First constructive proof of the REAL Mean Value Theorem in human history.

[John Gabriel]

https://youtu.be/tlp590sHgw8

The most important aspect of the mean value theorem is that it is an arithmetic mean of the ordinates of a function f, which is stated in terms of the difference of ordinates of F, where F is the primitive of f. It is completely unremarkable that a c exists in a given interval such that f'(c) = [ f(b) - f(a) ] / (b-a).

Proof using mainstream calculus and a patch (positional derivative) I created to make it possible:

https://drive.google.com/open?id=0B-mOEooW03iLZG1pNlVIX2RTR0E

Proof using the rigorous New Calculus:

https://drive.google.com/open?id=0B-mOEooW03iLblJNLWJUeGxqV0E

You won't find either of these proofs published anywhere else because if they were, this would amount to plagiarism and copyright infringement.

[Ross A. Finlayson]

But, that uses the derivative to try to
establish the MVT, but, the derivative
requires the MVT (or "IVT"), as via the
Fundamental Theorem(s) of Calculus.

It seems somewhat hypocritical,
to each of: use, swap, and ignore,
but it's not clear whether that is
purposeful on your part or consequence
the tools at the disposal.

Then, one might remark on some utility
of fleshing out useful formalisms for
the ordinates or otherwise establishing
sweep as it were over the range between
distinct, discrete bounds of a system,
but, it's rather in the realm of the
applied then so much at all "primary".

Also, the Intermediate Value Theorem is
built from the sweep principle as in the
establishment of a continuous domain, so,
it's not just hypocritical, but misplaced,
to do it over within itself and marvel
at the consistency.

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

i.v.t, yeah;
j.g is like a character out of Marvel comicS,
if not a central character, WanderaboutboY

[John Gabriel]

On Thursday, 15 December 2016 17:45:09 UTC-8, Ross A. Finlayson wrote:
> On Thursday, December 15, 2016 at 5:21:36 PM UTC-8, John Gabriel wrote:
> > https://youtu.be/tlp590sHgw8
> >
> > The most important aspect of the mean value theorem is that it is an arithmetic mean of the ordinates of a function f, which is stated in terms of the difference of ordinates of F, where F is the primitive of f. It is completely unremarkable that a c exists in a given interval such that f'(c) = [ f(b) - f(a) ] / (b-a).
> >
> > Proof using mainstream calculus and a patch (positional derivative) I created to make it possible:
> >
> > https://drive.google.com/open?id=0B-mOEooW03iLZG1pNlVIX2RTR0E
> >
> > Proof using the rigorous New Calculus:
> >
> > https://drive.google.com/open?id=0B-mOEooW03iLblJNLWJUeGxqV0E
> >
> > You won't find either of these proofs published anywhere else because if they were, this would amount to plagiarism and copyright infringement.
>
> But, that uses the derivative to try to
> establish the MVT, but, the derivative
> requires the MVT (or "IVT"), as via the
> Fundamental Theorem(s) of Calculus.

The new calculus derivative does NOT require the MVT you moron.
And the MVT is not the IVT you idiot.

The IVT states that a function can take any value between f(a) and f(b). How is that at all similar to the MVT you imbecile?

The MVT states that a function has one value between f(a) and f(b) that is the arithemtic mean.

Get a clue idiot.

>
> It seems somewhat hypocritical,
> to each of: use, swap, and ignore,
> but it's not clear whether that is
> purposeful on your part or consequence
> the tools at the disposal.
>
> Then, one might remark on some utility
> of fleshing out useful formalisms for
> the ordinates or otherwise establishing
> sweep as it were over the range between
> distinct, discrete bounds of a system,
> but, it's rather in the realm of the
> applied then so much at all "primary".
>
> Also, the Intermediate Value Theorem is

The IMV is completely inconsequential, but how could an idiot like you know, eh?

[burs...@gmail.com]

Better watch MTV than JGs videos.

[Dan Christensen]

On Friday, December 16, 2016 at 9:14:56 AM UTC-5, John Gabriel wrote:

>
> The IMV is completely inconsequential...

Another result that evades you, eh, Troll Boy? (Like the derivative at points of inflection, even 2+2=4? HA, HA!!) Without irrational numbers, of course, you can't get the Intermediate Value Theorem (IMV) to work. The function y=x^2, for example, never attains the value 2 in your goofy system.

Not very useful in applications, I'm afraid. You can forget about finding roots of polynomials, solving equations and the like. But you were never much concerned with building anything that actually works, were you, Troll Boy?


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

[John Gabriel]

On Friday, 16 December 2016 07:40:56 UTC-8, Dan Christensen wrote:
> On Friday, December 16, 2016 at 9:14:56 AM UTC-5, John Gabriel wrote:
>
> >
> > The IMV is completely inconsequential...
>
> Another result that evades you, eh, Troll Boy? (Like the derivative at points of inflection, even 2+2=4? HA, HA!!) Without irrational numbers, of course, you can't get the Intermediate Value Theorem (IMV) to work. The function y=x^2, for example, never attains the value 2 in your goofy system.
>
> Not very useful in applications, I'm afraid. You can forget about finding roots of polynomials, solving equations and the like. But you were never much concerned with building anything that actually works, were you, Troll Boy?
>

Shut up troll. Anyone reading and studying my work knows you are a cowardly libeler and a liar. Shut up moron. Nothing uyou write is true and everyone knows it. That's why they ignore you now. Only your ilk respond to your shit. You are toast.

[burs...@gmail.com]

If you want toast you get it

in the cheese cake factory

together with tea

[Dan Christensen]

On Friday, December 16, 2016 at 12:02:01 PM UTC-5, John Gabriel wrote:
> On Friday, 16 December 2016 07:40:56 UTC-8, Dan Christensen wrote:
> > On Friday, December 16, 2016 at 9:14:56 AM UTC-5, John Gabriel wrote:
> >
> > >
> > > The IMV is completely inconsequential...
> >
> > Another result that evades you, eh, Troll Boy? (Like the derivative at points of inflection, even 2+2=4? HA, HA!!) Without irrational numbers, of course, you can't get the Intermediate Value Theorem (IMV) to work. The function y=x^2, for example, never attains the value 2 in your goofy system.
> >
> > Not very useful in applications, I'm afraid. You can forget about finding roots of polynomials, solving equations and the like. But you were never much concerned with building anything that actually works, were you, Troll Boy?
> >
>
> Shut up troll. Anyone reading and studying my work knows you are a cowardly libeler and a liar.

Only in your wet dreams, Troll Boy.

Perhaps you can answer this question: What are points of intersection in the x-y plane of the curves given by y=2 and y=x^2. Nowhere, you say??? What a useless system!

[John Gabriel]

12/16/16

[burs...@gmail.com]

Am Freitag, 16. Dezember 2016 20:13:15 UTC+1 schrieb John Gabriel:
> 12/16/16

The size of your bird brain?

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

yes, a sixth is 0.17;
he has at least a sixth of a b.b

[John Gabriel]

On Thursday, 15 December 2016 17:21:36 UTC-8, John Gabriel wrote:

12/16/16 2:01

[burs...@gmail.com]

Am Freitag, 16. Dezember 2016 23:00:59 UTC+1 schrieb John Gabriel:
> 12/16/16 2:01

The number of times you did math?

[Ross A. Finlayson]

You see, he's link-farming all the
search results of all the copies of
all the archives of sci.math.

(That's one of his measures as
a google-bot, that he games.)

Don't feed the troll.

[John Gabriel]

On Thursday, 15 December 2016 17:21:36 UTC-8, John Gabriel wrote:

12/16/16 4:02pm

[last...@gmail.com]

Proving it by a bisection method is also constructive.

[John Gabriel]

On Friday, 16 December 2016 19:08:20 UTC-8, last...@gmail.com wrote:
> Proving it by a bisection method is also constructive.

Wrong. Shut up moron. Shut up. You now have at least aliases:

Jan (?)
Jan Burse
Harry Stoteles
last...@gmail.com
in..@

Anyone seeing these should mark them as abusive.

[burs...@gmail.com]

Lost your marbles again, completly paranoid?

Why not add more:
Kanye West
Jaden Smith
...

What are you?
https://www.youtube.com/watch?v=PV3_UHG73oQ

[John Gabriel]

On Thursday, 15 December 2016 17:21:36 UTC-8, John Gabriel wrote:

12/18/16

[john-c%...@gtempaccount.com]

Your definition of derivative is worthless because you do not
specify m and n.

If it is supposed to be true for all m and n, then it is even more
worthless because it gives different answers for different values of
m and n. Try it with f(x) = x^2, m = 5, n = 13. What do you get
for f'(1) ? Try it again with m = .07, n = -.3. What do you get for
f'(1) ?

So you have left something out.

J.

[John Gabriel]

On Sunday, 18 December 2016 16:07:05 UTC-8, john-c%...@gtempaccount.com wrote:
> On Thursday, December 15, 2016 at 7:21:36 PM UTC-6, John Gabriel wrote:
> > https://youtu.be/tlp590sHgw8
> >
> > The most important aspect of the mean value theorem is that it is an arithmetic mean of the ordinates of a function f, which is stated in terms of the difference of ordinates of F, where F is the primitive of f. It is completely unremarkable that a c exists in a given interval such that f'(c) = [ f(b) - f(a) ] / (b-a).
> >
> > Proof using mainstream calculus and a patch (positional derivative) I created to make it possible:
> >
> > https://drive.google.com/open?id=0B-mOEooW03iLZG1pNlVIX2RTR0E
> >
> > Proof using the rigorous New Calculus:
> >
> > https://drive.google.com/open?id=0B-mOEooW03iLblJNLWJUeGxqV0E
> >
> > You won't find either of these proofs published anywhere else because if they were, this would amount to plagiarism and copyright infringement.
>
> Your definition of derivative is worthless because you do not
> specify m and n.

Of course I do! m and n are related by the auxiliary equation for that function. The auxiliary equation is not possible in your bogus calculus. Chuckle.

>
> If it is supposed to be true for all m and n, then it is even more

It is true for any pair (m,n) where m and n are ***RELATED*** through the auxiliary equation.

>
> worthless because it gives different answers for different values of
> m and n. Try it with f(x) = x^2, m = 5, n = 13. What do you get
> for f'(1) ? Try it again with m = .07, n = -.3. What do you get for
> f'(1) ?

No, no.

The auxiliary equation for f(x) = x^2 is given by n-m=0 because f'(x)=2x+n-m.

So n=m. You can't just choose any m and n. You must choose either and then find the other by use of the auxiliary equation. So if you choose m=5, then n must also be equal to 5 because n=m. Get it? Now, the auxiliary equation is not the same for all functions. It is in most cases different.

The following article shows the auxiliary equation for f(x)=x^3 :

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

>
> So you have left something out.

No. You didn't study it properly!

>
> J.

[John Gabriel]

There are many videos on my channel that explain the auxiliary equation and the 8th grade derivation which is written for moron professors of mythmatics and teachers of mythmatics.

[john-c%...@gtempaccount.com]

On Monday, December 19, 2016 at 8:52:29 AM UTC-6, John Gabriel wrote:
> On Sunday, 18 December 2016 16:07:05 UTC-8, john-c%...@gtempaccount.com wrote:
> > On Thursday, December 15, 2016 at 7:21:36 PM UTC-6, John Gabriel wrote:
> > > https://youtu.be/tlp590sHgw8
> > >
> > > The most important aspect of the mean value theorem is that it is an arithmetic mean of the ordinates of a function f, which is stated in terms of the difference of ordinates of F, where F is the primitive of f. It is completely unremarkable that a c exists in a given interval such that f'(c) = [ f(b) - f(a) ] / (b-a).
> > >
> > > Proof using mainstream calculus and a patch (positional derivative) I created to make it possible:
> > >
> > > https://drive.google.com/open?id=0B-mOEooW03iLZG1pNlVIX2RTR0E
> > >
> > > Proof using the rigorous New Calculus:
> > >
> > > https://drive.google.com/open?id=0B-mOEooW03iLblJNLWJUeGxqV0E
> > >
> > > You won't find either of these proofs published anywhere else because if they were, this would amount to plagiarism and copyright infringement.
> >
> > Your definition of derivative is worthless because you do not
> > specify m and n.
>
> Of course I do! m and n are related by the auxiliary equation for that function. The auxiliary equation is not possible in your bogus calculus. Chuckle.
>
> >
> > If it is supposed to be true for all m and n, then it is even more
>
> It is true for any pair (m,n) where m and n are ***RELATED*** through the auxiliary equation.
>
> >
> > worthless because it gives different answers for different values of
> > m and n. Try it with f(x) = x^2, m = 5, n = 13. What do you get
> > for f'(1) ? Try it again with m = .07, n = -.3. What do you get for
> > f'(1) ?
>
> No, no.
>
> The auxiliary equation for f(x) = x^2 is given by n-m=0 because f'(x)=2x+n-m.
>

Your 'definition' of derivative says

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

That appears to be a self-contained definition. But clearly it's not.
If you need an 'auxiliary' equation, why didn't you include it? Why do
we have to go looking for it elsewhere?

John C.

[John Gabriel]

It is a complete definition if that's what you mean by "self-contained" definition.

> If you need an 'auxiliary' equation, why didn't you include it?

You don't need it unless you try to calculate (m,n) pairs.

> Why do we have to go looking for it elsewhere?

There is no elsewhere.

We can prove that if f(x) is a function with tangent line equation t(x)=kx+b and a parallel secant line equation
s(x)=[{f(x+n)-f(x-m)}/(m+n)] x + p, then f'(x)={f(x+n)-f(x-m)}/(m+n).

Proof:

Let t(x)=kx+b be the equation of the tangent line to the function f(x).

Then a parallel secant line is given by s(x)=[{f(x+n)-f(x-m)}/(m+n)] x + p.

So, k={f(x+n)-f(x-m)}/(m+n) because the secant lines are all parallel to the tangent line.

But the required derivative f'(x) of f(x), is given by the slope of the tangent line t(x).

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

Q.E.D.

Also, m+n is a factor of the expression f(x+n)-f(x-m).

Proof:

From k(m+n)=f(x+n)-f(x-m), it follows that m+n is a factor of the LHS. But since m+n is a factor of the left hand side, it follows that m+n must also be a factor of the RHS exactly. Hence, m+n is a factor of the expression f(x+n)-f(x-m). This means that if we divide f(x+n)-f(x-m) by m+n, the expression so obtained must be equal to k. This is only possible if the sum of all the terms in m and n are 0.

Q.E.D.

The previous two proofs hold for any function f.

[Ross A. Finlayson]

The "if f is a function...",
that's only for a line.

y = mx + b

That's a typical howler,
plugging in variables that
are only cancelled and
calling that the action.

https://en.wikipedia.org/wiki/Howler_(error)#Mathematics_as_a_special_case_of_terminology

Yes, the tangent is a line
incident to the smooth curve,
that's it's definition.

It's enough that Newton-Cotes simplifies
some function's analysis in numerical
resource terms, and that there are
even refinements for less functions,
of various forms (that not even all
classical functions have).

https://en.wikipedia.org/wiki/Newton%E2%80%93Cotes_formulas

[John Gabriel]

Incoherent babble again?

[John Gabriel]

Are you trying to make some point in your babble? Clearly it's a lot of irrelevant crap as usual.

[john-c%...@gtempaccount.com]

OK, your method works for positive integer powers of x, i.e., f(x) = x^n.

Please show how it works for either of the following:

f(x) = x^(.5) or

f(x) = ln(x).

Note that your derivation should not make use of what we
already know -

John C.

[John Gabriel]

It works for any continuous and smooth function. I did not need your confirmation that it works. I know it does.

>
> Please show how it works for either of the following:
>
> f(x) = x^(.5) or

f(x) = x^(1/2)
f'(x) = { (x+n)^(1/2) - (x-m)^(1/2) } / (m+n)
f'(x) = { (x^(1/2)+0.5 x^(-1/2)n + ... + n^(1/2)
(x^(1/2)-0.5 x^(-1/2)m + ... + m^(1/2) }/ (m+n)

f'(x) = { 0.5x^(-1/2)*(m+n) + Q(x,m,n) } / (m+n)
f'(x) = 1/(2sqrt(x)) + Q(x,m,n) / (m+n)

Since Q(x,m,n) equals to 0,

we have
f'(x) = 1/(2sqrt(x))

>
> f(x) = ln(x).
>
> Note that your derivation should not make use of what we
> already know -

Rubbish. My derivation can make use of any mathematics that is valid. What it does not make use of is nonsense such as limits, infinity and infinitesimals.

This example as well as all others have been covered in multiple examples.
>
> John C.

[burs...@gmail.com]

You must be joking, since when is the series for

sqrt(x+n) finite? It is, but shouldn't exist in your world:

sqrt(x+n) = sqrt(n) + x/(2 sqrt(n))
- x^2/(8 n^(3/2))
+ x^3/(16 n^(5/2))
- (5 x^4)/(128 n^(7/2)) +- etc...

Is this again cow deases Gabrieloconfusion and Johnostupidics?

[burs...@gmail.com]

JG wrote:
> What it does not make use of is nonsense
> such as limits, infinity and infinitesimals

Then explain your sqrt(x+n). Anyway we see clearly
the nature of every crank. He is a cheater.

[burs...@gmail.com]

Oops wrong series, switch the role of x and
n, to make it work for your derive.

Which also confirms, that besides the use
infinite series, youre method doesn't work,

since it doesn't include a method to choose
the right series in the first place,

BTW using n=1/25 and x=49/25 you get a
series for sqrt(2):

sqrt(x+n) = sqrt(x) + n/(2 sqrt(x))
- n^2/(8 x^(3/2))
+ n^3/(16 x^(5/2))
- (5 n^4)/(128 x^(7/2)) +- etc...

= sqrt(x) * (1 + 1/2 * n/x
- 1/8 * (n/x)^2
+ 1/16 * (n/x)^3
- 5/128 * (n/x)^4 +- etc...)

Then we have:

sqrt(2) = 7/5 * (1 + 1/2 * 1/49
- 1/8 * 1/49^2
+ 1/16 * 1/49^3
- 5/128 * 1/49^4 +- etc...)

Here are some partial sums in decimal
rounded to a fixed amount of digits:

1.4000000000
1.4142857143
1.4142128280
1.4142135717
1.4142135622

And the limes rounded:

1.4142135623

[John Gabriel]

On Thursday, 15 December 2016 17:21:36 UTC-8, John Gabriel wrote:
> https://youtu.be/tlp590sHgw8
>
> The most important aspect of the mean value theorem is that it is an arithmetic mean of the ordinates of a function f, which is stated in terms of the difference of ordinates of F, where F is the primitive of f. It is completely unremarkable that a c exists in a given interval such that f'(c) = [ f(b) - f(a) ] / (b-a).
>
> Proof using mainstream calculus and a patch (positional derivative) I created to make it possible:
>
> https://drive.google.com/open?id=0B-mOEooW03iLZG1pNlVIX2RTR0E
>
> Proof using the rigorous New Calculus:
>
> https://drive.google.com/open?id=0B-mOEooW03iLblJNLWJUeGxqV0E
>
> You won't find either of these proofs published anywhere else because if they were, this would amount to plagiarism and copyright infringement.

You can't rubbish the new calculus. It is the first and only rigorous formulation in human history. If more intelligent aliens were to visit earth and they learned of the bogus calculus, they would no doubt be surprised at the stupidity of humans.

[burs...@gmail.com]

By bogus you mean your hocus pocus?

[john-c%...@gtempaccount.com]

OK, right there ... how do you end up with n^(1/2) and m^(1/2) ... with
no hint of what goes in between? Of course the expansion of, for example,
(x + n)^(1/2) depends on Newton's binomial series ... which, in case you
don't know, does NOT end in n^(1/2). There is a big gap here. Which
means in fact your 'auxiliary equation' is not defined. Please fix.

John C.

[John Gabriel]

How does Newton end up with n^(1/2) and m^(1/2) ... with no hint of what goes in between?

In fact, I don't even care what happens at the end of that series because I have proved in the second part that m+n is a factor of every term in the numerator f(x+n)-f(x-m).

Of course the expansion of, for example,
> (x + n)^(1/2) depends on Newton's binomial series ... which, in case you
> don't know, does NOT end in n^(1/2).

It doesn't matter.

> There is a big gap here. Which means in fact your 'auxiliary equation' is not defined.

Wrong. The auxiliary equation is very well defined: Q(x,m,n)=0.

> Please fix.

Nothing to fix. Look John, if you want me to help you, watch your tone! Ask a question rather than shoot off your mouth in ignorance.

[John Gabriel]

From k(m+n)=f(x+n)-f(x-m), it follows that m+n divides the LHS exactly. But since m+n divides the left hand side exactly, it follows that m+n must also divide the RHS exactly. Hence, m+n is a factor of the expression f(x+n)-f(x-m). This means that if we divide f(x+n)-f(x-m) by m+n, the expression so obtained must be equal to k. This is only possible if the sum of all the terms in m and n are 0.

[burs...@gmail.com]

Am Freitag, 30. Dezember 2016 01:53:53 UTC+1 schrieb John Gabriel:
> From k(m+n)=f(x+n)-f(x-m), it follows that m+n divides the LHS exactly.
> But since m+n divides the left hand side exactly, it follows that
> m+n must also divide the RHS exactly.

I don't think your method works for this f(x) at x=0,
since there isn't a taylor expansion, so you cannot
do your polynom division as you usually do:

f(x) = e^(−1/x^2)

[John Gabriel]

Shut up moron.

[Python]

Chickening out, Mr Gabriel? Again?

[John Gabriel]

Shut up moron.

[Python]

So, yes, you're chickening out. Thanks for confirmation, Mr Gabriel.

[John Gabriel]

12/30/16

[john-c%...@gtempaccount.com]

This is a circular argument. First you say f'(x) can be written as

f'(x) = (f(x + n) - f(x - m))/(m + n),

where m and n are to be determined. Then you substitute k for f'(x).
The argument then is, k * (m + n) = f(x + n) - f(x - m). Then you say
since (m + n) divides the left side, it must also divide the right side.
Then you let the 'auxiliary' Q(x; n, m) equal the quotient. You set
Q(x; n, m) = 0, choose an arbitrary value for n or m, and then solve
for the other one. This works when f(x) = x^p where p is a positive integer,
because you can compute Q(x; m, n) explicitly, and then solve for m
in terms of n. But it doesn't work here because you have not derived
an explicit expression for Q(x; n, m). The expression for (x + n)^(1/2)
is an infinite series; this means that Q(x; n, m) is the difference of
two infinite series, and setting it equal to zero does not result in a
simple solution for m in terms of n.

John C.

[John Gabriel]

Nothing circular about it.

> First you say f'(x) can be written as
>
> f'(x) = (f(x + n) - f(x - m))/(m + n),
>
> where m and n are to be determined. Then you substitute k for f'(x).

Right. m and n can be determined but they don't have to be. No. I don't substitute k for f'(x). If k is the slope of the tangent line, then it will also be the slope of any parallel secant line. There is nothing special about f'(x) - it is only a notation. Newton wrote x with a dot on top of it. Other academics denoted it with different symbols.

> The argument then is, k * (m + n) = f(x + n) - f(x - m). Then you say
> since (m + n) divides the left side, it must also divide the right side.
> Then you let the 'auxiliary' Q(x; n, m) equal the quotient.

No. I do not. Q(x,m,n) is only part of the quotient and it is equal to 0.

> You set Q(x; n, m) = 0,

Wrong. I do not set it equal to 0. I proved in the second part that it is 0.

> choose an arbitrary value for n or m, and then solve for the other one. This works when f(x) = x^p where p is a positive integer,
> because you can compute Q(x; m, n) explicitly, and then solve for m
> in terms of n. But it doesn't work here because you have not derived
> an explicit expression for Q(x; n, m).

It works for ANY continuous and smooth function. Of course you cannot have a closed form for x^(1/2) unless you are using my Gabriel polynomial. But it does not matter!!!!! Whatever the expression in Q(x,m,n), I KNOW it must be ZERO. I have proved it.

> The expression for (x + n)^(1/2) is an infinite series;

Rubbish. It is not an infinite series, because there is no such thing. It's just a partial sum with an ellipsis following it. There is no such thing as infinity.

> this means that Q(x; n, m) is the difference of two infinite series,

No. It does not mean that! Q(x,m,n) is just the sum of all the terms in m and n.

> and setting it equal to zero does not result in a simple solution for m in terms of n.

True. In this case it doesn't. But that's not a problem because one does not need to find m and n. You can however use the "infinite" series to iterate for m or n or determine an approximation formula.

[John Gabriel]

The following is a jpeg showing the Gabriel Polynomial(*). You will notice that it does not matter what is the function f.

https://drive.google.com/open?id=0B-mOEooW03iLNHd4VXRrNGlvSG8


(*) The Gabriel polynomial is like a closed form Taylor expansion. You can't do this with the Taylor series because the finite differences used in the Taylor derivation are not accurate. Those used in the Gabriel polynomial are 100% accurate. You would have to study the Gabriel Polynomial first:

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

However, you don't need to know this because it is easily proved (as I have done) that Q(x,m,n) = 0.

Now if you don't think (x+n)^1/2 is valid when expanded by a binomial, that is another story. In truth, it is never exact, only an approximation. But then you are not arguing with me, but with Newton.

[john-c%...@gtempaccount.com]

But if you don't determine n and m, i.e., you merely claim that they
exist, then you do not have an effective way to compute the derivative.

Moreover, consider the function

f(x) = 1 - x^2 for x between 0 and 1,
= -(x - 1)^2 for x > 1.

This is continuous and differentiable for all x >= 0.

However, at x = 1, there are no n and m such that

f'(x) = (f(x + n) - f(x - m)) / (m + n).

So why does your definition fail for this example, whereas the
usual definition in terms of a limit does just fine?

John C.

[John Gabriel]

I never merely claim they exist. It is by virtue of sound geometry that they exist. Look, m and n are horizontal distances from the point of tangency to the endpoints of a *parallel* secant line. There is NOTHING to prove. They exist as long as there is a *parallel* secant line.

>
> Moreover, consider the function
>
> f(x) = 1 - x^2 for x between 0 and 1,
> = -(x - 1)^2 for x > 1.

Nope. That is not one function, but more than one function. Calculus only works if the function is both continuous and smooth. Your faux function is not smooth at x=1.

>
> This is continuous and differentiable for all x >= 0.

NO. It's not smooth at x=1, therefore not differentiable there.

>
> However, at x = 1, there are no n and m such that
>
> f'(x) = (f(x + n) - f(x - m)) / (m + n).
>
> So why does your definition fail for this example, whereas the
> usual definition in terms of a limit does just fine?

It doesn't fail. You have failed. See previous responses.

>
> John C.

[John Gabriel]

Also, your function is not differentiable at x=1 using mainstream calculus. What limit are you talking about? It's continuous but not differentiable.

[john-c%...@gtempaccount.com]

Simply wrong. This is a well-defined function and it has a derivative
at x = 1. Unless, perhaps, you have your own definition of function. Would
you care to state what that is?

John C.

[John Gabriel]

Excuse me??? There is NO derivative at x=1 when

f(x) = 1 - x^2 for x between 0 and 1,
= -(x - 1)^2 for x > 1.

Chuckle.

[john-c%...@gtempaccount.com]

John,

I'll make this even simpler. Define f(x) as

f(x) = x^3 - 2x^2 + x + 1.

This is clearly continuous and differentiable everywhere. The derivative
at x = 1 is zero. But there are no values n and m such that for x = 1,

f'(x) = 0 = (f(x + n) - f(x - m))/(m + n).

Hint: try graphing it.

John C.

[John Gabriel]

Only in the bogus mainstream calculus. In the new calculus, there is no derivative at a point of inflection.

> The derivative
> at x = 1 is zero.

Rubbish. There is no tangent line there. It is a point of inflection and tangent lines never cross a curve in the New Calculus, only in your bogus calculus.

> But there are no values n and m such that for x = 1,
>
> f'(x) = 0 = (f(x + n) - f(x - m))/(m + n).

That is not true. There is the common pair (m,n)=(0,0) which will yield the derivative in this case and EVERY other case. But the NC is very rigorous and will not allow a derivative at any point where a tangent line cannot be constructed.

You are an amateur. You need to study a lot more.

[john-c%...@gtempaccount.com]

The derivative of f(x) = x^3 - 2x^2 + x + 1 at x = 1 is zero, period.

You claim to have a definition of derivative. It should work in this
case. It does not.

Further, letting m = n = 0 does not help; in that case, for ANY
function, you get

f'(x) = (f(x + 0) - f(x -0)) / (0 + 0 ) = 0/0,

which is indeterminate.

John C.

[John Gabriel]

What? Are you trying to do math by ultimatum? Chuckle. No. There is no period. Moreover, there is NO derivative at x=1 because the derivative is the slope of a tangent line AND wait for it ... there is no tangent line at x=1, only half tangent lines.

>
> You claim to have a definition of derivative. It should work in this
> case. It does not.

It works very well in every situation. Of course if there is no derivative at a point, then it must not produce a derivative - as it should and so it does!

>
> Further, letting m = n = 0 does not help; in that case, for ANY
> function, you get
>
> f'(x) = (f(x + 0) - f(x -0)) / (0 + 0 ) = 0/0,

Rubbish. Neither m nor n is ever equal to 0 except at the point of tangency (the tangent line owns this pair), but f'(x) = f(x+n)-f(x-m) / (m+n) is not defined for the tangent line, only for parallel secant lines. So you want to force it to define the tangent line slope? Chuckle.

In the NC, nothing is forced. It is based on sound logic - something you are not accustomed to.

[konyberg]

It is the tangent to a point at a curve. Not the tangent to the curve at a point??

KON

[John Gabriel]

A derivative is the SLOPE of a tangent line to a given curve at a given point.

[John Gabriel]

Don't be misled by the local pus bag Dan Christensen. There is no bigger imbecile than he. In the NC, one can also find a derivative at x=0 for x^3, but the NC does not settle for vague definitions. It is more strict and will not allow for a derivative when there is no tangent line. The bogus calculus can have tangent lines and no derivative. This is not possible in the New Calculus.

[konyberg]

This no tangent of a curve (or half tangent) is your construction. Tangent to a point of a curve is well defined. Whether the tangent cut the curve at some other point don't matter.

Can you give me one book where the tangent (and tehrefor the derivative) is undefined because the tangent line cuts the function?

KON

[John Gabriel]

The original definition of tangent line:

https://www.merriam-webster.com/dictionary/tangent

And just think, if Archimedes and Newton thought of it in the same erratic way as you and the BIG STUPID, there would not have been any calculus at all!

[John Gabriel]

On Friday, 30 December 2016 09:25:59 UTC-8, konyberg wrote:

Yes. (x-1)^(1/3) + 1.

I have referred to it many times but you do not pay attention!

>
> KON

[john-c%...@gtempaccount.com]

It was you, not I who (above) said:

"There is the common pair (m,n)=(0,0) which will yield the derivative
in this case and EVERY other case."

I interpret your response to mean: "Yes, I said that

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

for some n and m for any smooth, continuous function f(x)", but now
you say there are exceptions. Not to mention the fact that, for

f(x) = x^(1/2),

you have not been able to specify what n and m should be - so what
good is your definition, if (unlike the usual limit definition), you
cannot use it to do calculations? Another example:

f(x) = sqrt(2x- x^2) for 0 <= x <= 1, and

1 for x > 1.

This is a well-defined function. It is smooth. Graph it. The tangent
line at x = 1 is horizontal. x = 1 is not an inflection point. However,
there are no positive values n and m such that for x = 1,

f'(x) = (f(x + n) - f(x - m))/(m + n).

John C.

[konyberg]

It is like saying that he derivative of tan(x) is undefined at infinity.

However, it just means that the derivative going to infinity; the vertical. And that is defined!

KON

[konyberg]

At 90 deg

[John Gabriel]

There are no exceptions.

> Not to mention the fact that, for
>
> f(x) = x^(1/2),
>
> you have not been able to specify what n and m should be

You are lying. One can find approximations.

> - so what good is your definition, if (unlike the usual limit definition), you cannot use it to do calculations?

You cannot do anything with the limit definition that you can't do with the NC. The limit definition is a bunch of crap. And of course you can use it in any calculation. You just can't do stupid things with it like you do in the bogus calculus.

> Another example:

You can't even handle the simple examples. I suggest you first try to understand the simple examples before further complicating matters for yourself.

>
> f(x) = sqrt(2x- x^2) for 0 <= x <= 1, and
>
> 1 for x > 1.
>
> This is a well-defined function. It is smooth. Graph it. The tangent
> line at x = 1 is horizontal. x = 1 is not an inflection point. However,
> there are no positive values n and m such that for x = 1,

LIAR!!!!! Grrrr. Do you enjoy being an idiot? Of course you can find m and n but they are approximations. Tell me stupid, do you actually think about what you write? It's very clear to see that you can find m and n. If you are asking me for a formula where you can simply find either without iteration, then you are just being a moron.

[John Gabriel]

Kony, watch this video:

https://youtu.be/cSKHOxbDxYQ

[john-c%...@gtempaccount.com]

So if it's so clear, why don't you tell us what they are?

> If you are asking me for a formula where you can simply find either without iteration, then you are just being a moron.

The standard definition of derivative gives a clear, simple formula. Your
definition does not. Further, in some cases it doesn't give any answer at all,
even theoretically, for a simple function like f(x) = x^3, when x = 0.

Moreover: the Merriam-Webster definition of tangent has two parts: one
for the adjective 'tangent' and one for the noun 'tangent'. In both cases
they state it in terms of a LIMIT. For the noun, the definition is
the LIMIT of secants, which leads immediately to ... the standard definition
of derivative. Which unlike your method, (1) is explicitly computable, and (2) does not have special exceptions like inflection points. In what way is
your definition superior?

John C.


>
>
> >

[burs...@gmail.com]

john-c wrote:
> f(x) = sqrt(2x- x^2) for 0 <= x <= 1, and
> 1 for x > 1.

> there are no positive values n and m such that for x = 1,

JG wrote:
> Of course you can find m and n but they are approximations.

But you dont say so in your new calculus. You
write k*(m+n) = f(x+n)-f(x-m), which is impossible
for the above f at x=1.

Go home JG, you are a sick bird brain John Gabriel
birdbrain spammer of nonsense videos.

[John Gabriel]

In fact, any (m,n) pair of the form (k,k) with k < radius will work.

Do you enjoy being obtuse John?

[John Gabriel]

Moron. Any (m,n) pair of the form (k,k) with k < radius will work. You are a fool. If you continue to be an idiot, I am going to plonk you. I have ZERO tolerance for stupid people.

[burs...@gmail.com]

Am Freitag, 30. Dezember 2016 20:02:48 UTC+1 schrieb John Gabriel:
> In fact, any (m,n) pair of the form (k,k) with k < radius will work.

Not for this (piecewise) function:

john-c wrote:
> f(x) = sqrt(2x- x^2) for 0 <= x <= 1, and
> 1 for x > 1.

> there are no positive values n and m such that for x = 1,

bird brain John Gabriel birdbrain go home,
you are a sick nonsense videos spammer.

[John Gabriel]

It's a flawed definition for many reasons. There are no real numbers and the theory of limits requires real numbers. Also, the definition is circular and full of errors.

> Your definition does not.

It does. Gee, moron, let's take a look and see which definition is simpler:

Bogus Calculus:

f'(x)= lim {h -> 0} f(x+h)-f(x) / h

New Calculus:

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

Hmm, obvious which is clearer.

> Further, in some cases it doesn't give any answer at all,

FALSE.

> even theoretically, for a simple function like f(x) = x^3, when x = 0.

Of course you cannot find an m and n there you idiot!!! That's because there is NO tangent line there. The NC produces the correct result.

>
> Moreover: the Merriam-Webster definition of tangent has two parts: one
> for the adjective 'tangent' and one for the noun 'tangent'. In both cases
> they state it in terms of a LIMIT. For the noun, the definition is
> the LIMIT of secants, which leads immediately to ... the standard definition
> of derivative.

You also have difficulty reading I see. There is nothing about limit in the Webster definition.

> Which unlike your method, (1) is explicitly computable, and (2) does not have special exceptions like inflection points.

My method IS explicitly computable. Your bogus calculus is NOT. The inflection points are not special exceptions stupid. There is no derivative at inflection points because there is no tangent line at a point of inflection. The NC is consistent. Your bogus calculus is NOT.

x^3 : you have a tangent line and derivative.

(x-1)^(1/3)+1 : you have a tangent line but no derivative.

A derivative is the SLOPE of a tangent line you IMBECILE!!!

Can you see a problem with your thinking? Nah. You're an idiot.

> In what way is your definition superior?

It is rigorous and has no ill-formed concepts.

[burs...@gmail.com]

By IMBECILE you are talking about yourself who
cannot take an example given by john-c and

show how the wonder new calculus works?
BTW: JG they need you in the cheese cake factory.

[John Gabriel]

Not to mention its many features such as the auxiliary equation which is not possible in the Bogus calculus.

[burs...@gmail.com]

Why shouldn't the aux eq not also exist in ordinary
calculus? Its even much easier to define there

since f'(x) is available in the first place. Go
home JG, you are ultra sick bird brain John

Gabriel birdbrain, chief spammer nonsense videos.

[John Gabriel]

On Friday, 30 December 2016 11:45:53 UTC-8, idiot bastard burs...@gmail.com wrote:
> Why shouldn't the aux eq not also exist in ordinary
> calculus? Its even much easier to define there

Then define it you fucking idiot and nuisance.

Shut up because you are an idiot and make a fool of yourself with every comment you write. Stop spamming the forum you dumb bastard!

[john-c%...@gtempaccount.com]

Really? In which case, for f(x) = x^(1/2), you should be able to
tell me explicitly what n and m are. But so far you have not.

>
> > even theoretically, for a simple function like f(x) = x^3, when x = 0.
>
> Of course you cannot find an m and n there you idiot!!! That's because there is NO tangent line there.

Only because you have your own personal special definition of
tangent line.

The NC produces the correct result.
>

Except when it doesn't ...

Demonstrate that it does for say f(x) = x^(1/2) or f(x) = ln(x).
I want to see the explicit details, not some vague references to
approximations. Computable results, like the standard limit
definition of derivative gives. Can you do that?

> >
> > Moreover: the Merriam-Webster definition of tangent has two parts: one
> > for the adjective 'tangent' and one for the noun 'tangent'. In both cases
> > they state it in terms of a LIMIT. For the noun, the definition is
> > the LIMIT of secants, which leads immediately to ... the standard definition
> > of derivative.
>
> You also have difficulty reading I see. There is nothing about limit in the Webster definition.
>

From https://www.merriam-webster.com/dictionary/tangent:

2. [definition of the noun]: a line that is tangent; specifically : a straight line that is the limiting position of a secant of a curve through a fixed point and a variable point on the curve as the variable point approaches the fixed point

See that? LIMITING position of a secant. You are trying to get around
the limiting part. But it doesn't work - cannot work - for some functions
at inflection points. And you cannot give me explicit values for n and m
for functions like f(x) = x^(1/2), even for non-inflection points.

> > Which unlike your method, (1) is explicitly computable, and (2) does not have special exceptions like inflection points.
>
> My method IS explicitly computable.

Show me that for f(x) = ln(x).

> Your bogus calculus is NOT.

On the contrary. I have no trouble at all giving explicit derivatives
for f(x) = x^(1/2) or f(x) = ln(x). No trouble at all at inflection points.
I cannot see any advantages to your definition. Can you give an example
where your definition works, and the standard definition does not?

> The inflection points are not special exceptions stupid. There is no derivative at inflection points because there is no tangent line at a point of inflection.

See definition above straight from Merriam-Webster.

The NC is consistent. Your bogus calculus is NOT.
>

Give an example of where the standard definition fails to be 'consistent'.

> x^3 : you have a tangent line and derivative.
>
> (x-1)^(1/3)+1 : you have a tangent line but no derivative.
>
> A derivative is the SLOPE of a tangent line you IMBECILE!!!
>

Correct.

> Can you see a problem with your thinking? Nah. You're an idiot.
>
> > In what way is your definition superior?
>
> It is rigorous and has no ill-formed concepts.
>

It fails to be computable in many cases. This is a fatal flaw. And
as long as you cannot explicitly compute n and m, it has ill-formed
concepts.

John C.

[John Gabriel]

On Friday, 30 December 2016 12:03:07 UTC-8, john-c%...@gtempaccount.com wrote:
Really? In which case, for f(x) = x^(1/2), you should be able to
tell me explicitly what n and m are. But so far you have not.

I can't help it that you are unable to read, but I told you already, any (m,n) pair of the form (k,k) with k < sqrt(2x-x^2) will work.

> > even theoretically, for a simple function like f(x) = x^3, when x = 0.
>
> Of course you cannot find an m and n there you idiot!!! That's because there is NO tangent line there.

Only because you have your own personal special definition of tangent line.

Nonsense. The definition of tangent line I use is the same as that which Newton and Archimedes used.

The NC produces the correct result.
>

Except when it doesn't ...

It always produces the correct result. Your BOGUS calculus does not produce correct results at inflection points.

Demonstrate that it does for say f(x) = x^(1/2) or f(x) = ln(x).
I want to see the explicit details, not some vague references to
approximations. Computable results, like the standard limit
definition of derivative gives. Can you do that?

Read idiot!!! The references to approximations are not vague. Are you such an imbecile that you are asking me to give you a formula solely in terms of m or n? Please tell me you are not. What are you continuing to babble about? You can in this case only approximate one in terms of the other. However, it is easy to see that any pair (k,k) with k < sqrt(2x-x^2) will work from the geometry.

> >
> > Moreover: the Merriam-Webster definition of tangent has two parts: one
> > for the adjective 'tangent' and one for the noun 'tangent'. In both cases
> > they state it in terms of a LIMIT. For the noun, the definition is
> > the LIMIT of secants, which leads immediately to ... the standard definition
> > of derivative.
>
> You also have difficulty reading I see. There is nothing about limit in the Webster definition.
>

From https://www.merriam-webster.com/dictionary/tangent:

2. [definition of the noun]: a line that is tangent; specifically : a straight line that is the limiting position of a secant of a curve through a fixed point and a variable point on the curve as the variable point approaches the fixed point

BULLSHIT! You chose the wrong one. The correct one is:


Definition of tangent
1
a : meeting a curve or surface in a single point if a sufficiently small interval is considered <straight line tangent to a curve>

<no time for excrement>

You will limit your future responses to one question at a time. I will not tolerate your stupidity any longer. In fact I will probably start ignoring you because you are sounding more and more like the local idiot Dan Christensen.

[John Gabriel]

On Friday, 30 December 2016 12:03:07 UTC-8, john-c%...@gtempaccount.com wrote:

> 2. [definition of the noun]: a line that is tangent; specifically : a straight line that is the limiting position of a secant of a curve through a fixed point and a variable point on the curve as the variable point approaches the fixed point

So tell me idiot, how do you get a secant line on a straight line? Stupid much? Chuckle. Couldn't resist this because the stupid dog christensen always carries on about a straight line having a derivative.

Bwaaaa haaaa haaaa.

[genm...@gmail.com]

A secant line intersects only two points on a curve.

Quick, quick! Run to the Wikipedia Moronica and edit the article!!!

[john-c%...@gtempaccount.com]

On Friday, December 30, 2016 at 2:22:13 PM UTC-6, John Gabriel wrote:
> On Friday, 30 December 2016 12:03:07 UTC-8, john-c%...@gtempaccount.com wrote:
>
> > 2. [definition of the noun]: a line that is tangent; specifically : a straight line that is the limiting position of a secant of a curve through a fixed point and a variable point on the curve as the variable point approaches the fixed point
>
> So tell me idiot, how do you get a secant line on a straight line?

Just like always. Connect two points on the line. From that, compute
the slope. You will get the same answer that you get by taking the standard
definition of derivative.

But you have failed again to answer my questions.

John C.

[john-c%...@gtempaccount.com]

Wikipedia says that a secant line intersects two points on a curve.
Note that it does NOT say "only" two points. That's your addition.

Meanwhile, you continue not to answer my questions:

1. Specify n and m for f(x) = x^(1/2). Or at least show explicitly
what the 'auxiliary function' Q(x; n, m) is.

2. Show how your definition works for f(x) = ln(x).

3. Give an example f(x) where the standard definition fails, but
yours works correctly.

John C.

John C.

John C.

[john-c%...@gtempaccount.com]

On Friday, December 30, 2016 at 2:13:25 PM UTC-6, John Gabriel wrote:
> On Friday, 30 December 2016 12:03:07 UTC-8, john-c%...@gtempaccount.com wrote:
> Really? In which case, for f(x) = x^(1/2), you should be able to
> tell me explicitly what n and m are. But so far you have not.
>
> I can't help it that you are unable to read, but I told you already, any (m,n) pair of the form (k,k) with k < sqrt(2x-x^2) will work.
>
> > > even theoretically, for a simple function like f(x) = x^3, when x = 0.
> >
> > Of course you cannot find an m and n there you idiot!!! That's because there is NO tangent line there.
>
> Only because you have your own personal special definition of tangent line.
>
> Nonsense. The definition of tangent line I use is the same as that which Newton and Archimedes used.
>
> The NC produces the correct result.
> >
>
> Except when it doesn't ...
>
> It always produces the correct result. Your BOGUS calculus does not produce correct results at inflection points.
>
> Demonstrate that it does for say f(x) = x^(1/2) or f(x) = ln(x).
> I want to see the explicit details, not some vague references to
> approximations. Computable results, like the standard limit
> definition of derivative gives. Can you do that?
>
> Read idiot!!! The references to approximations are not vague. Are you such an imbecile that you are asking me to give you a formula solely in terms of m or n? Please tell me you are not.

No ... it should be in terms of x, n, and m. It should be a solution to
your 'auxiliary equation' Q(x; n, m).

> What are you continuing to babble about? You can in this case only approximate one in terms of the other. However, it is easy to see that any pair (k,k) with k < sqrt(2x-x^2) will work from the geometry.

Let's try this with x = .5. Take n = m = k = .1.

f'(x) = (f(x + n) - f(x - m)) / (m + n).

So if x = .5, f(x + .1) = .91652, and f(x - .1) = .8,

and you conclude

f'(.5) = .58258.

Whereas the usual computation gives:

f'(.5) = .70711.

Which one do you think is correct?

And how about answering my other questions:

1. What are n and m for your computation of the derivative of

f(x) = sqrt(x) ?

2. Similar question for f(x) = ln(x). Or, f(x) = cos(x) if you
prefer.

3. Can you give one example - just one - where your definition of
derivative works, but the standard limit definition fails?

I am not Dan Christenson.

I have asked these questions previously, but you have failed to answer
them in every case.

John C.

[Ross A. Finlayson]

It's your own choice to joust the
windmills of troll-bots, but I think
you'll find this one quite recalcitrant
and I think often you'd find more interest
in some trivia or fact as you find true
of mathematics than the innumerati here
with their flash-cards and one-track spite.

[john-c%...@gtempaccount.com]

No doubt good advice, but you're right, it's my own choice.

But you have spent time here also, no?

John G. is a real person, not any kind of bot.

Again you are right, he is recalcitrant.

You want real mathematics? Let N = 200. Find the smallest
subset S of the numbers {0, 1, 2, ..., N} such that every
integer k between 1 and N can be written in the form

k = i + j,

where i and j are in S. Hint: cardinality(S) < 100.

John C.

[John Gabriel]

On Friday, 30 December 2016 12:31:14 UTC-8, john-c%...@gtempaccount.com wrote:
> On Friday, December 30, 2016 at 2:22:13 PM UTC-6, John Gabriel wrote:
> > On Friday, 30 December 2016 12:03:07 UTC-8, john-c%...@gtempaccount.com wrote:
> >
> > > 2. [definition of the noun]: a line that is tangent; specifically : a straight line that is the limiting position of a secant of a curve through a fixed point and a variable point on the curve as the variable point approaches the fixed point
> >
> > So tell me idiot, how do you get a secant line on a straight line?
>
> Just like always. Connect two points on the line.

No.

A secant line, also simply called a secant, is a line passing through two points of a curve.

http://mathworld.wolfram.com/SecantLine.html

It does not pass through all of them.

> From that, compute the slope. You will get the same answer that you get by taking the standard definition of derivative.
>
> But you have failed again to answer my questions.

So far you've not really asked any question that makes sense and if it did, there was no relevance.

[John Gabriel]

On Friday, 30 December 2016 12:49:23 UTC-8, john-c%...@gtempaccount.com wrote:
> On Friday, December 30, 2016 at 2:30:36 PM UTC-6, genm...@gmail.com wrote:
> > On Friday, 30 December 2016 12:22:13 UTC-8, John Gabriel wrote:
> > > On Friday, 30 December 2016 12:03:07 UTC-8, john-c%...@gtempaccount.com wrote:
> > >
> > > > 2. [definition of the noun]: a line that is tangent; specifically : a straight line that is the limiting position of a secant of a curve through a fixed point and a variable point on the curve as the variable point approaches the fixed point
> > >
> > > So tell me idiot, how do you get a secant line on a straight line? Stupid much? Chuckle. Couldn't resist this because the stupid dog christensen always carries on about a straight line having a derivative.
> > >
> > > Bwaaaa haaaa haaaa.
> >
> > A secant line intersects only two points on a curve.
> >
> > Quick, quick! Run to the Wikipedia Moronica and edit the article!!!
>
> Wikipedia says that a secant line intersects two points on a curve.
> Note that it does NOT say "only" two points. That's your addition.

Wikipedia is a load of crap and I was being sarcastic. A secant line intersects another curve in ONLY two points. If it intersected the other curve in every point, then it can't be called a secant because curves are NOT secants to themselves. But how could a stupid moron like you be expected to realise this...

> Meanwhile, you continue not to answer my questions:
>
> 1. Specify n and m for f(x) = x^(1/2).

It has been answered at least 3 times you BABOON!!!!

> Or at least show explicitly what the 'auxiliary function' Q(x; n, m) is.

Also shown already.

> 2. Show how your definition works for f(x) = ln(x).
>
> 3. Give an example f(x) where the standard definition fails, but
> yours works correctly.

Be careful reptile! I never claimed the bogus definition does not work. Only that it does not work at inflection points.

[burs...@gmail.com]

Am Samstag, 31. Dezember 2016 00:00:54 UTC+1 schrieb John Gabriel:

> I never claimed the bogus definition does not work.
> Only that it does not work at inflection points.

Is the above something else than an english sentence?
What does the IT refer to? bogus calculus aka old

calculus I guess. Whats wrong with old calculus bird
brain John Gabriel birdbrain nonsense video spammer?

[John Gabriel]

On Friday, 30 December 2016 13:47:31 UTC-8, john-c%...@gtempaccount.com wrote:
> On Friday, December 30, 2016 at 2:13:25 PM UTC-6, John Gabriel wrote:
> > On Friday, 30 December 2016 12:03:07 UTC-8, john-c%...@gtempaccount.com wrote:
> > Really? In which case, for f(x) = x^(1/2), you should be able to
> > tell me explicitly what n and m are. But so far you have not.
> >
> > I can't help it that you are unable to read, but I told you already, any (m,n) pair of the form (k,k) with k < sqrt(2x-x^2) will work.
> >
> > > > even theoretically, for a simple function like f(x) = x^3, when x = 0.
> > >
> > > Of course you cannot find an m and n there you idiot!!! That's because there is NO tangent line there.
> >
> > Only because you have your own personal special definition of tangent line.
> >
> > Nonsense. The definition of tangent line I use is the same as that which Newton and Archimedes used.
> >
> > The NC produces the correct result.
> > >
> >
> > Except when it doesn't ...
> >
> > It always produces the correct result. Your BOGUS calculus does not produce correct results at inflection points.
> >
> > Demonstrate that it does for say f(x) = x^(1/2) or f(x) = ln(x).
> > I want to see the explicit details, not some vague references to
> > approximations. Computable results, like the standard limit
> > definition of derivative gives. Can you do that?
> >
> > Read idiot!!! The references to approximations are not vague. Are you such an imbecile that you are asking me to give you a formula solely in terms of m or n? Please tell me you are not.
>
> No ... it should be in terms of x, n, and m.

Huh? It is only in terms of x, m and n. What rubbish are you talking about?

> It should be a solution to your 'auxiliary equation' Q(x; n, m).

Yes, any pair (k,k) is a solution to the auxiliary equation. To wit,

f'(1) = f(1+k)-f(1-k) / (1+1)

f'(1) = { 2(1+k)-(1+k)^2 -2(1-k)+(1-k)^2 } / (1+1)

f'(1) = { 2+2k -1-2k-k^2 -2+2k+1-2k+k^2 } / (2)

f'(1) = { -1 +1 } / (2) = 0 as expected.

>
> > What are you continuing to babble about? You can in this case only approximate one in terms of the other. However, it is easy to see that any pair (k,k) with k < sqrt(2x-x^2) will work from the geometry.
>
> Let's try this with x = .5. Take n = m = k = .1.

Idiot. You were calculating it at x=1. Now you are trying x = 0.5 ?

In this case, it is not any pair (k,k) and you have to use an approximation technique to find m and n.

[burs...@gmail.com]

So we witness below that JG cannot read a function definition,
that includes sqrt and a piecewise condition. We will take k=0.1
and refute JGs silly claim.

Here is again the function from john_c post 5 hours ago:

f(x) = sqrt(2x- x^2) for 0 <= x <= 1, and

1 for x > 1.

Now lets compute, we know already f'(1)=0, but lets see
what f(1-0.1) and f(1+0.1) give:

f(1-0.1) = sqrt(2*0.9-0.9^2) = 3*sqrt(11)/10 ≈ 0.994987

f(1+0.1) = 1

Now check JGs claim:

f'(1) = (f(1+0.1)-f(1-0.1))/(0.1+0.1)

= 5 - 3*sqrt(11)/2

≈ 0.0250628

!= 0

Yes its wrong, no zero!

Witness also the error in JGs own use of his own new
calculus. Why is there below /(1+1) and not /(k+k).
Because he is bird brain John Gabriel birdbrain?

[John Gabriel]

On Friday, 30 December 2016 15:25:19 UTC-8, burs...@gmail.com wrote:
> So we witness below that JG cannot read a function definition,
> that includes sqrt and a piecewise condition. We will take k=0.1
> and refute JGs silly claim.
>
> Here is again the function from john_c post 5 hours ago:
>
> f(x) = sqrt(2x- x^2) for 0 <= x <= 1, and
>
> 1 for x > 1.
>
> Now lets compute, we know already f'(1)=0, but lets see
> what f(1-0.1) and f(1+0.1) give:
>
> f(1-0.1) = sqrt(2*0.9-0.9^2) = 3*sqrt(11)/10 ≈ 0.994987
>
> f(1+0.1) = 1
>
> Now check JGs claim:
>
> f'(1) = (f(1+0.1)-f(1-0.1))/(0.1+0.1)
>
> = 5 - 3*sqrt(11)/2
>
> ≈ 0.0250628
>
> != 0
>
> Yes its wrong, no zero!

Try again you stupid little twerp!

ANY pair (k,k) is a solution to the auxiliary equation when x=1. To wit,

f'(1) = f(1+k)-f(1-k) / (1+1)

f'(1) = { 2(1+k)-(1+k)^2 -2(1-k)+(1-k)^2 } / (1+1)

f'(1) = { 2+2k -1-2k-k^2 -2+2k+1-2k+k^2 } / (2)

f'(1) = { -1 +1 } / (2) = 0 as expected.

See idiot? In this case it does not matter what is k. Chuckle.

Shut up moron. Shut up.

[burs...@gmail.com]

Am Samstag, 31. Dezember 2016 00:43:52 UTC+1 schrieb John Gabriel:
> See idiot? In this case it does not matter what is k.

Is this again Gabrieloconfusion and Johnostupidics,
nobody is discussing this case. john_c didn't ask

for this function. bird brain John Garbiel birdbrain,
chief spammer of nonsense videos on sci.math

[john-c%...@gtempaccount.com]

On Friday, December 30, 2016 at 5:00:54 PM UTC-6, John Gabriel wrote:
> On Friday, 30 December 2016 12:49:23 UTC-8, john-c%...@gtempaccount.com wrote:
> > On Friday, December 30, 2016 at 2:30:36 PM UTC-6, genm...@gmail.com wrote:
> > > On Friday, 30 December 2016 12:22:13 UTC-8, John Gabriel wrote:
> > > > On Friday, 30 December 2016 12:03:07 UTC-8, john-c%...@gtempaccount.com wrote:
> > > >
> > > > > 2. [definition of the noun]: a line that is tangent; specifically : a straight line that is the limiting position of a secant of a curve through a fixed point and a variable point on the curve as the variable point approaches the fixed point
> > > >
> > > > So tell me idiot, how do you get a secant line on a straight line? Stupid much? Chuckle. Couldn't resist this because the stupid dog christensen always carries on about a straight line having a derivative.
> > > >
> > > > Bwaaaa haaaa haaaa.
> > >
> > > A secant line intersects only two points on a curve.
> > >
> > > Quick, quick! Run to the Wikipedia Moronica and edit the article!!!
> >
> > Wikipedia says that a secant line intersects two points on a curve.
> > Note that it does NOT say "only" two points. That's your addition.
>
> Wikipedia is a load of crap and I was being sarcastic. A secant line intersects another curve in ONLY two points.

Who says? Is this just semantics?

> If it intersected the other curve in every point, then it can't be called a secant because curves are NOT secants to themselves.

Who says? Do you just declare this as a fact?

But how could a stupid moron like you be expected to realise this...
>
> > Meanwhile, you continue not to answer my questions:
> >
> > 1. Specify n and m for f(x) = x^(1/2).
>
> It has been answered at least 3 times you BABOON!!!!
>

Humor me and show the solution for n and m, which so far you have
not done.

> > Or at least show explicitly what the 'auxiliary function' Q(x; n, m) is.
>
> Also shown already.
>

Not true. You waved your hands at it. But to really write it
down, you need to expand Newton's infinite series for the binomial
when the exponent is not an integer. You did not do that.

> > 2. Show how your definition works for f(x) = ln(x).
> >

Still waiting on this one.

> > 3. Give an example f(x) where the standard definition fails, but
> > yours works correctly.
>

And this one as well.

> Be careful reptile! I never claimed the bogus definition does not work. Only that it does not work at inflection points.

But it works just fine at inflection points. Your definition does not;
you are forced to claim there is no derivative at inflection points. But
there is.

John C.

[Ross A. Finlayson]

It seems clear that posters in
"agreement" with the troll-bot
are pretty much just sock-puppets
of same, i.e., 7777 is J.G., then
that it is pretty much just another
delegitimization attempt of the medium.

You present an example problem, then
suggest computing a solution. The
mathematics then isn't just the
computation, it's the arrival at
a solution. The corresponding solution
has intermediate developments that are
parameterizable and reusable for similar
problems, that's the mathematics. Here
it's for "find subsets of an n-set such that
for each 1 <= k <= N there exists i, j
s.t. i + j = k", then "find the smallest
such subset". Properties of laws of
arithmetic have that the odd numbers
plus zero are so a set. A Goldbach conjecture
would have it that the even numbers are
sums of (exactly two) primes, then for
figuring out where between log N[+] and 0.5 N
there are enough for also the odd numbers,
then for what those are, and otherwise as
of what additive partitions of size two,
among additive partitions, there would be
enough. Where Goldbach's prime conjecture
is so, then the primes and successors of
primes should suffice, on the order of log N
many, but that's an external fact and also
a conjecture.


[+] or pi(N)/N, see https://en.wikipedia.org/wiki/Prime-counting_function

The count of additive partitions (of a non-
negative integer) grows rapidly, but the
number of additive 2-partitions doesn't so
necessarily, as between zero and N they are
exhausted or on the order of N/2 many. (This
is similar to some other examples of the "curse
of dimensions" or real difference between 2-many
and 3- or more-many, including in radix representation
where otherwise people don't see a difference as
about binary versus decimal or Ramsey numbers and
the completeness or computability of the 2-many
but not the more-many, or solving 2-body versus
3-body problems et cetera.) That said, the
mathematics includes not just arriving at laser-
like precision to a reduced problem tractable to
linear means, but then boxing out furthermore
what solutions there are under re-parameterizations
of the problem.

Then, the smallest such subset is some intersection
of the additive 2-partitions about that being of
among the N^2/2 many instead of the partition-count
summed over N many.

https://en.wikipedia.org/wiki/Partition_(number_theory)

Having reduced the problem, or defined it sufficiently
and referred to related concerns as establish its
means, then there are mathematics about it.


Now, about the troll-bot and sock-puppet coterie's
refusal to refine definition and putting the cart
before the horse and so on, first you put the horse
before the cart, then imagine it might so proceed.
Don't go beating the horse because it's hitched wrongly.

Then, we observe your jousting with the windmills, but,
keep in mind that's just a troll-bot's blustery windmill,
not the giant that as mathematicians we would want to
either grow to, climb to, or supplant altogether.
Chasing the phantoms of truth or verity is Quixotic
in a sense, then that disambiguating the phantasmal
and really getting to the true is the mathematically
valorous (as it begins as the Quixotic).

Then really if you want to solve the subset sum problem,
why don't you just give your research grant to somebody
else and have them solve it for you. Now you can excuse
me if that's rather barbed but you can keep breaking down
the problems to build them back up, then about that you
should find that there is the extra- and super-classical
(and not the non-classical except as so follows in the
extra). The very means or tractability are involved as
concrete numerical resources then about how simple careful
organization suffices to begin and proceed about then the
resulting richer organization of the entirety and about
otherwise the exhaustion of such simple means and then about
how the deductive inference about these limits and the
surpassing them then so follows again with simple careful
organization so as to begin, proceed, and finish.

[John Gabriel]

On Friday, 30 December 2016 17:32:37 UTC-8, john-c%...@gtempaccount.com wrote:

> > Wikipedia is a load of crap and I was being sarcastic. A secant line intersects another curve in ONLY two points.

> Who says?

That is the definition of a secant line.

> > If it intersected the other curve in every point, then it can't be called a secant because curves are NOT secants to themselves.

> Who says? Do you just declare this as a fact?

It's common sense - something you don't have.

> Humor me and show the solution for n and m, which so far you have
> not done.

You have been told how to find it but you are an idiot, so you just keep repeating yourself like an immature child.

> > > Or at least show explicitly what the 'auxiliary function' Q(x; n, m) is.
> >
> > Also shown already.

> Not true. You waved your hands at it. But to really write it
> down, you need to expand Newton's infinite series for the binomial
> when the exponent is not an integer. You did not do that.

Nonsense. You cannot do this because it is impossible. You have to approximate the values of m or n using some kind of iterative technique, the most common being the root approximation technique.

> > Be careful reptile! I never claimed the bogus definition does not work. Only that it does not work at inflection points.
>
> But it works just fine at inflection points.

It does NOT work at inflection points because there is NO tangent line there. Moooorooooon!

> Your definition does not;

My definition works.

> you are forced to claim there is no derivative at inflection points. But
> there is.

You're just an idiot. "Nah Uh" makes you look stupid.

[john-c%...@gtempaccount.com]

On Friday, December 30, 2016 at 11:30:12 AM UTC-6, John Gabriel wrote:
> On Friday, 30 December 2016 09:25:59 UTC-8, konyberg wrote:

> > > > > > > > > f'(x) = (f(x + n) - f(x - m)) / (m + n).
> > > > > > > > >

> > > In the NC, nothing is forced. It is based on sound logic - something you are not accustomed to.
> > >
> > > >
> > > > which is indeterminate.
> > > >
> > > > John C.
> >

> > This no tangent of a curve (or half tangent) is your construction. Tangent to a point of a curve is well defined. Whether the tangent cut the curve at some other point don't matter.
> >
> > Can you give me one book where the tangent (and tehrefor the derivative) is undefined because the tangent line cuts the function?
> >
> > KON
>
> The original definition of tangent line:
>
> https://www.merriam-webster.com/dictionary/tangent

That definition says the tangent line meets the curve in one point.

Assume the curve is defined by f(x) = x^3. The standard-definition
of derivative implies f'(0) = 0. The horizontal line y = 0 goes
through (0, 0) [the ONLY point where it goes through the curve] and has slope zero. So it satisfies the Merriam-Webster definition of tangent line.

Thus there IS a tangent line, according to the definition you have
cited, which goes through an inflection point. But the slope of that
tangent line is never equal to the slope of any secant line.

Again, I am citing the definition that you selected.

In general, by your definition, there ARE tangent lines at points of inflection. Another inconsistency in the 'New Calculus'.

John C.



>
> And just think, if Archimedes and Newton thought of it in the same erratic way as you and the BIG STUPID, there would not have been any calculus at all!

[john-c%...@gtempaccount.com]

On Friday, December 30, 2016 at 12:49:06 PM UTC-6, John Gabriel wrote:
> On Friday, 30 December 2016 09:39:38 UTC-8, john-c%...@gtempaccount.com wrote:

> > It was you, not I who (above) said:
> >

> > "There is the common pair (m,n)=(0,0) which will yield the derivative
> > in this case and EVERY other case."
> >

> > I interpret your response to mean: "Yes, I said that
> >
> > f'(x) = (f(x + n) - f(x -m) / (m + n)
> >
> > for some n and m for any smooth, continuous function f(x)", but now
> > you say there are exceptions.
>
> There are no exceptions.
>

> > Not to mention the fact that, for
> >
> > f(x) = x^(1/2),
> >

> > you have not been able to specify what n and m should be
>
> You are lying. One can find approximations.
>
> > - so what good is your definition, if (unlike the usual limit definition), you cannot use it to do calculations?
>
> You cannot do anything with the limit definition that you can't do with the NC. The limit definition is a bunch of crap. And of course you can use it in any calculation. You just can't do stupid things with it like you do in the bogus calculus.
>
> > Another example:
>
> You can't even handle the simple examples. I suggest you first try to understand the simple examples before further complicating matters for yourself.
>
> >

> > f(x) = sqrt(2x- x^2) for 0 <= x <= 1, and
> >
> > 1 for x > 1.
> >

> > This is a well-defined function. It is smooth. Graph it. The tangent
> > line at x = 1 is horizontal. x = 1 is not an inflection point. However,
> > there are no positive values n and m such that for x = 1,
>
> LIAR!!!!! Grrrr. Do you enjoy being an idiot? Of course you can find m and n but they are approximations.

So, your definition works only approximately?

In fact, for this function it doesn't work at all. There is NO secant
line defined by two points on this curve which has slope 0 when x = 1.
The point (1, 1) is not an inflection point on this curve (not that that
matters); the tangent line through (1, 1) has slope 0.

>Tell me stupid, do you actually think about what you write? It's very clear to see that you can find m and n.

Clearly you cannot. There is no secant line defined by two points on
this curve which has slope 0. You can see this if you draw the graph.

John C.

If you are asking me for a formula where you can simply find either without iteration, then you are just being a moron.
>
>
> >

> > f'(x) = (f(x + n) - f(x - m))/(m + n).
> >
> > John C.

[John Gabriel]

It does not satisfy the Webster definition. You did not understand it much like you have not understood most of what has been discussed.

>
> Thus there IS a tangent line, according to the definition you have
> cited, which goes through an inflection point.

FALSE. According to Webster, there is NO tangent line at x=0.

> But the slope of that tangent line is never equal to the slope of any secant line.
>
> Again, I am citing the definition that you selected.
>
> In general, by your definition, there ARE tangent lines at points of inflection. Another inconsistency in the 'New Calculus'.

Nonsense. There are no tangent lines at points of inflection in the NC. The inconsistencies are in your inability to think logically and clearly.

You've been making mistakes right from the beginning.

[John Gabriel]

It works 100%, not approximately but you are too stupid to realise that when you are dealing with a radical and the binomial theorem, you are never going to get exact results. There is a way to describe m in terms of n exactly, but it involves very heavy algebra and you are not up to the task evidently so I won't discuss with you. I mean, you are such a moron that you can't grasp even the simplest of concepts. Tsk, tsk.

>
> In fact, for this function it doesn't work at all. There is NO secant
> line defined by two points on this curve which has slope 0 when x = 1.

Idiot. I proved to you that any pair (k,k) will satisfy slope 0 when x=1.

You're a fucking idiot and that can't be helped. Do you realise you are being laughed at for being so inept? Chuckle.

<too much excrement and too little time.>