - f'(x)
>
> Placing a "df" in front of my formula doesn't mean shit. There is no record or publication anywhere before I revealed the theorem.
>
> Do not fall for this plagiarism and libel of my honourable and historic work.
>
> The bastards on this forum and their fellow scum in the mainstream will do everything they can to minimise and belittle my work.
>
>
> Early in January 2020, I revealed the historic geometric theorem which provides the general derivative for any smooth function:
>
> https://drive.google.com/open?id=1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
>
> I also showed how it fixes the broken mainstream formulation of integral:
>
> https://drive.google.com/open?id=1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y
>
> I'll post a reminder and refresh this information so that new visitors don't get to miss it.
>
> Also, download and study the most important mathematics book ever written:
>
> https://drive.google.com/file/d/1CIul68phzuOe6JZwsCuBuXUR8X-AkgEO/view
>
> It's free because it's priceless!

Very sorry to disappoint you all, but Redmond's bullshit (the Lipschitzian Condition) has NOTHING to do with my theorem. It is an exercise in stupidity and completely unremarkable.