No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory. NO MORE BULLSHIT.

[Eram semper recta]

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

No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.

This theorem provides the rigour you need in mainstream calculus but have never had.

The historic theorem:

https://lnkd.in/gxBsQf7

How it fixes your bogus mainstream formulation:

https://lnkd.in/g6q4CJA

This FIX is brought to you by the FIRST and ONLY rigorous formulation of calculus in human history - The New Calculus, courtesy of its author John Gabriel:

Download the free ebook here:

https://lnkd.in/gXwPzy4

[math...@gmail.com]

On Monday, February 24, 2020 at 7:55:39 AM UTC-5, Eram semper recta wrote:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)

Use this "derivative formula" to show that the Weierstrass function has no derivatives on RR. Do it in the space provided below.

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[math...@gmail.com]

On Monday, February 24, 2020 at 7:55:39 AM UTC-5, Eram semper recta wrote:

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

[Mostowski Collapse]

Where was it again, that you lost all
your marbles? In south africa?

LMAO!

[Mostowski Collapse]

So you say starting from f1, f2, h where:

f1 = Q(x,h) * h = f(x+h) - f(x) - f'(x) * h

f2 = f'(x) * h

using some formulas:

f'(x)= A(f2,h)

Q(x,h) = B(f1,h)

You get:

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

Woa! Thats amazing. Hint:

You already have A(f2,h), rather pointless:

f2 f'(x) * h
f'(x) = ----- = ----------- = f'(x)
h h

And your sin B, sin G juggling is also pointless,
why not simply try for B(f1,h) this here:

f1 Q(x,h) * h
Q(x,h) = ---- = ----------- = Q(x,h)
h h

It would much more show your Quack.

[Python]

Le 24/02/2020 à 13:55, Eram semper recta a écrit :
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities,
> no ultimate ratios (LMAO at poor Newton!), no limit theory (which
> states unremarkable properties disguised as "proof").

No logic, no mental sanity, no shame, complete delusion, cognitive
dissonance.

When it comes to state unremarkable properties disguised as "proofs"
you'd better shut the f* up, John, the genius who "discovered" that
(f(x+h)-f(x))/h - p(x) is a function of x,h if p = f' and pretend
to deduce f' from that... :-D

Your "work" is complete bullshit from start to finish.

[Eram semper recta]

On Monday, 24 February 2020 10:07:56 UTC-5, Mostowski Collapse wrote:
> So you say starting from f1, f2, h where:
>
> f1 = Q(x,h) * h = f(x+h) - f(x) - f'(x) * h

And you ONLY know this because I told YOU!!! Idiot.

You wouldn't have known ANY of this Jan Burse had I not already revealed to you that [f(x+h)-f(x)]/h = f'(x) + Q(x,h) from the geometric theorem.

You poor, stupid, COPYCAT clown!!! Rewriting my formula doesn't make you smart idiot!!!!

LMAO.

<drivel>

[Eram semper recta]

You fucking psycho crank jean pierre messager!

I am still waiting for you to provide a counter-example. If I hadn't proved that theorem and REVEALED it to you, you dishonest bastard, you wouldn't know SHIT about it.

[Eram semper recta]

So now we have the ultra-stupid birdbrain Jan Burse trying to make out as if the theorem is of no significance because the moron figured out another way to rewrite it USING the assumption that I had to PROVE.

What a clown!!!

You can't say any of

f1 = Q(x,h) * h = f(x+h) - f(x) - f'(x) * h

f2 = f'(x) * h

UNLESS you assume my theorem, you most ignorant Swiss MORON!!!! You didn't prove shit. I proved the theorem.

How do you know that f2 = f'(x) * h you fucking piece of shit?????/

Right moron. I told you when I revealed the theorem to you!!!!!!!

Eat shit and die you dishonest troll cunt.

[Python]

Proven and revealed to the world that f' is a function of x.

Good John, very well down. You could'nt have lived without this
tremendous new knowledge.

Now please let the men in white coats enter your place, it's about
time, you know?

[Eram semper recta]

On Monday, 24 February 2020 17:36:33 UTC-5, jean pierre messager aka Python wrote:
> John Gabriel, aka Eram semper recta wrote:
> > On Monday, 24 February 2020 11:42:24 UTC-5, Python wrote:
> >> Le 24/02/2020 à 13:55, Eram semper recta a écrit :
> >>> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
> >>>
> >>> No infinity, no infinitesimals, no ghosts of departed quantities,
> >>> no ultimate ratios (LMAO at poor Newton!), no limit theory (which
> >>> states unremarkable properties disguised as "proof").
> >>
> >> No logic, no mental sanity, no shame, complete delusion, cognitive
> >> dissonance.
> >>
> >> When it comes to state unremarkable properties disguised as "proofs"
> >> you'd better shut the f* up, John, the genius who "discovered" that
> >> (f(x+h)-f(x))/h - p(x) is a function of x,h if p = f' and pretend
> >> to deduce f' from that... :-D
> >>
> >> Your "work" is complete bullshit from start to finish.
> >
> > You fucking psycho crank jean pierre messager!
> >
> > I am still waiting for you to provide a counter-example. If I hadn't proved that theorem and REVEALED it to you, you dishonest bastard, you wouldn't know SHIT about it.
>
> Proven and revealed to the world that f' is a function of x.

What's your problem idiot? You've been going on non-stop about this not being the case in my theorem. I proved these facts in my article and also in the applet.

A high school student can prove the same without too much trouble, but somehow you CANNOT! Because you are a pscyhotic moron.

Proof:

We can prove that f'(x) = [f(x+h)-f(x)]/h - Q(x,h) as follows.

Let t(x) be the equation of the tangent line which we don't yet know.

Then [t(x+h)-t(x)]/h = f2/h = f'(x) from the geometry theorem.

This means that f'(x) contains no terms in h because t(x) is a straight line.

But f1/h = [f(x+h)-f(x)-f2]/h and so f2/h = [f(x+h)-f(x)]/h - f1/h

Thus, f'(x)= [f(x+h)-f(x)]/h - f1/h which implies that f1/h = Q(x,h).

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

Since the secant line slope [f(x+h)-f(x)]/h contains the sum of f'(x) and Q(x,h), it follows that Q(x,h) has terms with factors of h because f'(x) consists of terms that don't contain h.


Gee, you couldn't do this on your own? Instead you kept talking shit about
sin x + 2x + h - sin x NOT being equal to 2x + h and crap about being able to choose f1/h and f2/h. In any case, you have not been able to produce even ONE counter-example.

You filthy dog! I am lambasting you because you are a psychotic, dishonest and vile piece of shit.

[Zelos Malum]

uh huh, and how do we determine Q for an unknown function? How do we determine that any suppsoed Q is the correct Q?

[Eram semper recta]

My word! You are fucking delusional! You CANNOT do ANYTHING without being given a FUNCTION, you incorrigible moron!

Wow! Just when I think you can't get any more stupid, you surprise me all over again.

[Ross A. Finlayson]

You can't use rise/run with zero run,
but the derivative is only in terms
of h for h = 0.

It's like hooks, line, and sinkers,
it's a strange medical condition that
the howler troll gobbles them all the time,
X-ray shows it's mostly full of lead and spit.
Unfortunately it's taken this as also
recommending the (bad) diet to others.

Maybe this is what Burse kept mentioning about
"the idiot quack troll divides by zero wrongly".

The limit is the sum, in this sense:
the series that defines the limit is
no different than the sum, or the limit.


It's much better to approach various approaches
about finite differences and, minimization without
(standard, integral) calculus even, or the riddle
of the contour integral and line integrals generally
and defining rates and relations in rates over a
continuous time-like domain, constructively,
where then the real existence of infinitesimals
about other standard models of integers and for
example a standard complete ordered field for
real numbers, that the mathematics the troll
claims as booty really has its own home and place,
and, just because the troll sings its praises,
it can be confiscated.

Berkeley's treatise on infinitesimals, if you've
read it, details the features of standard ideas of
infinitesimals, and is more a lecture on rigor (and,
good-natured, at that), than actual denial.

Divide by zero.



Well-ordering the reals is a usual consideration in
modern mathematics after set theory, that many
intelligent people have followed as part of
usual mathematical curriculum, which can't lean
too much on just teaching simple tricks to avoid
analysis, that aren't general and are simple enough
as that instead there are calculators for them.

The derivative there is a function, defined as a
limit in terms of the differential, which is not a
finite difference, and is standardly no different
than zero, which is always different than the
denominator of any non-vertical line (which has none,
except for that one line to have rise/run = infinity,
in a convenient notation that isn't stupid though
is sharp, pointy, and dangerous for people who
don't know what they're doing or aren't fully
conscientious in obeying the laws of numbers.

Which is why it's kept usual simple and in
terms of the discrete and finite, as possible,
but that only continuous functions on continuous
domains have necessary properties for a mathematics,
not just a toy, or a trick, or a tool (or a turd).

[Zelos Malum]

we have f given, but we have many possible p and Qs, the issue is how do we determine Q?

[Eram semper recta]

You DO NOT have many possible f'(x)s and Q(x,h)s, you fucking moron!!

It has been told to you clearly how you determine which is f'(x) and Q(x,h). And it has been PROVED in my article and here on sci.math.

You simply repeating your drivel will not make it any more credible. Get it idiot?

Q(x,h) consists of the terms that contain h.
f'(x) consists of the terms that DO NOT contain h.

Very easy.

Proof for high school students:

[Zelos Malum]

>You DO NOT have many possible f'(x)s and Q(x,h)s, you fucking moron!!

Are you noticing the fact I say p, but you say f'? Rather dishonest!

>Q(x,h) consists of the terms that contain h.
>f'(x) consists of the terms that DO NOT contain h.

Those are rules :)

and then I can get f'(x)=0 for f(x)=x^2, after all, 2xh/h has h in it :)

[Eram semper recta]

On Thursday, 27 February 2020 01:34:11 UTC-5, Zelos Malum wrote:
> >You DO NOT have many possible f'(x)s and Q(x,h)s, you fucking moron!!
>
> Are you noticing the fact I say p, but you say f'?

Because it is f'(x) and you saying p is drivel.

> Rather dishonest!

Monkey, you are the last one who should talk about honesty because you have no honesty or integrity.

>
> >Q(x,h) consists of the terms that contain h.
> >f'(x) consists of the terms that DO NOT contain h.
>
> Those are rules :)

No idiot. They are PROVED in my article and also in this thread.

>
> and then I can get f'(x)=0 for f(x)=x^2, after all, 2xh/h has h in it :)

FACEPALM. I suppose in your syphilitic and diseased brain any drivel is possible.

[Eram semper recta]

There are NO rules in mathematics, only facts, logic and reason. The proof that Q(x,h) has factors of h and f'(x) does not has been given several times, only masters mythmatics graduates from Sweden have difficulty understanding:

An easy proof of the theorem (you need to refer to the geometry to see what f1 and f2 mean):

We can prove that f'(x) = [f(x+h)-f(x)]/h - Q(x,h) as follows.

Let t(x) be the equation of the tangent line which we don't yet know.

Then [t(x+h)-t(x)]/h = f2/h = f'(x) from the geometry theorem.

This means that f'(x) contains no terms in h because t(x) is a straight line and we know that h is a factor of every term in t(x+h)-t(x).

[Zelos Malum]

>Because it is f'(x) and you saying p is drivel.

Once again you are using "drivel" to just mean "things I do not understand"

>Monkey, you are the last one who should talk about honesty because you have no honesty or integrity.

I've never lied here, unlike you whom constantly alter quotes to fit you.

>No idiot. They are PROVED in my article and also in this thread.

You cannot prove those given that I can with basic algebra force them to be violated.

>FACEPALM. I suppose in your syphilitic and diseased brain any drivel is possible.

The issue is you are trying to set up rules and such without even understanding the issue. This is how ti goes when one is stuck doing rudimentary highschool mathematics.

>There are NO rules in mathematics, only facts, logic and reason. The proof that Q(x,h) has factors of h and f'(x) does not has been given several times, only masters mythmatics graduates from Sweden have difficulty understanding:

Yet you keep conjuring rules constantly.

[Eram semper recta]

On Friday, 28 February 2020 01:36:11 UTC-5, Zelos Malum wrote:
> >Because it is f'(x) and you saying p is drivel.
>
> Once again you are using "drivel" to just mean "things I do not understand"

Address the mathematics idiot. I am not interested in your whimsical opinions.

Oops, sorry you can't address the math because you don't any.

[Zelos Malum]

Comes from the one refusing to adress anything beyond highschool stuff ebcause he cannot understand it.

But lets have fun, a new thread, genuine discussion, address all mathematics, both of us, at every post. You are not allowed to cut anything out, say its drivel or anything, and I am neither allowed it.

[Eram semper recta]

It is very unfortunate that the jealous bastards in mainstream academia refuse to admit their ignorance, arrogance, incompetence, grand stupidity, dishonesty and jealousy.

You can't grow in knowledge when truth is rejected. What happens is that you recede further back into the mindset of the Middle Ages where beliefs reign supreme and all manner of disease, filth and darkness is the norm.

[Dan Christensen]

On Monday, February 24, 2020 at 7:55:39 AM UTC-5, Eram semper RECTUM (formerly "John Gabriel" and "Jew") wrote:

> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.
>
> This theorem provides the rigour you need in mainstream calculus but have never had.
>

Sadly, this "theorem" of yours cannot handle functions as simple as f(x)=|x|, largely because you are unable to precisely define your terms--derivatives in this case. You really must establish the necessary and sufficient conditions for the existence of the derivative at any point in the domain of the function in question.


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 the credentials of their membership.


Interested readers should see: “About the spamming troll John Gabriel in his own words...” (lasted updated December 2019) 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

[Zelos Malum]

Are you man enough to have an actual discussion where we both adress everything, or are you the coward I know you are?

[Eram semper recta]

On Monday, February 24, 2020 at 7:55:39 AM UTC-5, Eram semper recta wrote:

Lim_{h->0} [f(x+h)-f(x)]/h is monkey business for many reasons but the most pertinent is that it directly implies there is a NON-PARALLEL secant line with the same slope as the tangent line, a sort of ultimate ratio.

The finite difference quotients are those for the NON-PARALLEL secant line and there is no valid difference quotient when h=0.

[Zelos Malum]

Are you man enough to actually have a conversation where EVERYTHING is adressed, you're NOT allowed to say anything is drivel and skip it, or are you a coward?

[konyberg]

Then why does it give the correct answer?
KON

[Sergio]

On 2/24/2020 8:27 AM, math...@gmail.com wrote:
> On Monday, February 24, 2020 at 7:55:39 AM UTC-5, Eram semper recta wrote:
>> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>

> Use this "derivative formula" to show that the Weierstrass function has no derivatives on RR. Do it in the space provided below.
>
> .
> .
> .
> .
JG => show us how, using your method, we find and/or confirm the
derivative of f(x)=ln(x).

Show us NOW. If you do not, we declare your math phony, fake,
unworkable, discard it totally.

SO show us up, and find f'(x) for f'(x) = ln(x)
and show all the steps.

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> .

[Eram semper recta]

Your methods may give the correct answer but not for the reasons you think!

You are 100% pure moron.

> KON

[konyberg]

Shall we compare?
You give me your way of finding (ln(x))' and my way of doing it. I think your is very simplified, but completely wrong!
KON

[Eram semper recta]

Shut up moron.

> KON

[Zelos Malum]

do it Gerbil, whats the matter? Does it hurt because you know you can't? That you're not man enough to ever back up your claims?

[Eram semper recta]

I showed him how to do it in another thread and he called it garbage. What the moron didn't realise is that the exact same method is used to do it in the mainstream:

x = e^y (means y = ln(x))

dx/dy = e^y which is found from f(x,n)=(1+xn)^(1/n), that is, e^y=f(y,0).

=> dy/dx = 1/x

This is all valid in the New Calculus, but in your bogus calculus, you have to close your eyes when you flip dx/dy because you don't believe in your ignorance that differentials are numbers. In fact, you have no fucking clue what is a differential.

[Me]

Psycho asshole wrote:

> ... from f(x,n) = (1+xn)^(1/n), that is, e^y = f(y,0).

No, you retearded cunt,
if f(x,n) := (1+xn)^(1/n), then f(y,0) is simply *not defined*.

See: https://www.wolframalpha.com/input/?i=%281%2Bxn%29%5E%281%2Fn%29%2C+n%3D0

[Eram semper recta]

On Wednesday, 4 March 2020 12:56:27 UTC-5, Me wrote:
> Psycho asshole wrote:
>
> > ... from f(x,n) = (1+xn)^(1/n), that is, e^y = f(y,0).
>
> No, you retearded cunt,
> if f(x,n) := (1+xn)^(1/n), then f(y,0) is simply *not defined*.

It is defined you stupid cunt.

If you expand the binomial, then you can set n=0.

It's the same story with sin(x)/x when x=0 or (x-3)(x+3)/(x-3) when x=3.

If an expression is not in its irreducible form, then you can't say anything about it being undefined.

As you know moron, baboons like you like to multiply by x/x imagining it is 1 and then when you try to go back the other way, you balk in your syphilitic and diseased brain by thinking that x/x might not be 1 for some unknown reason.

You are 100% RETARD.

[Me]

On Thursday, March 5, 2020 at 2:30:08 AM UTC+1, Eram semper recta wrote:
> On Wednesday, 4 March 2020 12:56:27 UTC-5, Me wrote:
> > Psycho asshole wrote:
> >
> > > ... from f(x,n) = (1+xn)^(1/n), that is, e^y = f(y,0).
> >

> > No, you retarded cunt, if f(x,n) := (1+xn)^(1/n), then f(y,0) is simply *not defined*.
> >
> It is defined

No, it isn't, you psychotic asshole full of shit.

Hint: Since "1/n" isn't defined for n = 0, ANY expression containing the sub-expression "1/n" isn't defined for n = 0 as well.

> If <bla bla bal>

Nope.

Hint: Since "1/n" isn't defined for n = 0, ANY expression containing the sub-expression "1/n" ist defined as well.

> It's the same story with sin(x)/x when x=0

Exactly. Since <real number>/0 *ISN'T* defined, "sin(x)/x" isn't defined for x = 0.

Check: https://www.wolframalpha.com/input/?i=sin%28x%29%2Fx%2C+x+%3D+0

What's the matter with you, you psychotic asshole? Did you REALLY loose all your marbles, due to your illness?

> x/x

Hint: x/x = 1 for all x e IR\{0} .

[Eram semper recta]

On Wednesday, 4 March 2020 20:44:24 UTC-5, Me wrote:
> On Thursday, March 5, 2020 at 2:30:08 AM UTC+1, Eram semper recta wrote:
> > On Wednesday, 4 March 2020 12:56:27 UTC-5, Me wrote:
> > > Psycho asshole wrote:
> > >
> > > > ... from f(x,n) = (1+xn)^(1/n), that is, e^y = f(y,0).
> > >
> > > No, you retarded cunt, if f(x,n) := (1+xn)^(1/n), then f(y,0) is simply *not defined*.
> > >
> > It is defined
>

> No, it isn't, <shit>

It is defined idiot otherwise you couldn't multiply by x/x in the first place.

What do you think imbecile? That you can have x not equal to 0 in x/x when you multiply by x/x and then worry about x equal to 0 when you divide by x/x?

No idiot. You cannot undefine an expression by punching a hole in it. You need it for your bogus calculus, but in sound mathematics, there are no holes. LMAO.

[Zelos Malum]

f(y,0)=(1+0x)^(1/0)

divisino by zero much? Your thing is invalid then.

[Me]

Psycho asshole wrote:

> It is defined

No, it isn't.

> otherwise you couldn't multiply by x/x

We can always multiply by x/x (with x =/= 0).

> What do you think? That you can have x not equal to 0 in x/x when you multiply by x/x

Nope. x has to be =/= 0.

> in sound mathematics, there are no holes.

Sure there are.

Hint: 1/x is not defined for x = 0, but everywhere else (on IR).

[Eram semper recta]

No. It is valid because if you expand (1+xn)^(1/n), you will see that there is NEVER any attempted division by 0.

The same thing applies to sin(x)/x - if you reduce it, you end up with

1 - x^2/3! + x^4/5! - ... which is defined when x=0.

You obviously can't evaluate (1+xn)^(1/n) by plugging in n=0, no more than you can evaluate sin(x)/x by plugging in x=0 because when you multiplied 1 - x^2/3! + x^4/5! - ... by x/x to get sin(x)/x, x/x was not equal to 0/0 (which is nonsense), but it was understood to be ONE.

You can't change the value of an expression by multiplying it by 1. If x/x is not 1, then it is NOT a number and hence wrong.

The psychosis is on your end.

[Me]

Psycho troll wrote:

> x/x was not equal to 0/0 (which is nonsense), but it was understood to be ONE.

For all x e IR\{0}: x/x = 1.

Hence if f is a function defined on IR, for all x e IR\{0}: f(x) * x/x = f(x). Actually, f(x) * x/x is not defined for x = 0.

> You can't change the value of an expression by multiplying it by 1.

Right.

Hint: x/x isn't "just" 1. It's 1 *for all x e IR, x =/= 0*.

[Eram semper recta]

On Thursday, 5 March 2020 07:29:18 UTC-5, Me wrote:
> Psycho troll wrote:
>
> > x/x was not equal to 0/0 (which is nonsense), but it was understood to be ONE.
>
> For all x e IR\{0}: x/x = 1.

No, no, no my little stupid. None of your drivel in set theory changes the facts. Want to try again? Set theory is NOT mathematics.

<SHIT>

[Eram semper recta]

And you cannot punch holes into any function with your monkey business *(x/x).

This bullshit was invented to support your bogus limit definition:

|x-c|<delta => |f(x)-L|<epsilon

which is the requirement for CONTINUITY.

To incorporate the limit without necessarily having continuity at c, you add delta > 0:

0 < |x-c|<delta => |f(x)-L|<epsilon

If a function has a limit at x=k, then it is CONTINUOUS at k. Your bullshit does not hold. Rules are not part of mathematics, but of games.To wit, you can't even name one function that has a limit and is not continuous at a given point.

[Zelos Malum]

>No. It is valid because if you expand (1+xn)^(1/n), you will see that there is NEVER any attempted division by 0.

It is invalid when n=0 because you get division by 0.

>The same thing applies to sin(x)/x - if you reduce it, you end up with

Which only works when
1: You alraedy got the derivative (hence cannot be used to prove it)
2: when x is NOT 0

>You can't change the value of an expression by multiplying it by 1. If x/x is not 1, then it is NOT a number and hence wrong.

x/x=1, iff x~=0

>Set theory is NOT mathematics.

It is part of mathematics, get over it.

[Eram semper recta]

On Friday, 6 March 2020 01:23:43 UTC-5, Zelos Malum wrote:
> >No. It is valid because if you expand (1+xn)^(1/n), you will see that there is NEVER any attempted division by 0.
>
> It is invalid when n=0 because you get division by 0.

It is NOT invalid, because (1+xn)^(1/n) is in a form where you can't replace n with 0. Expanding the binomial allows you to replace n by 0.

3/4 is a valid number.

Now multiply it by x^2/x to get (3x^2)/4x.

You can't say (3x^2)/4x is no longer valid because you can't put x=0. You can't put x=0 in that expression because 0/0 is nonsense. BUT you can put x=0 into 3x/4.

AND anyone with half a brain knows that (3x^2)/4x = 3x/4

<drivel follows>

[Zelos Malum]

>It is NOT invalid, because (1+xn)^(1/n) is in a form where you can't replace n with 0. Expanding the binomial allows you to replace n by 0.

Given you said it is equal to (1+xn)^(1/n) then it is invalid, simple as that.

[Eram semper recta]

On Monday, March 9, 2020 at 2:42:52 AM UTC-4, Zelos Malum wrote:
> >It is NOT invalid, because (1+xn)^(1/n) is in a form where you can't replace n with 0. Expanding the binomial allows you to replace n by 0.
>
> Given you said it is equal to (1+xn)^(1/n) then it is invalid, simple as that.

Rubbish. (x-3)(x+3) / (x-3) = x + 3 but you can't just plug in x=3 in the form (x-3)(x+3) / (x-3).

[Zelos Malum]

that equality only holds if x is NOT 3, otherwise they are unequal.

[Eram semper recta]

The equality holds for any value of x you moron, because only an idiot thinks that 0/0 is valid.

[Zelos Malum]

you're the idiot that thinks 0/0 is valid, we don't!

Hence it is not valid for x=3!

[Eram semper recta]

Really idiot? I wrote:

"The equality holds for any value of x you moron, because only an idiot thinks that 0/0 is valid."

What part of this don't you understand moron?

>
> Hence it is not valid for x=3!

x/x is VALID for ALL numbers. 0 is NOT a number. There is no contradiction or inconsistency anywhere.

[Zelos Malum]

>x/x is VALID for ALL numbers. 0 is NOT a number. There is no contradiction or inconsistency anywhere.

It is a fucking number you fucking moron.

[Me]

On Wednesday, March 11, 2020 at 3:10:51 PM UTC+1, Zelos Malum wrote:

> > 0 is NOT a number.
> >

> It is a fucking number you fucking moron.

C'mon guys, let's solve the problem in a constructive way!

Def.:

x is a /rectum number/ iff x is a number and x =/= 0.

From this we get:

1. 0 is NOT a rectum number ,

and

2. x/x = 1 for all rectum numbers x .

Note though, that the rectum numbers do not "form" a field. For example, the sum of the two rectum numbers 1 and -1 isn't a rectum number.

[Eram semper recta]

On Wednesday, 11 March 2020 14:54:18 UTC-4, Me wrote:

> C'mon guys, let's <crap>

You do stupid things like multiply by x/x or (x-k)/(x-k) because you need HOLES in your HOLEY calculus.

The effect of multiplying by (x-k)/(x-k) with some insignificant stipulations from set theory when we are not even dealing with sets (they don't exist because ZFC is a bunch of crap) is the mainstream approach to soothing the following verifinition:

0 < |x-c| < delta => |f(x)-f'(c)|<epsilon

In order to have delta > 0, you need a hole at x=c. This is according to you accomplished through multiplication by (x-c)/(x-c).

Never mind, you are too fucking stupid to understand these things. I mean you can't even understand simple things like series and sequences.

[Zelos Malum]

wanna try again?

There is nothing of "needing holes", it is just a natural consequence of the fact that 0 is not invertible.

[Eram semper recta]

ROTFLMAO! You mean like dan christensen's "annihiliator" 0 x p = 0 ?

Sheesh, you are one blundering moron.

The idiocy in mainstream theory is a natural consequence of ignorance and stupidity which is like the corona virus of academia - very infectious.

[Zelos Malum]

That property is the one that makes 0 not invertible, too difficult to grasp?

The issue is you do not understand mathematics, always have been

[Eram semper recta]

Stupid, driveling snot ball who has never been educated in mathematics describes you exactly.

[Zelos Malum]

Given I got a degree in mathematics and you don't, the one never educated is you.

[Eram semper recta]

You don't know what I got and what I don't got moron. You're just a stupid, uneducated CRANK and a psychopathic troll. Soon you will be more notorious than AP (ludwig poehlman).

"Mathematics is not about measure or number" - Zelos Malum (claims to have a degree in mathematics. LMAO)

[Zelos Malum]

I know you don't have a masters in mathematics.

Crank? COmmon, we've been over this, you must DISAGREE with mainstream to be a crank!

You keep citing that thing like it is meanignful, anyone who nkows mathematics knwos you are wrong.

[Eram semper recta]

You don't know anything unless I have told you. It is common knowledge that I do not have a master degree in ANYTHING. And it makes ZERO difference.

One look at what an uneducated moron you are is enough to know that degrees do not mean you are educated. Donald Trump has a degree. George Bush has a degree. Many mainstream cranks have degrees. It doesn't mean anything.

Your real education always begins once you leave school. You have this obsession with degrees which are meaningless because they are granted to you by fools who have accomplished NOTHING all their lives and have never actually worked in the real world.

>
> Crank? COmmon, we've been over this, you must DISAGREE with mainstream to be a crank!

That is NOT the main property of a crank. The main property of a crank is that he CANNOT be convinced that his views are WRONG. It has NOTHING to do with the views of the mainstream because the mainstream has often been shown to be WRONG.

>
> You keep citing that thing like it is meanignful,

Because it is!

> anyone who nkows mathematics knwos you are wrong.

Anyone who knows mathematics knows that I am RIGHT and you are mentally ill.

[Zelos Malum]

>You don't know anything unless I have told you. It is common knowledge that I do not have a master degree in ANYTHING. And it makes ZERO difference.

You said it long ago so I know, and yes it makes a difference because it demonstrates, along with your inability to understand anyhing here, that I am more educated in mathematics than you.

>Your real education always begins once you leave school

Typical crank

>That is NOT the main property of a crank.

Actually, it is

"Crank is a pejorative term used for a person who holds an unshakable belief that most of his or her contemporaries consider to be false.[1] "

"contemporaries consider to be false", that is disagreeing with mainstream.

>It has NOTHING to do with the views of the mainstream because the mainstream has often been shown to be WRONG.

Yet here for the last 120 years in amthematics, it has not.

>Because it is!

it isn't, get over it.

>Anyone who knows mathematics knows that I am RIGHT and you are mentally ill.

They don't, that is why you're a laughing stock.

[Eram semper recta]

On Tuesday, 17 March 2020 02:41:56 UTC-4, Zelos Malum wrote:


"Mathematics is not about measure or number" - Zelos Malum, psychotic crank from Sweden.

[Zelos Malum]

Den tisdag 17 mars 2020 kl. 13:55:36 UTC+1 skrev Eram semper recta:
> On Tuesday, 17 March 2020 02:41:56 UTC-4, Zelos Malum wrote:
>
>
> "Mathematics is not about measure or number" - Zelos Malum, psychotic crank from Sweden.

Yet mathematicians rarely deal with numbers, why is that?

[Eram semper recta]

Hardly surprising that you make such a statement. Why, you have no clue what it means to be a "number". Too funny!

[Zelos Malum]

Yes because your little pet idea of what a number is is SOOOOOO important. Grow the fuck up

[Eram semper recta]

On Monday, February 24, 2020 at 7:55:39 AM UTC-5, Eram semper recta wrote:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.
>
> This theorem provides the rigour you need in mainstream calculus but have never had.
>
> The historic theorem:
>
> https://lnkd.in/gxBsQf7
>
> How it fixes your bogus mainstream formulation:
>
> https://lnkd.in/g6q4CJA
>
> This FIX is brought to you by the FIRST and ONLY rigorous formulation of calculus in human history - The New Calculus, courtesy of its author John Gabriel:
>
> Download the free ebook here:
>
> https://lnkd.in/gXwPzy4

Please do not comment on anything other than the topic at hand. Certain people (Zelos Malum) can't read.

[Mostowski Collapse]

New calculoose, the only mathematical network that
is a closed loop consisting of one person. It connects
John Gabbermonkeys bird brain directly with his rectum.

LMAO!

[Eram semper recta]

On Thursday, March 19, 2020 at 9:51:07 AM UTC-4, Swiss retard and crank Jan Burse aka Mostowski Collapse driveled:
> New calculoose, the only mathematical ...

No one cares what you think birdbrain. You are and always will be a moron.

[Zelos Malum]

I can read just fine, you are just stupid

[Eram semper recta]

On Monday, 24 February 2020 07:55:39 UTC-5, Eram semper recta wrote:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.
>
> This theorem provides the rigour you need in mainstream calculus but have never had.
>
> The historic theorem:
>
> https://lnkd.in/gxBsQf7
>
> How it fixes your bogus mainstream formulation:
>
> https://lnkd.in/g6q4CJA
>
> This FIX is brought to you by the FIRST and ONLY rigorous formulation of calculus in human history - The New Calculus, courtesy of its author John Gabriel:
>
> Download the free ebook here:
>
> https://lnkd.in/gXwPzy4

The consequences of my new theorem are enormous and course altering in so many respects.

The new definitions are a result of the well-formed concepts in my New Calculus, the first and only rigorous formulation of calculus in human history. What you are about to learn is historic! It has never been realised before. It has never been published anywhere in any form whatsoever and it is almost certain that no other human even came close to realising this knowledge. There is ONE differentiation formula for all functions and the implications are many, but here are just a few:

1. No need to learn limit theory or real analysis and solid proof that the mainstream formulation of calculus is a kludge based on ill-formed concepts. This knowledge reveals without any doubt that limit theory is neither required in calculus, nor is it rigorous. The mainstream calculus was never rigorous.

2. A rigorous and complete geometric derivation that refutes Cauchy’s claim that a derivative cannot be defined by any means other than algebra (*). Cauchy is the main reason that mainstream calculus was never rigorous.

(*) He believed a combination of algebra and limits would be rigorous. He felt that he had to remove algebra by itself as an approach to calculus. You'll see shortly that Cauchy was wrong because algebra is an extension of geometry.

3. Easy to learn using only high school geometry and trigonometry.

4. No need to learn many differentiation rules and techniques. The ONE formula works on any function.

This ingenious idea came to me during my research on how to produce a complete rigorous geometric formulation. Well, the New Calculus is such a formulation, but the most recent revelation is even more primitive in that it could have been realised by my brilliant Ancestors - the Ancient Greeks."genius

No doubt the English will not welcome this newly realised knowledge because it shows that their national "genius" Newton was not the world's greatest genius as they imagined. I suppose the same applies to the Germans and their wrong ideas about Leibniz's "genius".

Read more here:

https://www.linkedin.com/pulse/general-derivative-formula-all-functions-john-gabriel

[Mostowski Collapse]

New calculoose, the only mathematical network that
is a closed loop consisting of one person. It connects

John Gabbermonkeys bird brain directly with his rectum vagina.

[Me]

On Monday, March 23, 2020 at 1:57:23 PM UTC+1, Eram semper recta wrote:

> The ONE formula works on any function.

Nope. It does not even work for all SMOOTH functions.

See: https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic

[Eram semper recta]

On Tuesday, 24 March 2020 10:12:50 UTC-4, Me wrote:
> On Monday, March 23, 2020 at 1:57:23 PM UTC+1, Eram semper recta wrote:
>
> > The ONE formula works on any function.
>
> Nope. It does not even work for all SMOOTH functions.

It works on ALL smooth functions.

You have no counter-example.

[Me]

On Tuesday, March 24, 2020 at 6:02:47 PM UTC+1, Eram semper recta wrote:
> On Tuesday, 24 March 2020 10:12:50 UTC-4, Me wrote:

> > Your "method" does not even work for all SMOOTH functions.

> >
> It works on ALL smooth functions.
>
> You have no counter-example.

Sure, I do.

See: https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic

[Eram semper recta]

Sure you don't.

That is not a counter example moron.

[Me]

On Tuesday, March 24, 2020 at 9:16:02 PM UTC+1, Eram semper recta wrote:

> > See: https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic

> That is not a counter example

Ok, so please DETERMINE f' for this function using your "method". :-)

[Eram semper recta]

I told you numerous times you moron: My method works on ALL smooth functions.

[Mostowski Collapse]

tanh(x) is smooth. But it has an inflection point
at x=0. So you are contradicting yourself.

[Zelos Malum]

stating it works on all smooth functions is quite circular.

"Derivation works on derivable functions", that is some great claims!

[Eram semper recta]

On Wednesday, March 25, 2020 at 2:51:40 AM UTC-4, Zelos Malum wrote:
> Den tisdag 24 mars 2020 kl. 18:02:47 UTC+1 skrev Eram semper recta:
> > On Tuesday, 24 March 2020 10:12:50 UTC-4, Me wrote:
> > > On Monday, March 23, 2020 at 1:57:23 PM UTC+1, Eram semper recta wrote:
> > >
> > > > The ONE formula works on any function.
> > >
> > > Nope. It does not even work for all SMOOTH functions.
> >
> > It works on ALL smooth functions.
> >
> > You have no counter-example.
>
> stating it works on all smooth functions is quite circular.

Nope. If there is a tangent line, then there is a slope (derivative).

>
> "Derivation works on derivable functions", that is some great claims!

Apparently one you've never understood.

No tangent line => no derivative.

[Zelos Malum]

>Nope. If there is a tangent line, then there is a slope (derivative).

Smooth means it is infinitely derivable, so saying derivation works on smooth functions is circular.

>No tangent line => no derivative.

You are so pathetic when all you can think in is images

[Eram semper recta]

On Thursday, 26 March 2020 03:32:37 UTC-4, Zelos Malum wrote:
> >Nope. If there is a tangent line, then there is a slope (derivative).
>
> Smooth means it is infinitely derivable,

Nonsense. That is ambiguous bullshit. If f has a derivative at every point in a given interval (except inflection points), then it is automatically smooth.

But in order to be differentiable at every point, f must have exactly ONE tangent line possible at every point in the interval.

> so saying derivation works on smooth functions is circular.

Look crank, what you know about circular is circular!

Your bogus mainstream calculus is circular. LMAO. If you say a f is differentiable over a given interval, then it is already SMOOTH because it can't be differentiable without being SMOOTH first.


Very simple: NO tangent line => NO derivative.

[Dan Christensen]

On Thursday, March 26, 2020 at 8:55:12 AM UTC-4, Eram semper recta wrote:
> On Thursday, 26 March 2020 03:32:37 UTC-4, Zelos Malum wrote:
> > >Nope. If there is a tangent line, then there is a slope (derivative).
> >
> > Smooth means it is infinitely derivable,
>
> Nonsense. That is ambiguous bullshit. If f has a derivative at every point in a given interval (except inflection points), then it is automatically smooth.
>

Derivatives at points of inflection are easily handled in real-world calculus. Yeah, it ACTUALLY works, John. Go figure.

In your goofy little "system," however, it seems to be an intractable problem. Better fix that. Better still, just put a match to the whole thing and get on with your life.

> But in order to be differentiable at every point, f must have exactly ONE tangent line possible at every point in the interval.
>

So, what is the derivative of f(x)=|x|? (Hint: In the real world, f'(x) is -1 if x<0, +1 if x>0 and undefined if x=x.)

> > so saying derivation works on smooth functions is circular.
>
> Look crank, what you know about circular is circular!
>
> Your bogus mainstream calculus is circular. LMAO. If you say a f is differentiable over a given interval, then it is already SMOOTH because it can't be differentiable without being SMOOTH first.
>
>
> Very simple: NO tangent line => NO derivative.

Sorry, not good enough, John. Given a function, it is not enough to draw pretty little pictures with with wiggly line approximations of curves and tangent lines, etc. Since you do not allow limits, you will have prove the existence (or non-existence) of the derivative at any given point, using ONLY algebraic methods. Apply your silly little "theorem" if it's good for anything. (It isn't but you can try.)


Dan

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

[Me]

On Wednesday, March 25, 2020 at 12:53:39 AM UTC+1, Eram semper recta wrote:
> On Tuesday, 24 March 2020 17:31:50 UTC-4, Me wrote:
> > On Tuesday, March 24, 2020 at 9:16:02 PM UTC+1, Eram semper recta wrote:
> > > >
> > > > See: https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic
> > >
> > > That is not a counter example

Sure it is.

> I told you <bla>

Please DETERMINE f' for this function using your "method" to PROVE your point. :-)

[Eram semper recta]

Repeating your bullshit that this is a counter-example won't make it so.

Once again: my theorem works for ALL *smooth* functions.

[Me]

On Thursday, March 26, 2020 at 5:21:10 PM UTC+1, Eram semper recta wrote:
> On Thursday, 26 March 2020 10:18:52 UTC-4, Me wrote:
> > On Wednesday, March 25, 2020 at 12:53:39 AM UTC+1, Eram semper recta wrote:
> > > On Tuesday, 24 March 2020 17:31:50 UTC-4, Me wrote:
> > > > On Tuesday, March 24, 2020 at 9:16:02 PM UTC+1, Eram semper recta wrote:
> > > > > >
> > > > > > See: https://en.wikipedia.org/wiki/Non-analytic_smooth_function#A_smooth_function_which_is_nowhere_real_analytic
> > > > >
> > > > > That is not a counter example
> >
> > Sure it is.
> >
> > > I told you <bla>
> >
> > Please DETERMINE f' for this function using your "method" to PROVE your point. :-)
> >

> Repeating that [...]

[Eram semper recta]

On Monday, 24 February 2020 07:55:39 UTC-5, Eram semper recta wrote:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.
>
> This theorem provides the rigour you need in mainstream calculus but have never had.
>
> The historic theorem:
>
> https://lnkd.in/gxBsQf7
>
> How it fixes your bogus mainstream formulation:
>
> https://lnkd.in/g6q4CJA
>
> This FIX is brought to you by the FIRST and ONLY rigorous formulation of calculus in human history - The New Calculus, courtesy of its author John Gabriel:
>
> Download the free ebook here:
>
> https://lnkd.in/gXwPzy4

No matter how much jealous and ignorant mainstream academics hate my work, they will soon become irrelevant.

[Zelos Malum]

>Nonsense. That is ambiguous bullshit

https://en.wikipedia.org/wiki/Smoothness

Study mathematics.

>If f has a derivative at every point in a given interval (except inflection points), then it is automatically smooth.

The fact you need to write "except inflection points", demonstrates how weak your derivative is.

and no, you can have functions that are only derivable once.

>Look crank, what you know about circular is circular!

You don't know what circular even is.

>Your bogus mainstream calculus is circular. LMAO

Not at all, it is not. I challange you to find it and it is not just your misunderstanding (it usually is)

>If you say a f is differentiable over a given interval, then it is already SMOOTH because it can't be differentiable without being SMOOTH first.

Nope, read the link.

>Very simple: NO tangent line => NO derivative.

When will you grow up adn stop thinking like a toddler?

[Eram semper recta]

On Monday, 24 February 2020 07:55:39 UTC-5, Eram semper recta wrote:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.
>
> This theorem provides the rigour you need in mainstream calculus but have never had.
>
> The historic theorem:
>
> https://lnkd.in/gxBsQf7
>
> How it fixes your bogus mainstream formulation:
>
> https://lnkd.in/g6q4CJA
>
> This FIX is brought to you by the FIRST and ONLY rigorous formulation of calculus in human history - The New Calculus, courtesy of its author John Gabriel:
>
> Download the free ebook here:
>
> https://lnkd.in/gXwPzy4

The church of academia is trembling with fear because their day is coming. Sooner rather than later, they will all have to learn the New Calculus otherwise their students will shame them.

Fasten your seat belts cranks! I am only getting started. Chuckle.

[Eram semper recta]

On Monday, 24 February 2020 07:55:39 UTC-5, Eram semper recta wrote:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.
>
> This theorem provides the rigour you need in mainstream calculus but have never had.
>
> The historic theorem:
>
> https://lnkd.in/gxBsQf7
>
> How it fixes your bogus mainstream formulation:
>
> https://lnkd.in/g6q4CJA
>
> This FIX is brought to you by the FIRST and ONLY rigorous formulation of calculus in human history - The New Calculus, courtesy of its author John Gabriel:
>
> Download the free ebook here:
>
> https://lnkd.in/gXwPzy4

To learn about the first rigorous formulation of calculus in human history:

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

My most recently revealed geometric theorem places the final nail in the coffin of bogus mainstream calculus:

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

Being the kind and caring genius that I am, I show you how to fix your bogus mainstream calculus formulation:

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

[Mostowski Collapse]

Bird brain, can you show:

than'(x) = 1 - tanh(x)^2

Whats the tangent at the inflection point?

LMAO!

[Eram semper recta]

On Monday, 24 February 2020 07:55:39 UTC-5, Eram semper recta wrote:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.
>
> This theorem provides the rigour you need in mainstream calculus but have never had.
>
> The historic theorem:
>
> https://lnkd.in/gxBsQf7
>
> How it fixes your bogus mainstream formulation:
>
> https://lnkd.in/g6q4CJA
>
> This FIX is brought to you by the FIRST and ONLY rigorous formulation of calculus in human history - The New Calculus, courtesy of its author John Gabriel:
>
> Download the free ebook here:
>
> https://lnkd.in/gXwPzy4

Euler's Blunder has infected every aspect of mainstream mathematics:

1. Arithmetic: morons believe that there are infinite series which can actually be summed to produce a limit. e.g. {0.3; 0.03; 0.003; ...}

2. Differentiation: The fallacious f'(x)=lim_{h->0} [f(x+h)-f(x)]/h definition is in fact the limit of finite difference quotients, ALL of which NEVER represent the derivative. This is similar to the arithmetic idea in (1) except it is a sequence of finite differences:

[f(x+h_1)-f(x)]/h_1; [f(x+h_2)-f(x)]/h_2; ...; [f(x+h_n)-f(x)]/h_n; ...

The idea here is that somehow at infinity, that is, as h becomes infinitely small, the "infiniteth" finite difference will suddenly become the derivative, even though a 4 years old knows that there is no finite difference possible when h=0.

3. Integration: In this case, the orangutans of mainstream academia are looking at the limit of areas or products which take the form of rectangles. The irony here is that in order to arrive at the limit, each rectangle would have to become a 'line' which has no area. The baboons imagine summing up all these "infinitely many" lines or areas (?) to get the total area. e.g. 0+0+0+0+... = Area.

Fascinating how this Swiss cunt changed the history of mainstream mathematics. He looked like a moron and he was one. No offense to those who weren't as ugly as Euler. This blunder alone dulls ALL the shine out of his remaining accomplishments.

[Dan Christensen]

On Tuesday, March 31, 2020 at 8:14:48 AM UTC-4, Eram semper recta wrote:

>
> Euler's Blunder has infected every aspect of mainstream mathematics:
>

The "blunder" was YOURS, John. Be a man for once and own up to your mistake.

Admit that it was YOU who defaced a copy of Euler's original article by writing in big red letters, the nonsensical "S = Lim S." The modern limit notation was not invented until several decades after his death.

> 1. Arithmetic: morons believe that there are infinite series which can actually be summed to produce a limit. e.g. {0.3; 0.03; 0.003; ...}
>

There actually are. 0.3333... = 1/3. Yes, I know, you never did master fractions. (See below.)

> 2. Differentiation: The fallacious f'(x)=lim_{h->0} [f(x+h)-f(x)]/h definition is in fact the limit of finite difference quotients, ALL of which NEVER represent the derivative.

Nothing "fallacious" about it. It makes a trivial case of functions that present an intractable problem for your goofy little, dead-end system (e.g f(x)=|x|). Fix it or scrap it, John. Preferably scrap it, cut your losses and get on with your life. You aren't getting any younger.


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

“Numbers don't exist.”
--March 19, 2020

"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

[Eram semper recta]

On Monday, 24 February 2020 07:55:39 UTC-5, Eram semper recta wrote:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No infinity, no infinitesimals, no ghosts of departed quantities, no ultimate ratios (LMAO at poor Newton!), no limit theory (which states unremarkable properties disguised as "proof"). NO MORE BULLSHIT.
>
> This theorem provides the rigour you need in mainstream calculus but have never had.
>
> The historic theorem:
>
> https://lnkd.in/gxBsQf7
>
> How it fixes your bogus mainstream formulation:
>
> https://lnkd.in/g6q4CJA
>
> This FIX is brought to you by the FIRST and ONLY rigorous formulation of calculus in human history - The New Calculus, courtesy of its author John Gabriel:
>
> Download the free ebook here:
>
> https://lnkd.in/gXwPzy4

Crank Alert! Psycho troll Dan Christensen took a dump on this thread.

[Zelos Malum]

You're the crank still , again you're the one opposing mainstream and no one is fearing you

[Eram semper recta]

You were gone for a couple of days? I was hoping covid19 got you. Well, I guess I can still keep on hoping...