On Thursday, 29 July 2021 at 19:35:35 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:

> ...
> >>>> Oh, and that is NOT true for any function. But how could a psycho crank like you know this...
> >>>
> >>> Saying it is true for any function is saying that any function has a tangent line at every point. Tsk, tsk.
> >
> >> No, it's not because \exists f'(x) : [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
> >
> >> express NOTHING about f'(x) being the slope of a tangent line at every
> >> point.
> >
> > It does. [f(x+h)-f(x)]/h = f'(x) + Q(x,h) <==> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
> >
> >
> >> It's just stating that [f(x+h)-f(x)]/h - f'(x) is a function of
> >> (x,h), which is TRUE for ANY function, not only f'.
> >
> > No. It's not true for ANY function, only for SMOOTH functions. If this were not the case, then you couldn't even write \exists f'(x) : [f(x+h)-f(x)]/h = f'(x) + Q(x,h) because [f(x+h)-f(x)]/h requires a smooth function.
> Huh? What prevents you to compute [f(x+h)-f(x)]/h for ANY function
> smooth or not? NOTHING John. You are more brain damaged each day.
>
> ...
> >> Compare with proper calculus which put restrictions on Q, while you don't.
> >
> > What are you babbling about

> [f(x+h)-f(x)]/h = f'(x) + Q(x,h), lim_(h->0) Q(x,h) = 0

What?! LMAO. NO, NO, NO.

There is only [f(x+h)-f(x)]/h = f'(x) + Q(x,h). Nothing about bullshit like lim_(h->0) Q(x,h) = 0 over there. That's just your wishful thinking! LMAO. Sorry, but Q(x,h) is a constant and if you take the limit if a constant, it is the constant itself.

>
> usually expressed as :
>
> f(x+h) = f(x) + h*f'(x) + h*Q(x,h), lim_(h->0) Q(x,h) = 0

Bullshit. You are indeed losing your last few marbles.

>
> which actually express that f'(x) is the slope of the tangent line
> at (x,f(x)). Something you are UNABLE to do.

Chuckle. The identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) does it just fine:

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

No hand waving limit bullshit there.

>
> You could try to learn calculus once, John.

I learned the flawed mainstream calculus before you were born, you fucking shithead!

Crank John Gabriel, aka Eram semper recta wrote:
....
>> which actually express that f'(x) is the slope of the tangent line
>> at (x,f(x)). Something you are UNABLE to do.
>
> Chuckle. The identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) does it just fine:
>
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
>
> No hand waving limit bullshit there.

+ or - Q(x,h), it makes no difference as Q is arbritary. How can you be
THAT stupid John?

100% hand waving and meaningless you provide ZERO constraint on
Q. For any function g one can pick Q so that

g(x) = [f(x+h)-f(x)]/h - Q(x,h), it is as simple as :

[f(x+h)-f(x)]/h - g(x)

true for g=f' but TRUE ALSO for ANY function.

SO 100% irrelevance when it comes to derivative.

Learn calculus, then die, John.

On Thursday, 29 July 2021 at 19:44:01 UTC-4, Eram semper recta wrote:
> On Thursday, 29 July 2021 at 19:35:35 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:
>
> > ...
> > >>>> Oh, and that is NOT true for any function. But how could a psycho crank like you know this...
> > >>>
> > >>> Saying it is true for any function is saying that any function has a tangent line at every point. Tsk, tsk.
> > >
> > >> No, it's not because \exists f'(x) : [f(x+h)-f(x)]/h = f'(x) + Q(x,h)
> > >
> > >> express NOTHING about f'(x) being the slope of a tangent line at every
> > >> point.
> > >
> > > It does. [f(x+h)-f(x)]/h = f'(x) + Q(x,h) <==> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
> > >
> > >
> > >> It's just stating that [f(x+h)-f(x)]/h - f'(x) is a function of
> > >> (x,h), which is TRUE for ANY function, not only f'.
> > >
> > > No. It's not true for ANY function, only for SMOOTH functions. If this were not the case, then you couldn't even write \exists f'(x) : [f(x+h)-f(x)]/h = f'(x) + Q(x,h) because [f(x+h)-f(x)]/h requires a smooth function.
> > Huh? What prevents you to compute [f(x+h)-f(x)]/h for ANY function
> > smooth or not? NOTHING John. You are more brain damaged each day.
> >
> > ...
> > >> Compare with proper calculus which put restrictions on Q, while you don't.
> > >
> > > What are you babbling about
>
>
> > [f(x+h)-f(x)]/h = f'(x) + Q(x,h), lim_(h->0) Q(x,h) = 0
> What?! LMAO. NO, NO, NO.
>
> There is only [f(x+h)-f(x)]/h = f'(x) + Q(x,h). Nothing about bullshit like lim_(h->0) Q(x,h) = 0 over there. That's just your wishful thinking! LMAO. Sorry, but Q(x,h) is a constant and if you take the limit if a constant, it is the constant itself.
> >
> > usually expressed as :
> >
> > f(x+h) = f(x) + h*f'(x) + h*Q(x,h), lim_(h->0) Q(x,h) = 0
> Bullshit. You are indeed losing your last few marbles.
> >
> > which actually express that f'(x) is the slope of the tangent line
> > at (x,f(x)). Something you are UNABLE to do.
> Chuckle. The identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) does it just fine:
> f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
> No hand waving limit bullshit there.
> >
> > You could try to learn calculus once, John.
> I learned the flawed mainstream calculus before you were born, you fucking shithead!

Students:

lim_(h->0) Q(x,h) = 0 is the wish of mainstream cranks.

But watch what happens.

Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h

Now if x = 1 and h=4, we have:

[f(5)-f(1)]/4 = 6 = 2(1)+4 = 6

Now shithead Jean Pierre Messager (aka Python) does not like the constant 4 on the RHS of the identity. So he believes that lim_(h->0) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:

[f(5)-f(1)]/4 = 6 = 2(1)+0=2

So 6 = 2 ? LMAO!!!! Huge Lols.

That particular secant line slope is 6. It cannot change to 2!

See, this is the kind of fundamentalist mindset and lack of attention to detail one finds in the many morons of mainstream academia.

I do know better than ALL of you fucking idiots because I am a genius whose knowledge of mathematics is unequaled by any other human.

You will be crushed if you try to talk bullshit. Be warned from the example I am making of Python. His colleagues and students will be laughing at idiot.

Crank John Gabriel, aka Eram semper recta wrote:
> ... Sorry, but Q(x,h) is a constant and if you take the limit if a constant,
> it is the constant itself.

Q(x,h) can have different values for different values taken by h. This
is what is meant by not being a constant.

Q, is general, is definitely NOT A CONSTANT for different values of h.
If so, why don't you call it Q(x) instead...

You're such a psychotic kind of dork, John. Look for medical help.

On Thursday, 29 July 2021 at 19:50:38 UTC-4, Python wrote:
> Crank John Gabriel, aka Eram semper recta wrote:
> ...
> >> which actually express that f'(x) is the slope of the tangent line
> >> at (x,f(x)). Something you are UNABLE to do.
> >
> > Chuckle. The identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) does it just fine:
> >
> > f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
> >
> > No hand waving limit bullshit there.

> + or - Q(x,h), it makes no difference as Q is arbritary.

?? Wake up moron!!!!!! Q(x,h) is not arbitrary, it is a CONSTANT. Go back and study the previous comment, you dipstick idiot!

> How can I be THAT stupid?

Right!
>
> 100% hand waving and meaningless you provide ZERO constraint on
> Q.

You mean Q(x,h). Does it occur to you idiot, that Q has special meaning?

No, no constraints are needed. The identity is a result of the geometric **theorem**.

> For any function g one can pick Q so that

NO. f must be SMOOTH.

On Thursday, 29 July 2021 at 19:56:09 UTC-4, Eram semper recta wrote:
> On Thursday, 29 July 2021 at 19:50:38 UTC-4, Python wrote:
> > Crank John Gabriel, aka Eram semper recta wrote:
> > ...
> > >> which actually express that f'(x) is the slope of the tangent line
> > >> at (x,f(x)). Something you are UNABLE to do.
> > >
> > > Chuckle. The identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) does it just fine:
> > >
> > > f'(x) = [f(x+h)-f(x)]/h - Q(x,h)
> > >
> > > No hand waving limit bullshit there.
>
> > + or - Q(x,h), it makes no difference as Q is arbritary.
> ?? Wake up moron!!!!!! Q(x,h) is not arbitrary, it is a CONSTANT. Go back and study the previous comment, you dipstick idiot!
>
> > How can I be THAT stupid?
>
> Right!
> >
> > 100% hand waving and meaningless you provide ZERO constraint on
> > Q.
> You mean Q(x,h). Does it occur to you idiot, that Q has special meaning?
>
> No, no constraints are needed. The identity is a result of the geometric **theorem**.
> > For any function g one can pick Q so that
> NO. f must be SMOOTH.

And Q(x,h) is never picked. It is fixed for any (x,h) pair. Picking your nose is what you do in your bogus calculus! LOL.

Crank Psycho John Gabriel, recta wrote:
....
> But watch what happens.
>
> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
>
> Now if x = 1 and h=4, we have:

Note that here *h=4*

> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
>
> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:

Note that now *h->0* John is now making 4 to tend to 0.

> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
>
> So 6 = 2 ? LMAO!!!! Huge Lols.

Well, right. Sure. Ask for medical help John. You're not going well.

On Thursday, 29 July 2021 at 19:57:55 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:

> ...
> > But watch what happens.
> >
> > Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
> >
> > Now if x = 1 and h=4, we have:
> Note that here *h=4*
> > [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
> >
> > Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
>
> Note that now *h->0* John is now making 4 to tend to 0.

What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!

> > [f(5)-f(1)]/4 = 6 = 2(1)+0=2
> >
> > So 6 = 2 ? LMAO!!!! Huge Lols.

> Well, right. Sure.

Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.

Man, you are so boring. I've kicked your arse countless times and you still don't know that I am your master?

Crank John Gabriel, aka Eram semper recta wrote:
> On Thursday, 29 July 2021 at 19:57:55 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:
>
>> ...
>>> But watch what happens.
>>>
>>> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
>>>
>>> Now if x = 1 and h=4, we have:
>> Note that here *h=4*
>>> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
>>>
>>> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
>>
>> Note that now *h->0* John is now making 4 to tend to 0.
>
> What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!
>
>>> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
>>>
>>> So 6 = 2 ? LMAO!!!! Huge Lols.
>
>> Well, right. Sure.
>
> Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.

Compute yourself:

f(x) = x^2
f'(x) = 2x [taken from good old real calculus btw]

Now write down Q(x,h) and compute Q(0,1/2) and Q(0,1/4), are they equal?

They're not, so Q(x,h), is NOT a constant, not even Q(0,h) is.

John, you are loosing the meaning of the most basic concepts such as
being a constant. You are really mentally damaged. You definitely need
medical help.

Q(x,h) =

On Thursday, 29 July 2021 at 20:10:25 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:

> >
> >> ...
> >>> But watch what happens.
> >>>
> >>> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
> >>>
> >>> Now if x = 1 and h=4, we have:
> >> Note that here *h=4*
> >>> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
> >>>
> >>> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
> >>
> >> Note that now *h->0* John is now making 4 to tend to 0.
> >
> > What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!
> >
> >>> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
> >>>
> >>> So 6 = 2 ? LMAO!!!! Huge Lols.
> >
> >> Well, right. Sure.
> >
> > Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.
> Compute yourself:
>
> f(x) = x^2
> f'(x) = 2x [taken from good old real calculus btw]
>
> Now write down Q(x,h) and compute Q(0,1/2) and Q(0,1/4), are they equal?

Hello Moron!

Of course they will not be equal, you silly fool! So what?!!!!!!

>
> They're not, so Q(x,h), is NOT a constant, not even Q(0,h) is.

Q(0,1/2)=1/2 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/2, 1/4).
Q(0,1/4)=1/4 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/4, 1/16).

If you change h, you are no longer talking about the same secant line. Get it idiot?

>
> John, you are loosing the meaning of the most basic concepts such as
> being a constant. You are really mentally damaged. You definitely need
> medical help.

Oh, the irony! How can you be such an obstinate ass?!

On Friday, 30 July 2021 at 07:24:02 UTC-4, Eram semper recta wrote:
> On Thursday, 29 July 2021 at 20:10:25 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:
>
> > >
> > >> ...
> > >>> But watch what happens.
> > >>>
> > >>> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
> > >>>
> > >>> Now if x = 1 and h=4, we have:
> > >> Note that here *h=4*
> > >>> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
> > >>>
> > >>> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
> > >>
> > >> Note that now *h->0* John is now making 4 to tend to 0.
> > >
> > > What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!
> > >
> > >>> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
> > >>>
> > >>> So 6 = 2 ? LMAO!!!! Huge Lols.
> > >
> > >> Well, right. Sure.
> > >
> > > Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.
> > Compute yourself:
> >
> > f(x) = x^2
> > f'(x) = 2x [taken from good old real calculus btw]
> >
> > Now write down Q(x,h) and compute Q(0,1/2) and Q(0,1/4), are they equal?
> Hello Moron!
>
> Of course they will not be equal, you silly fool! So what?!!!!!!
> >
> > They're not, so Q(x,h), is NOT a constant, not even Q(0,h) is.
> Q(0,1/2)=1/2 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/2, 1/4).
> Q(0,1/4)=1/4 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/4, 1/16).
>
> If you change h, you are no longer talking about the same secant line. Get it idiot?
> >
> > John, you are loosing the meaning of the most basic concepts such as
> > being a constant. You are really mentally damaged. You definitely need
> > medical help.
> Oh, the irony! How can you be such an obstinate ass?!

Students:

See, this is what I mean when I tell you that mainstream math academics are incorrigible morons who cannot be fixed.

Jean Pierre Messager claims to have a master degree in mathematics.

One of two things is happening here:

i. Messager knows he is wrong and is being a stubborn ass.
ii. Messager really does not understand.

I would bet on (i) because I am not convinced Messager is that stupid. All the same, the level of dishonesty, jealousy and all the other undesirable human emotions can be very annoying.

On Friday, 30 July 2021 at 07:24:02 UTC-4, Eram semper recta wrote:
> On Thursday, 29 July 2021 at 20:10:25 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:
>
> > >
> > >> ...
> > >>> But watch what happens.
> > >>>
> > >>> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
> > >>>
> > >>> Now if x = 1 and h=4, we have:
> > >> Note that here *h=4*
> > >>> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
> > >>>
> > >>> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
> > >>
> > >> Note that now *h->0* John is now making 4 to tend to 0.
> > >
> > > What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!
> > >
> > >>> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
> > >>>
> > >>> So 6 = 2 ? LMAO!!!! Huge Lols.
> > >
> > >> Well, right. Sure.
> > >
> > > Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.
> > Compute yourself:
> >
> > f(x) = x^2
> > f'(x) = 2x [taken from good old real calculus btw]
> >
> > Now write down Q(x,h) and compute Q(0,1/2) and Q(0,1/4), are they equal?
> Hello Moron!
>
> Of course they will not be equal, you silly fool! So what?!!!!!!
> >
> > They're not, so Q(x,h), is NOT a constant, not even Q(0,h) is.
> Q(0,1/2)=1/2 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/2, 1/4).
> Q(0,1/4)=1/4 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/4, 1/16).

What mainstream math baboons want you to believe is that you can do something like this:

Q(0,1/2)=1/2
Q(0,1/4)=1/4
Q(0,1/16)=1/16
Q(0,1/32)=1/32

Until Q(0,h) gets very close to 0 so that 2x + h = 2x. LMAO!

But as you can see (if you have two brain cells at least!), NONE of the Q(x,h) change for any given secant line. There is a ONE-TO-ONE correspondence between each secant line and Q(x,h).

YOU CANNOT USE A DIFFERENT Q(X,H) FOR THE SAME FUCKING SECANT LINE, MORONSSSS!!!!!!!! WHY?

Because then the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) is NO LONGER TRUE!!!!! Stupid fuckwads!!!!!

Those of you who know I am right and are keeping silent, I hope you get COVID19 and DIE, you vile scumbags!!!!!

Your mainstream calculus was NEVER rigorous. It was developed by minds far inferior to mine. I am a genius and ALL of you without any exception are fucking retards! Get it? CHUCKLE.

Eat shit and die every one of you!!!!

Grrr. I hate dishonest and vile, lying academics. Speak up or die, you scum!!!!!
>
> If you change h, you are no longer talking about the same secant line. Get it idiot?
> >
> > John, you are loosing the meaning of the most basic concepts such as
> > being a constant. You are really mentally damaged. You definitely need
> > medical help.
> Oh, the irony! How can you be such an obstinate ass?!

On Friday, 30 July 2021 at 07:35:51 UTC-4, Eram semper recta wrote:
> On Friday, 30 July 2021 at 07:24:02 UTC-4, Eram semper recta wrote:
> > On Thursday, 29 July 2021 at 20:10:25 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:
> >
> > > >
> > > >> ...
> > > >>> But watch what happens.
> > > >>>
> > > >>> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
> > > >>>
> > > >>> Now if x = 1 and h=4, we have:
> > > >> Note that here *h=4*
> > > >>> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
> > > >>>
> > > >>> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
> > > >>
> > > >> Note that now *h->0* John is now making 4 to tend to 0.
> > > >
> > > > What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!
> > > >
> > > >>> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
> > > >>>
> > > >>> So 6 = 2 ? LMAO!!!! Huge Lols.
> > > >
> > > >> Well, right. Sure.
> > > >
> > > > Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.
> > > Compute yourself:
> > >
> > > f(x) = x^2
> > > f'(x) = 2x [taken from good old real calculus btw]
> > >
> > > Now write down Q(x,h) and compute Q(0,1/2) and Q(0,1/4), are they equal?
> > Hello Moron!
> >
> > Of course they will not be equal, you silly fool! So what?!!!!!!
> > >
> > > They're not, so Q(x,h), is NOT a constant, not even Q(0,h) is.
> > Q(0,1/2)=1/2 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/2, 1/4).
> > Q(0,1/4)=1/4 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/4, 1/16).
> What mainstream math baboons want you to believe is that you can do something like this:
>
> Q(0,1/2)=1/2
> Q(0,1/4)=1/4
> Q(0,1/16)=1/16
> Q(0,1/32)=1/32
>
> Until Q(0,h) gets very close to 0 so that 2x + h = 2x. LMAO!
>
> But as you can see (if you have two brain cells at least!), NONE of the Q(x,h) change for any given secant line. There is a ONE-TO-ONE correspondence between each secant line and Q(x,h).
>
> YOU CANNOT USE A DIFFERENT Q(X,H) FOR THE SAME FUCKING SECANT LINE, MORONSSSS!!!!!!!! WHY?
>
> Because then the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) is NO LONGER TRUE!!!!! Stupid fuckwads!!!!!
>
> Those of you who know I am right and are keeping silent, I hope you get COVID19 and DIE, you vile scumbags!!!!!
>
> Your mainstream calculus was NEVER rigorous. It was developed by minds far inferior to mine. I am a genius and ALL of you without any exception are fucking retards! Get it? CHUCKLE.
>
> Eat shit and die every one of you!!!!
>
> Grrr. I hate dishonest and vile, lying academics. Speak up or die, you scum!!!!!
> >
> > If you change h, you are no longer talking about the same secant line. Get it idiot?
> > >
> > > John, you are loosing the meaning of the most basic concepts such as
> > > being a constant. You are really mentally damaged. You definitely need
> > > medical help.
> > Oh, the irony! How can you be such an obstinate ass?!

The cat has been out of the bag a long time. Admit that I am the greatest mathematician ever, you vile bastards!

On Friday, July 30, 2021 at 10:43:03 AM UTC-4, I am Super Rectum (aka John Gabriel) wrote:

> The cat has been out of the bag a long time. Admit that I am the greatest mathematician ever, you vile bastards!

Mr. Rectum here, the self-styled "greatest mathematician ever" has re-invented numbers but still cannot even prove the 2+2=4 in his goofy little system. What a moron!

Dan

On 7/30/2021 10:42 AM, Eram semper recta wrote:

> The cat has been out of the bag a long time. Admit that I am the greatest mathematician ever, you vile bastards!

You're in the wrong place. This is Usenet, where we laugh at those who
call themselves "the greatest mathematician ever", "Genius", "King of
Science" etc.

If you want a bunch of fools calling you "the greatest mathematician
ever", you need to start an evil cult, be its messiah and brainwash your
minions into saying that. That might be rather difficult, brainwashing
people only goes so far.

On Friday, 30 July 2021 at 13:00:04 UTC-4, Michael MORONey wrote:

> > The cat has been out of the bag a long time. Admit that I am the greatest mathematician ever, you vile bastards!
> Yo...<drivel>

Oh, cute! MORONey thought I was addressing him! ROFLMFAO!!!

No, MORONey, No. If you praised me, then I would be very worried!!! Chuckle.

On 7/30/2021 1:11 PM, Eram semper recta wrote:
> On Friday, 30 July 2021 at 13:00:04 UTC-4, Michael MORONey wrote:
>
>>> The cat has been out of the bag a long time. Admit that I am the greatest mathematician ever, you vile bastards!
>> Yo...<drivel>
>
> Oh, cute! MORONey thought I was addressing him! ROFLMFAO!!!
>
Of course not! I was pointing out how you seem to be addressing some
nonexistent lurkers who would praise you. They're not here. They're just
as make-believe as the pretend "students" you often address in your
posts. I just point out this is Usenet where people laugh at you NPD
sufferers, and if you want praise, start a cult with yourself as
messiah. But as I pointed out, brainwashing only goes so far.

On Tuesday, 20 July 2021 at 13:56:53 UTC-4, Eram semper recta wrote:
> Telling students that the derivative definition they are given at university is fraudulent, is the TRUTH. The evidence is ALL here:
>
> https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
>
> https://drive.google.com/file/d/1520NjhgiakcrssQxtbxRCDXus_aHXpI9
>
> https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
>
> https://drive.google.com/file/d/1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y

There's a damn good reason Prof. Gilbert Strang (aka Port563), Prof. Ullrich, Prof. Joyce, etc no longer argue with me - they know I'll kick their arses.

Only the local dimwits still hang around to be ridiculed. They all know who they are.

On Friday, 30 July 2021 at 07:35:51 UTC-4, Eram semper recta wrote:
> On Friday, 30 July 2021 at 07:24:02 UTC-4, Eram semper recta wrote:
> > On Thursday, 29 July 2021 at 20:10:25 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:
> >
> > > >
> > > >> ...
> > > >>> But watch what happens.
> > > >>>
> > > >>> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
> > > >>>
> > > >>> Now if x = 1 and h=4, we have:
> > > >> Note that here *h=4*
> > > >>> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
> > > >>>
> > > >>> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
> > > >>
> > > >> Note that now *h->0* John is now making 4 to tend to 0.
> > > >
> > > > What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!
> > > >
> > > >>> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
> > > >>>
> > > >>> So 6 = 2 ? LMAO!!!! Huge Lols.
> > > >
> > > >> Well, right. Sure.
> > > >
> > > > Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.
> > > Compute yourself:
> > >
> > > f(x) = x^2
> > > f'(x) = 2x [taken from good old real calculus btw]
> > >
> > > Now write down Q(x,h) and compute Q(0,1/2) and Q(0,1/4), are they equal?
> > Hello Moron!
> >
> > Of course they will not be equal, you silly fool! So what?!!!!!!
> > >
> > > They're not, so Q(x,h), is NOT a constant, not even Q(0,h) is.
> > Q(0,1/2)=1/2 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/2, 1/4).
> > Q(0,1/4)=1/4 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/4, 1/16).
> What mainstream math baboons want you to believe is that you can do something like this:
>
> Q(0,1/2)=1/2
> Q(0,1/4)=1/4
> Q(0,1/16)=1/16
> Q(0,1/32)=1/32
>
> Until Q(0,h) gets very close to 0 so that 2x + h = 2x. LMAO!
>
> But as you can see (if you have two brain cells at least!), NONE of the Q(x,h) change for any given secant line. There is a ONE-TO-ONE correspondence between each secant line and Q(x,h).
>
> YOU CANNOT USE A DIFFERENT Q(X,H) FOR THE SAME FUCKING SECANT LINE, MORONSSSS!!!!!!!! WHY?
>
> Because then the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) is NO LONGER TRUE!!!!! Stupid fuckwads!!!!!
>
> Those of you who know I am right and are keeping silent, I hope you get COVID19 and DIE, you vile scumbags!!!!!
>
> Your mainstream calculus was NEVER rigorous. It was developed by minds far inferior to mine. I am a genius and ALL of you without any exception are fucking retards! Get it? CHUCKLE.
>
> Eat shit and die every one of you!!!!
>
> Grrr. I hate dishonest and vile, lying academics. Speak up or die, you scum!!!!!
> >
> > If you change h, you are no longer talking about the same secant line. Get it idiot?
> > >
> > > John, you are loosing the meaning of the most basic concepts such as
> > > being a constant. You are really mentally damaged. You definitely need
> > > medical help.
> > Oh, the irony! How can you be such an obstinate ass?!

Have to refresh this comment because it is so damning to mainstream math academics who are stupid beyond belief.

On Sunday, 1 August 2021 at 12:19:44 UTC-4, Eram semper recta wrote:
> On Friday, 30 July 2021 at 07:35:51 UTC-4, Eram semper recta wrote:
> > On Friday, 30 July 2021 at 07:24:02 UTC-4, Eram semper recta wrote:
> > > On Thursday, 29 July 2021 at 20:10:25 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:
> > >
> > > > >
> > > > >> ...
> > > > >>> But watch what happens.
> > > > >>>
> > > > >>> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
> > > > >>>
> > > > >>> Now if x = 1 and h=4, we have:
> > > > >> Note that here *h=4*
> > > > >>> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
> > > > >>>
> > > > >>> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
> > > > >>
> > > > >> Note that now *h->0* John is now making 4 to tend to 0.
> > > > >
> > > > > What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!
> > > > >
> > > > >>> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
> > > > >>>
> > > > >>> So 6 = 2 ? LMAO!!!! Huge Lols.
> > > > >
> > > > >> Well, right. Sure.
> > > > >
> > > > > Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.
> > > > Compute yourself:
> > > >
> > > > f(x) = x^2
> > > > f'(x) = 2x [taken from good old real calculus btw]
> > > >
> > > > Now write down Q(x,h) and compute Q(0,1/2) and Q(0,1/4), are they equal?
> > > Hello Moron!
> > >
> > > Of course they will not be equal, you silly fool! So what?!!!!!!
> > > >
> > > > They're not, so Q(x,h), is NOT a constant, not even Q(0,h) is.
> > > Q(0,1/2)=1/2 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/2, 1/4).
> > > Q(0,1/4)=1/4 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/4, 1/16).
> > What mainstream math baboons want you to believe is that you can do something like this:
> >
> > Q(0,1/2)=1/2
> > Q(0,1/4)=1/4
> > Q(0,1/16)=1/16
> > Q(0,1/32)=1/32
> >
> > Until Q(0,h) gets very close to 0 so that 2x + h = 2x. LMAO!
> >
> > But as you can see (if you have two brain cells at least!), NONE of the Q(x,h) change for any given secant line. There is a ONE-TO-ONE correspondence between each secant line and Q(x,h).
> >
> > YOU CANNOT USE A DIFFERENT Q(X,H) FOR THE SAME FUCKING SECANT LINE, MORONSSSS!!!!!!!! WHY?
> >
> > Because then the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) is NO LONGER TRUE!!!!! Stupid fuckwads!!!!!
> >
> > Those of you who know I am right and are keeping silent, I hope you get COVID19 and DIE, you vile scumbags!!!!!
> >
> > Your mainstream calculus was NEVER rigorous. It was developed by minds far inferior to mine. I am a genius and ALL of you without any exception are fucking retards! Get it? CHUCKLE.
> >
> > Eat shit and die every one of you!!!!
> >
> > Grrr. I hate dishonest and vile, lying academics. Speak up or die, you scum!!!!!
> > >
> > > If you change h, you are no longer talking about the same secant line. Get it idiot?
> > > >
> > > > John, you are loosing the meaning of the most basic concepts such as
> > > > being a constant. You are really mentally damaged. You definitely need
> > > > medical help.
> > > Oh, the irony! How can you be such an obstinate ass?!
> Have to refresh this comment because it is so damning to mainstream math academics who are stupid beyond belief.

Quiz for sci.math morons:

f(x)=x^2 and f'(x)=2x and Q(x,h)=2x+h

If x=0 and h =1/16, then which of the following will make the identity [f(0+(1/16))-f(0)]/(1/16) = f '(0) + Q(0,1/16) true?

[A] Q(0,1/2)=1/2
[B] Q(0,1/4)=1/4
[C] Q(0,1/16)=1/16
[D] Q(0,1/32)=1/32

You have a 25% chance of getting it right!

So how about it Jean Pierre Messager (aka Python aka JPM aka YBM)? You going to try my Quiz?

Chuckle.

On Sunday, 1 August 2021 at 16:58:07 UTC-4, Eram semper recta wrote:
> On Sunday, 1 August 2021 at 12:19:44 UTC-4, Eram semper recta wrote:
> > On Friday, 30 July 2021 at 07:35:51 UTC-4, Eram semper recta wrote:
> > > On Friday, 30 July 2021 at 07:24:02 UTC-4, Eram semper recta wrote:
> > > > On Thursday, 29 July 2021 at 20:10:25 UTC-4, Psychopath Jean Pierre Messager aka Python wrote:
> > > >
> > > > > >
> > > > > >> ...
> > > > > >>> But watch what happens.
> > > > > >>>
> > > > > >>> Let f(x)=x^2 and so [f(x+h)-f(x)]/h = 2x + h
> > > > > >>>
> > > > > >>> Now if x = 1 and h=4, we have:
> > > > > >> Note that here *h=4*
> > > > > >>> [f(5)-f(1)]/4 = 6 = 2(1)+4 = 6
> > > > > >>>
> > > > > >>> Now Python does not like the constant 4 on the RHS of the identity. So he believes that lim_(*h->0*) 4 = 0 ! Yes, folks! This is what the crank is telling you! If Q(x,h) were equal to 0, then you would have:
> > > > > >>
> > > > > >> Note that now *h->0* John is now making 4 to tend to 0.
> > > > > >
> > > > > > What are you saying crank? You are the one who claimed lim_(h->0) Q(x,h) = 0 !!!!!!
> > > > > >
> > > > > >>> [f(5)-f(1)]/4 = 6 = 2(1)+0=2
> > > > > >>>
> > > > > >>> So 6 = 2 ? LMAO!!!! Huge Lols.
> > > > > >
> > > > > >> Well, right. Sure.
> > > > > >
> > > > > > Exactly, it would be fucking absurd to insist on lim_(h->0) Q(x,h) = 0 when Q(x,h) is CONSTANT.
> > > > > Compute yourself:
> > > > >
> > > > > f(x) = x^2
> > > > > f'(x) = 2x [taken from good old real calculus btw]
> > > > >
> > > > > Now write down Q(x,h) and compute Q(0,1/2) and Q(0,1/4), are they equal?
> > > > Hello Moron!
> > > >
> > > > Of course they will not be equal, you silly fool! So what?!!!!!!
> > > > >
> > > > > They're not, so Q(x,h), is NOT a constant, not even Q(0,h) is.
> > > > Q(0,1/2)=1/2 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/2, 1/4).
> > > > Q(0,1/4)=1/4 IS a CONSTANT, you MORON! It applies to the secant line with endpoints (0,0) and (1/4, 1/16).
> > > What mainstream math baboons want you to believe is that you can do something like this:
> > >
> > > Q(0,1/2)=1/2
> > > Q(0,1/4)=1/4
> > > Q(0,1/16)=1/16
> > > Q(0,1/32)=1/32
> > >
> > > Until Q(0,h) gets very close to 0 so that 2x + h = 2x. LMAO!
> > >
> > > But as you can see (if you have two brain cells at least!), NONE of the Q(x,h) change for any given secant line. There is a ONE-TO-ONE correspondence between each secant line and Q(x,h).
> > >
> > > YOU CANNOT USE A DIFFERENT Q(X,H) FOR THE SAME FUCKING SECANT LINE, MORONSSSS!!!!!!!! WHY?
> > >
> > > Because then the identity [f(x+h)-f(x)]/h = f'(x) + Q(x,h) is NO LONGER TRUE!!!!! Stupid fuckwads!!!!!
> > >
> > > Those of you who know I am right and are keeping silent, I hope you get COVID19 and DIE, you vile scumbags!!!!!
> > >
> > > Your mainstream calculus was NEVER rigorous. It was developed by minds far inferior to mine. I am a genius and ALL of you without any exception are fucking retards! Get it? CHUCKLE.
> > >
> > > Eat shit and die every one of you!!!!
> > >
> > > Grrr. I hate dishonest and vile, lying academics. Speak up or die, you scum!!!!!
> > > >
> > > > If you change h, you are no longer talking about the same secant line. Get it idiot?
> > > > >
> > > > > John, you are loosing the meaning of the most basic concepts such as
> > > > > being a constant. You are really mentally damaged. You definitely need
> > > > > medical help.
> > > > Oh, the irony! How can you be such an obstinate ass?!
> > Have to refresh this comment because it is so damning to mainstream math academics who are stupid beyond belief.
> Quiz for sci.math morons:
>
> f(x)=x^2 and f'(x)=2x and Q(x,h)=2x+h
>
> If x=0 and h =1/16, then which of the following will make the identity [f(0+(1/16))-f(0)]/(1/16) = f '(0) + Q(0,1/16) true?
>
> [A] Q(0,1/2)=1/2
> [B] Q(0,1/4)=1/4
> [C] Q(0,1/16)=1/16
> [D] Q(0,1/32)=1/32
>
> You have a 25% chance of getting it right!
>
> So how about it Jean Pierre Messager (aka Python aka JPM aka YBM)? You going to try my Quiz?
>
> Chuckle.

Of course Jean will not take the quiz because he knows that doing so will incriminate him! LMAO.

On Friday, 6 August 2021 at 01:12:45 UTC-4, zelos...@gmail.com wrote:
> tisdag 20 juli 2021 kl. 19:56:53 UTC+2 skrev Eram semper recta:
> > Telling students that the derivative definition they are given at university is fraudulent, is the TRUTH. The evidence is ALL here:
> >
> > https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
> >
> > https://drive.google.com/file/d/1520NjhgiakcrssQxtbxRCDXus_aHXpI9
> >
> > https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
> >
> > https://drive.google.com/file/d/1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y
> It isn't because in standard calculus that is the definition.

It is and the nice thing is that students can check it without any problems whatsoever. In other words, your hand waving days of bullshit, are over Malum.

fredag 6 augusti 2021 kl. 15:15:42 UTC+2 skrev Eram semper recta:
> On Friday, 6 August 2021 at 01:12:45 UTC-4, zelos...@gmail.com wrote:
> > tisdag 20 juli 2021 kl. 19:56:53 UTC+2 skrev Eram semper recta:
> > > Telling students that the derivative definition they are given at university is fraudulent, is the TRUTH. The evidence is ALL here:
> > >
> > > https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
> > >
> > > https://drive.google.com/file/d/1520NjhgiakcrssQxtbxRCDXus_aHXpI9
> > >
> > > https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
> > >
> > > https://drive.google.com/file/d/1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y
> > It isn't because in standard calculus that is the definition.
> It is and the nice thing is that students can check it without any problems whatsoever. In other words, your hand waving days of bullshit, are over Malum.

None is handwaving, you are just too stupid to understand

On Tuesday, 20 July 2021 at 13:56:53 UTC-4, Eram semper recta wrote:
> Telling students that the derivative definition they are given at university is fraudulent, is the TRUTH. The evidence is ALL here:
>
> https://drive.google.com/file/d/1peL7lzXsmZ4AVsgAwDLvn8QIZbWy8Ebm
>
> https://drive.google.com/file/d/1520NjhgiakcrssQxtbxRCDXus_aHXpI9
>
> https://drive.google.com/file/d/1RDulODvgncItTe7qNI1d8KTN5bl0aTXj
>
> https://drive.google.com/file/d/1uIBgJ1ObroIbkt0V2YFQEpPdd8l-xK6y

There is no longer any question that the mainstream formulation of calculus is a fraud and that it's just a matter of time before students will help their "math educators" come to their senses, or so it is hoped.

What is the truth value of 3 =< 4, bird brain?

LMAO!