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Jack Huizenga and his "critique" of my New Calculus. One learns more about Huizenga than my new calculus...
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John Gabriel
Mar 28, 2017, 9:32:50 PM3/28/17
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9/21/2016 A Harvard alumnus comments! | The New Calculus
Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.
The question was:
“What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.
Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.
One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.
In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!
There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga
The great mathematics historian Carl Boyer had this to say:
"Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.
That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer
I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.
Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga
One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
don’t exist.
Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga
The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.
Needless to say, if calculus were already extremely well understood, we would not have other mathematics PhDs making statements as follows:
Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)
Knisley goes on to say:
However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)
As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.
There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga
It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga
Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.
In the following comment I debunk the concept of Dedekind cuts and Cauchy sequences:
https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU
Our buffoon continues:
The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga
It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.
In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.
His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga
Of course we know the slope of the tangent line, provided we know the slope of a parallel secant line. This is grade 8 mathematics! What he states in parenthesis is even more amusing. It demonstrates clearly that he, like many of his colleagues never understood calculus:
If a given function is not continuous and smooth, then any of the methods of calculus are null and void.
One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.
He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga
I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.
It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.
That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article
https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E
The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!
Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga
Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.
The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
ceased to continue, when he could no longer understand what he was reading.
The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.
(*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.
And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.
This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.
Comments are unwelcome and will be ignored.
Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
gils...@gmail.com (MIT)
huiz...@psu.edu (HARVARD)
and...@mit.edu (MIT)
david....@math.okstate.edu (David Ullrich)
djo...@clarku.edu
mar...@gmail.com
Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.
The question was:
“What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.
Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.
One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.
In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!
There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga
The great mathematics historian Carl Boyer had this to say:
"Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.
That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer
I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.
Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga
One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
don’t exist.
Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga
The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.
Needless to say, if calculus were already extremely well understood, we would not have other mathematics PhDs making statements as follows:
Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)
Knisley goes on to say:
However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)
As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.
There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga
It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga
Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.
In the following comment I debunk the concept of Dedekind cuts and Cauchy sequences:
https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU
Our buffoon continues:
The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga
It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.
In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.
His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga
Of course we know the slope of the tangent line, provided we know the slope of a parallel secant line. This is grade 8 mathematics! What he states in parenthesis is even more amusing. It demonstrates clearly that he, like many of his colleagues never understood calculus:
If a given function is not continuous and smooth, then any of the methods of calculus are null and void.
One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.
He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga
I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.
It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.
That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article
https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E
The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!
Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga
Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.
The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
ceased to continue, when he could no longer understand what he was reading.
The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.
(*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.
And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.
This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.
Comments are unwelcome and will be ignored.
Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
gils...@gmail.com (MIT)
huiz...@psu.edu (HARVARD)
and...@mit.edu (MIT)
david....@math.okstate.edu (David Ullrich)
djo...@clarku.edu
mar...@gmail.com
burs...@gmail.com
Mar 29, 2017, 5:08:10 AM3/29/17
to
John Gabriel thinks Cauchy did e-d Calculus:
Epsilonic Blues - Topologabriely of Mythnifolds
https://drive.google.com/file/d/0B-mOEooW03iLX1Z5eUpkdHZKaUE/view
Others think this is not historically adequate:
On the history of epsilontics, Sinkevich Galina
https://arxiv.org/abs/1502.06942
Epsilonic Blues - Topologabriely of Mythnifolds
https://drive.google.com/file/d/0B-mOEooW03iLX1Z5eUpkdHZKaUE/view
Others think this is not historically adequate:
On the history of epsilontics, Sinkevich Galina
https://arxiv.org/abs/1502.06942
burs...@gmail.com
Mar 29, 2017, 5:14:15 AM3/29/17
to
And relating to mean value theorem, as JG does:
f'(x) = (f(x+h)-f(x))/h
doesn't make sense. We would need to write:
f'(c) = (f(x+h)-f(x))/h
And say there is c in (x,x+h). The below arvix also
says something about mean value theorem and cauchy.
But how long will it take that JG gets anything right?
f'(x) = (f(x+h)-f(x))/h
doesn't make sense. We would need to write:
f'(c) = (f(x+h)-f(x))/h
And say there is c in (x,x+h). The below arvix also
says something about mean value theorem and cauchy.
But how long will it take that JG gets anything right?
7777777
Apr 14, 2017, 2:57:36 AM4/14/17
to
keskiviikko 29. maaliskuuta 2017 12.14.15 UTC+3 burs...@gmail.com kirjoitti:
> And relating to mean value theorem, as JG does:
>
> f'(x) = (f(x+h)-f(x))/h
>
> doesn't make sense.
it does not make sense as long as h is finite.
> And relating to mean value theorem, as JG does:
>
> f'(x) = (f(x+h)-f(x))/h
>
> doesn't make sense.
> We would need to write:
>
> f'(c) = (f(x+h)-f(x))/h
>
> And say there is c in (x,x+h)
j4n bur53
Apr 14, 2017, 7:49:25 AM4/14/17
to
h is from R, its not a set, its some urelement.
It does not have a set size property finite, infinite
countable or infinite uncountable.
I guess you mean "not infinitessimal small", i.e.
not outside of R, since R is archimedian, it doesn't
have anyway some infinitessimals in it.
Why you would have f'(x) = (f(x+h)-f(x))/h in an
other domain D than R, where h in D, you would need
to demonstrate. Can you?
7777777 schrieb:
It does not have a set size property finite, infinite
countable or infinite uncountable.
I guess you mean "not infinitessimal small", i.e.
not outside of R, since R is archimedian, it doesn't
have anyway some infinitessimals in it.
Why you would have f'(x) = (f(x+h)-f(x))/h in an
other domain D than R, where h in D, you would need
to demonstrate. Can you?
7777777 schrieb:
7777777
Apr 14, 2017, 9:20:34 AM4/14/17
to
perjantai 14. huhtikuuta 2017 14.49.25 UTC+3 j4n bur53 kirjoitti:
> h is from R, its not a set, its some urelement.
> It does not have a set size property finite, infinite
> countable or infinite uncountable.
>
> I guess you mean "not infinitessimal small", i.e.
> not outside of R, since R is archimedian, it doesn't
> have anyway some infinitessimals in it.
>
> Why you would have f'(x) = (f(x+h)-f(x))/h in an
> other domain D than R, where h in D, you would need
> to demonstrate. Can you?
>
I am not exactly sure what you want me to demonstrate, but I will try anyway.
> h is from R, its not a set, its some urelement.
> It does not have a set size property finite, infinite
> countable or infinite uncountable.
>
> I guess you mean "not infinitessimal small", i.e.
> not outside of R, since R is archimedian, it doesn't
> have anyway some infinitessimals in it.
>
> Why you would have f'(x) = (f(x+h)-f(x))/h in an
> other domain D than R, where h in D, you would need
> to demonstrate. Can you?
>
Consider first
f'(c) ={ f(c+n) - f(c-m) } / (m+n)
if m and n are finite then f'(c) is not equal to f'(x)
So lets use the infinitesimal dx = m = n
We end up with
f'(c) ={ f(c+dx) - f(c-dx) } / 2dx
And we need to demonstrate
f'(x) = (f(x+h)-f(x))/h = (f(x+dx)-f(x))/dx
so that f'(c) = f'(x) also
{ f(c+dx) - f(c-dx) } / 2dx = { f(x+dx) - f(x-dx) } / 2dx
in other words, we need to demonstrate
(f(x+dx)-f(x))/dx = { f(x+dx) - f(x-dx) } / 2dx
also the same as
f(x + dx) - f(x) = 1/2 f(x + dx) - 1/2 f(x - dx)
f(x) = 1/2 [ f(x + dx) + f(x - dx)]
if that's a valid formula remains to be seen...
John Gabriel
Apr 14, 2017, 9:40:23 AM4/14/17
to
On Friday, 14 April 2017 08:20:34 UTC-5, 7777777 wrote:
> perjantai 14. huhtikuuta 2017 14.49.25 UTC+3 j4n bur53 kirjoitti:
> > h is from R, its not a set, its some urelement.
> > It does not have a set size property finite, infinite
> > countable or infinite uncountable.
> >
> > I guess you mean "not infinitessimal small", i.e.
> > not outside of R, since R is archimedian, it doesn't
> > have anyway some infinitessimals in it.
> >
> > Why you would have f'(x) = (f(x+h)-f(x))/h in an
> > other domain D than R, where h in D, you would need
> > to demonstrate. Can you?
> >
> I am not exactly sure what you want me to demonstrate, but I will try anyway.
> Consider first
> f'(c) ={ f(c+n) - f(c-m) } / (m+n)
> if m and n are finite then f'(c) is not equal to f'(x)
Yes idiot. { f(c+n) - f(c-m) } / (m+n) is equal to f'(c). What is x? If you are talking about the x in f'(x) = (f(x+h)-f(x))/h, then these are two different things.
> perjantai 14. huhtikuuta 2017 14.49.25 UTC+3 j4n bur53 kirjoitti:
> > h is from R, its not a set, its some urelement.
> > It does not have a set size property finite, infinite
> > countable or infinite uncountable.
> >
> > I guess you mean "not infinitessimal small", i.e.
> > not outside of R, since R is archimedian, it doesn't
> > have anyway some infinitessimals in it.
> >
> > Why you would have f'(x) = (f(x+h)-f(x))/h in an
> > other domain D than R, where h in D, you would need
> > to demonstrate. Can you?
> >
> I am not exactly sure what you want me to demonstrate, but I will try anyway.
> Consider first
> f'(c) ={ f(c+n) - f(c-m) } / (m+n)
> if m and n are finite then f'(c) is not equal to f'(x)
f'(x) = (f(x+h)-f(x))/h is WRONG. You need Lim h->0 to make f'(x) = (f(x+h)-f(x))/h somewhat sensible.
<More Moron shit ignored>
7777777
Apr 14, 2017, 10:06:33 AM4/14/17
to
perjantai 14. huhtikuuta 2017 16.40.23 UTC+3 John Gabriel kirjoitti:
> f'(x) = (f(x+h)-f(x))/h is WRONG.
what happens if h = dx <-------- the infinitesimal ?
> f'(x) = (f(x+h)-f(x))/h is WRONG.
> You need Lim h->0 to make f'(x) = (f(x+h)-f(x))/h somewhat sensible.
> <More Moron shit ignored>
for how long time the moron keeps ignoring the infinitesimal dx ?
Dan Christensen
Apr 14, 2017, 10:41:17 AM4/14/17
to
On Tuesday, March 28, 2017 at 9:32:50 PM UTC-4, John Gabriel wrote:
> 9/21/2016 A Harvard alumnus comments! | The New Calculus
>
> Just a few weeks ago...
> 9/21/2016 A Harvard alumnus comments! | The New Calculus
>
When it comes to calculus, JG is utterly clueless. Using his Wacky New "Calclueless," he cannot even determine the derivative of functions as simple as y=x. He actually claims they are "undefined!" What a moron.
Neither can JG derive even the most elementary results of basic arithmetic. A proof of 2+2=4 is well beyond his meager capabilities.
JG's Wacky New Calclueless is a dead end and a complete waste of time. Interested readers should see my recent posting, "About the spamming troll John Gabriel in his own words (April 2017)."
Dan
burs...@gmail.com
Apr 14, 2017, 11:40:31 AM4/14/17
to
Am Freitag, 14. April 2017 15:20:34 UTC+2 schrieb 7777777:
> I am not exactly sure what you want me to demonstrate, but I will try anyway.
> Consider first
> f'(c) ={ f(c+n) - f(c-m) } / (m+n)
> if m and n are finite then f'(c) is not equal to f'(x)
Yes and no.
> I am not exactly sure what you want me to demonstrate, but I will try anyway.
> Consider first
> f'(c) ={ f(c+n) - f(c-m) } / (m+n)
> if m and n are finite then f'(c) is not equal to f'(x)
Yes:
If m and n are positive numbers, then the equation
f'(c) ={ f(c+n) - f(c-m) } / (m+n) does hold in general.
No:
But I think calling m and n finite doesn't make any
sense, they are not sets, they are reals.
For the No please consider the definition:
infinitesimal (plural infinitesimals)
(mathematics) A non-zero quantity whose magnitude is smaller than any positive number (by definition it is not a real number).
https://en.wiktionary.org/wiki/infinitesimal
So it reads smaller than any positive number and not smaller
than any finite number. You cannot use the adjective finite
with numbers, its not a set.
You can say:
{0, 1, 2} is finite
or
{2, 4, 6, ...} is not finite
But applying the adjective finite to real numbers doesn't
make any sense to me.
burs...@gmail.com
Apr 14, 2017, 11:42:04 AM4/14/17
to
Corr.:
If m and n are positive numbers, then the equation
f'(c) ={ f(c+n) - f(c-m) } / (m+n) doesn't hold in general.
John Gabriel
Apr 14, 2017, 12:00:59 PM4/14/17
to
On Friday, 14 April 2017 09:06:33 UTC-5, 7777777 wrote:
> perjantai 14. huhtikuuta 2017 16.40.23 UTC+3 John Gabriel kirjoitti:
>
> > f'(x) = (f(x+h)-f(x))/h is WRONG.
>
> what happens if h = dx <-------- the infinitesimal ?
There are NO infinitesimals dummy. For the thousandth time:
> perjantai 14. huhtikuuta 2017 16.40.23 UTC+3 John Gabriel kirjoitti:
>
> > f'(x) = (f(x+h)-f(x))/h is WRONG.
>
> what happens if h = dx <-------- the infinitesimal ?
THERE ARE NO INFINITESIMALS!!!!
Grow a brain and stop being a moron. You keep holding onto this shit even though I have shown you over and over again on STATU that there is no such thing as an infinitesimal.
Look you moron: It is either ZERO or it is NOT. What part of this do you not understand? Do you subscribe to the stupidity of the BIG STUPID?
If you claim infinitesimal, then what you are saying is that there is a number that follows ZERO. There isn't. Get it?!! No, don't give me your shit about ineffable or ourobouros.
>
>
> > You need Lim h->0 to make f'(x) = (f(x+h)-f(x))/h somewhat sensible.
>
> but do you need dx, the infinitesimal ?
h or dx is NEVER zero in the mainstream formulation except when they determine the derivative and then use it in the self-referential limit verifinition.
>
>
>
> > <More Moron shit ignored>
>
> for how long time the moron keeps ignoring the infinitesimal dx ?
John Gabriel
Apr 14, 2017, 12:01:42 PM4/14/17
to
burs...@gmail.com
Apr 14, 2017, 12:09:00 PM4/14/17
to
Compare with the following mismatch:
a) A menstruating woman is sitting in the kitchen.
b) A menstruating man is standing at the door.
Which one has an error?
Now:
a) the set of primes isn't finite
b) if h is finite then ...
On the other hand in the past it was indeed common to
use the adjective finite on quantities. But still it
sounds strange to me. But the 777777 usage, is for example
found in the introduction here:
The Method of Fluxions and Infinite Series: With
Its Application to the ... von Isaac Newton
I easily find the following passage:
.. and introduces non but infinitely little Quantities
that are relatively to; which he arrives at by beginning
with finite Quantities, and proceeding by a gradual
and necessary progress of diminution. His computations
always commence by finite and intelligible quantities, ...
So maybe thats why 77777 uses the adjective finite
for h, m and n, etc.. but its still sounds wrong to
me since in our modern times we have sets and set theory,
and it looks like a confusion error.
But without any clear apparatus given by 77777, where the
notion of finite for a number is exactly defined, and
also what it means for a number to be non-finite, what
77777 wrote doesn't make any sense. For example I asked:
> Why you would have f'(x) = (f(x+h)-f(x))/h in an
> other domain D than R, where h in D, you would need
> to demonstrate. Can you?
So 7777777, what D (?surreals etc..?) are you using?
a) A menstruating woman is sitting in the kitchen.
b) A menstruating man is standing at the door.
Which one has an error?
Now:
a) the set of primes isn't finite
b) if h is finite then ...
On the other hand in the past it was indeed common to
use the adjective finite on quantities. But still it
sounds strange to me. But the 777777 usage, is for example
found in the introduction here:
The Method of Fluxions and Infinite Series: With
Its Application to the ... von Isaac Newton
I easily find the following passage:
.. and introduces non but infinitely little Quantities
that are relatively to; which he arrives at by beginning
with finite Quantities, and proceeding by a gradual
and necessary progress of diminution. His computations
always commence by finite and intelligible quantities, ...
So maybe thats why 77777 uses the adjective finite
for h, m and n, etc.. but its still sounds wrong to
me since in our modern times we have sets and set theory,
and it looks like a confusion error.
But without any clear apparatus given by 77777, where the
notion of finite for a number is exactly defined, and
also what it means for a number to be non-finite, what
77777 wrote doesn't make any sense. For example I asked:
> Why you would have f'(x) = (f(x+h)-f(x))/h in an
> other domain D than R, where h in D, you would need
> to demonstrate. Can you?
burs...@gmail.com
Apr 14, 2017, 12:38:28 PM4/14/17
to
Although newtons booklet looks a little
shitty compared to the booklet of euler,
On page 51 he left handedly computes the
tangent to a spiral of archimedes, so that
I guess we nevertheless have a genius before us...
BTW: And funny thing, he writes "affirmative
integer" instead of "positive integer".
shitty compared to the booklet of euler,
On page 51 he left handedly computes the
tangent to a spiral of archimedes, so that
I guess we nevertheless have a genius before us...
BTW: And funny thing, he writes "affirmative
integer" instead of "positive integer".
burs...@gmail.com
Apr 14, 2017, 12:47:44 PM4/14/17
to
Its Application to the Geometry of Curve-lines. By ... Sir Isaac Newton, ... Translated from the Author's Latin Original Not Yet Made Publick. To which is Subjoin'd, a Perpetual Comment Upon the Whole Work, ... By John Colson, ...
Isaac Newton
Henry Woodfall; and sold by John Nourse, 1736 - 339 pages
Isaac Newton
Henry Woodfall; and sold by John Nourse, 1736 - 339 pages
7777777
Apr 14, 2017, 12:48:51 PM4/14/17
to
perjantai 14. huhtikuuta 2017 19.00.59 UTC+3 John Gabriel kirjoitti:
> On Friday, 14 April 2017 09:06:33 UTC-5, 7777777 wrote:
> > perjantai 14. huhtikuuta 2017 16.40.23 UTC+3 John Gabriel kirjoitti:
> >
> > > f'(x) = (f(x+h)-f(x))/h is WRONG.
> >
> > what happens if h = dx <-------- the infinitesimal ?
>
> There are NO infinitesimals dummy. For the thousandth time:
>
> THERE ARE NO INFINITESIMALS!!!!
>
> Grow a brain and stop being a moron. You keep holding onto this shit even though I have shown you over and over again on STATU that there is no such thing as an infinitesimal.
you have not shown anything. you have only shown that you are a moron.
> On Friday, 14 April 2017 09:06:33 UTC-5, 7777777 wrote:
> > perjantai 14. huhtikuuta 2017 16.40.23 UTC+3 John Gabriel kirjoitti:
> >
> > > f'(x) = (f(x+h)-f(x))/h is WRONG.
> >
> > what happens if h = dx <-------- the infinitesimal ?
>
> There are NO infinitesimals dummy. For the thousandth time:
>
> THERE ARE NO INFINITESIMALS!!!!
>
> Grow a brain and stop being a moron. You keep holding onto this shit even though I have shown you over and over again on STATU that there is no such thing as an infinitesimal.
It is here: dx <---------- the infinitesimal
>
> Look you moron: It is either ZERO or it is NOT.
>What part of this do you not understand? Do you subscribe to the stupidity of the BIG STUPID?
is a problem, a problem common to all, but people like you just pretend
that the problem simply does not exist. While at the same time, the rest of us
have to struggle solving the very hard problems. You just keep ridiculing us,
without understanding that it is you who is in error.
>
> If you claim infinitesimal, then what you are saying is that there is a number that follows ZERO. There isn't. Get it?!! No, don't give me your shit about ineffable or ourobouros.
convince anyone, because what exists and what doesn't are just your beliefs.
> > > You need Lim h->0 to make f'(x) = (f(x+h)-f(x))/h somewhat sensible.
> >
> > but do you need dx, the infinitesimal ?
>
> No stupid. In mainstream mythmatics, that is the LIMIT of the finite difference ratios as dx gets close to 0.
I am talking about dx. Do you ever pay attention to detail?
dx does not necessarily get close to anything, and if it does, then the situation is different than when h gets close to 0.
>
>
> h or dx is NEVER zero in the mainstream formulation except when they determine the derivative and then use it in the self-referential limit verifinition.
>
>
dx = 0.000...001 = 0
> >
> >
> > > <More Moron shit ignored>
> >
> > for how long time the moron keeps ignoring the infinitesimal dx ?
>
> I don't think that even a hundred slaps on the back of your head will help you to put some brains.
You can't have calculus based only on finite difference.
burs...@gmail.com
Apr 14, 2017, 12:52:30 PM4/14/17
to
Who is them? The deamons that hunt you in the night?
There is no them, you are responsible for yourself to
do what you want to do in math.
There is no them, you are responsible for yourself to
do what you want to do in math.
burs...@gmail.com
Apr 14, 2017, 12:55:20 PM4/14/17
to
So 7777777 dont blame others, you still didn't answer:
> Why you would have f'(x) = (f(x+h)-f(x))/h in an
> other domain D than R, where h in D, you would need
> to demonstrate. Can you?
> Why you would have f'(x) = (f(x+h)-f(x))/h in an
> other domain D than R, where h in D, you would need
> to demonstrate. Can you?
7777777
Apr 14, 2017, 1:16:50 PM4/14/17
to
perjantai 14. huhtikuuta 2017 19.55.20 UTC+3 burs...@gmail.com kirjoitti:
> So 7777777 dont blame others, you still didn't answer:
>
> > Why you would have f'(x) = (f(x+h)-f(x))/h in an
> > other domain D than R, where h in D, you would need
> > to demonstrate. Can you?
>
I am still wondering what exactly are you talking about. I already tried to answer it. And here I go again:
> So 7777777 dont blame others, you still didn't answer:
>
> > Why you would have f'(x) = (f(x+h)-f(x))/h in an
> > other domain D than R, where h in D, you would need
> > to demonstrate. Can you?
>
dx = 0.000...001
f'(x) = (f(x+dx)-f(x))/dx = (f(x+0.000...001)-f(x))/0.000...001
burs...@gmail.com
Apr 14, 2017, 2:06:30 PM4/14/17
to
How does this exactly work?
Take f(x) = x^2.
What is (x + 0.000...001) ?
Take f(x) = ln(x).
What is ln(x + 0.000...001) ?
Take f(x) = g(h(x)).
What is g(h(x + 0.000...001)) ?
I don't think it works so easily... You neeed to
define the arithmetic for infinitessimals and show us.
Basically the arithmetic of mixture your "finite" and
"non-finite" numbers. And transcendental functions as well.
Can you do that, do you have such a sound system?
Where is it written, or is in your mind only?
Take f(x) = x^2.
What is (x + 0.000...001) ?
Take f(x) = ln(x).
What is ln(x + 0.000...001) ?
Take f(x) = g(h(x)).
What is g(h(x + 0.000...001)) ?
I don't think it works so easily... You neeed to
define the arithmetic for infinitessimals and show us.
Basically the arithmetic of mixture your "finite" and
"non-finite" numbers. And transcendental functions as well.
Can you do that, do you have such a sound system?
Where is it written, or is in your mind only?
7777777
Apr 14, 2017, 2:33:19 PM4/14/17
to
perjantai 14. huhtikuuta 2017 21.06.30 UTC+3 burs...@gmail.com kirjoitti:
> How does this exactly work?
>
> Take f(x) = x^2.
> What is (x + 0.000...001) ?
>
>
f(x) = x^2
> How does this exactly work?
>
> Take f(x) = x^2.
> What is (x + 0.000...001) ?
>
>
f(x + dx) = (x + dx)^2 = x^2 + 2xdx + (dx)^2
now insert dx = 0.000...001
burs...@gmail.com
Apr 14, 2017, 2:49:15 PM4/14/17
to
And? What is the value of (0.000...001)^2, you
know for to obtain f'(x) it must vanish.
But you said 0.000...001 != 0.
This is a contradiction. Its similar to the contradiction
that BKK had, when he said 1.4142... != sqrt(2).
I am pretty sure we end up with:
Garbage in, Garbage out
know for to obtain f'(x) it must vanish.
But you said 0.000...001 != 0.
This is a contradiction. Its similar to the contradiction
that BKK had, when he said 1.4142... != sqrt(2).
I am pretty sure we end up with:
Garbage in, Garbage out
burs...@gmail.com
Apr 14, 2017, 3:01:57 PM4/14/17
to
More precisely it must exactly happen 0.000...001 == 0
to obtain f'(x). Please observe for f(x)=x^2, we have
f'(x) = 2*x
Now lets try your method of plugging in. First of all
you said f(x+dx)=x^2+2*x*dx+dx^2. So far so good.
But the question is for (f(x+dx)-f(x))/dx. Which is
now (for you "finite" quantities, and you didn't tell
me how to compute it for infinitessimals):
x^2 + 2*x*dx + dx^2 - x^2
------------------------- = 2*x + dx
dx
Now you said your idea is just plugging in infinitessimal
in pace of a finite quantity, so I do plugging in, i.e.
setting dx = 0.000...001:
g(x) = 2*x + 0.000...001
But since you said 0.000...001 != 0, we don't have g(x)
= f'(x). But sometimes you say is even both, 0.000...001 != 0
and 0.000...001 == 0.
This is very confusing, you need a clear framework how
everything works. The limes framework is much clearer,
we all know that "f(x)" and "lim x->a f(x)" are two different
expressions.
Please also show:
f(x) = ln(x)
f(x) = g(h(x))
to obtain f'(x). Please observe for f(x)=x^2, we have
f'(x) = 2*x
Now lets try your method of plugging in. First of all
you said f(x+dx)=x^2+2*x*dx+dx^2. So far so good.
But the question is for (f(x+dx)-f(x))/dx. Which is
now (for you "finite" quantities, and you didn't tell
me how to compute it for infinitessimals):
x^2 + 2*x*dx + dx^2 - x^2
------------------------- = 2*x + dx
dx
Now you said your idea is just plugging in infinitessimal
in pace of a finite quantity, so I do plugging in, i.e.
setting dx = 0.000...001:
g(x) = 2*x + 0.000...001
But since you said 0.000...001 != 0, we don't have g(x)
= f'(x). But sometimes you say is even both, 0.000...001 != 0
and 0.000...001 == 0.
This is very confusing, you need a clear framework how
everything works. The limes framework is much clearer,
we all know that "f(x)" and "lim x->a f(x)" are two different
expressions.
Please also show:
f(x) = ln(x)
f(x) = g(h(x))
7777777
Apr 14, 2017, 3:14:16 PM4/14/17
to
perjantai 14. huhtikuuta 2017 22.01.57 UTC+3 burs...@gmail.com kirjoitti:
> More precisely it must exactly happen 0.000...001 == 0
> to obtain f'(x). Please observe for f(x)=x^2, we have
>
> f'(x) = 2*x
>
> Now lets try your method of plugging in. First of all
> you said f(x+dx)=x^2+2*x*dx+dx^2. So far so good.
>
> But the question is for (f(x+dx)-f(x))/dx. Which is
> now (for you "finite" quantities, and you didn't tell
> me how to compute it for infinitessimals):
>
> x^2 + 2*x*dx + dx^2 - x^2
> ------------------------- = 2*x + dx
> dx
>
> Now you said your idea is just plugging in infinitessimal
> in pace of a finite quantity, so I do plugging in, i.e.
> setting dx = 0.000...001:
>
> g(x) = 2*x + 0.000...001
>
> But since you said 0.000...001 != 0, we don't have g(x)
> = f'(x). But sometimes you say is even both, 0.000...001 != 0
> and 0.000...001 == 0.
>
> This is very confusing
>
It is not confusing.
> More precisely it must exactly happen 0.000...001 == 0
> to obtain f'(x). Please observe for f(x)=x^2, we have
>
> f'(x) = 2*x
>
> Now lets try your method of plugging in. First of all
> you said f(x+dx)=x^2+2*x*dx+dx^2. So far so good.
>
> But the question is for (f(x+dx)-f(x))/dx. Which is
> now (for you "finite" quantities, and you didn't tell
> me how to compute it for infinitessimals):
>
> x^2 + 2*x*dx + dx^2 - x^2
> ------------------------- = 2*x + dx
> dx
>
> Now you said your idea is just plugging in infinitessimal
> in pace of a finite quantity, so I do plugging in, i.e.
> setting dx = 0.000...001:
>
> g(x) = 2*x + 0.000...001
>
> But since you said 0.000...001 != 0, we don't have g(x)
> = f'(x). But sometimes you say is even both, 0.000...001 != 0
> and 0.000...001 == 0.
>
> This is very confusing
>
You need to consider (dx)^2 = (0.000...001)^2 can be equal to 0 even though
0.000...001 is non-zero
And then you end up with f'(x) = 2x
burs...@gmail.com
Apr 14, 2017, 4:06:28 PM4/14/17
to
Thats actually not true. Can you please demonstrate
step by step? I didn't use (0.000...001)^2 = 0.
If you use (0.000...001)^2 = 0, you don't compute
(f(x+h)-f(x))/h. Instead you only compute f(x+h)-f(x).
And for this computation you get:
f(x+0.000...001)-f(x) = 2*x*0.000...001+0.000...001^2
Now if you alow 0.000...001^2 = 0, you only get:
g(x) = 2*x*0.000...001
Which is still not f'(x). You are completely confused
what f'(x) is and what it is not. Maybe you are inspired
by this paper:
An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf
step by step? I didn't use (0.000...001)^2 = 0.
If you use (0.000...001)^2 = 0, you don't compute
(f(x+h)-f(x))/h. Instead you only compute f(x+h)-f(x).
And for this computation you get:
f(x+0.000...001)-f(x) = 2*x*0.000...001+0.000...001^2
Now if you alow 0.000...001^2 = 0, you only get:
g(x) = 2*x*0.000...001
Which is still not f'(x). You are completely confused
what f'(x) is and what it is not. Maybe you are inspired
by this paper:
An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf
burs...@gmail.com
Apr 14, 2017, 4:12:05 PM4/14/17
to
For my taste setting e^2 = 0 is a cludge, which
leads to nowhere. It just covers up the division
by e, and the limes e = 0.
The results are similar to Edward Studys duals,
but one could probably do them without the
e^2 = 0 cludige.
leads to nowhere. It just covers up the division
by e, and the limes e = 0.
The results are similar to Edward Studys duals,
but one could probably do them without the
e^2 = 0 cludige.
John Gabriel
Apr 14, 2017, 4:14:02 PM4/14/17
to
On Friday, 14 April 2017 11:48:51 UTC-5, 7777777 wrote:
> you have not shown anything. you have only shown that you are a moron.
> It is here: dx <---------- the infinitesimal
!#@!&^*!#@ <---- This is Jesus! You imbecile!
> you have not shown anything. you have only shown that you are a moron.
> It is here: dx <---------- the infinitesimal
> > Look you moron: It is either ZERO or it is NOT.
>
> If it would be that simple someone would have already solved the problem a long time ago.
> What I have shown (and not just me, btw), over and over again, it maybe be both 0 and non-zero at the same time.
> There is no such thing. There is no such thing as "BIG STUPID".
> You just keep ridiculing us, without understanding that it is you who is in error.
> No-one takes you seriously, you just keep repeating the same nonsense over and over again, basically telling us what can exist and what can't, but you can't convince anyone, because what exists and what doesn't are just your beliefs.
> in the mainstream formulation dx is equal to 0 because according to them
> dx = 0.000...001 = 0
> are you talking to me? It seems to me that it is you who needs to grow brains. You can't have calculus based only on finite difference.
John Gabriel
Apr 14, 2017, 4:14:52 PM4/14/17
to
burs...@gmail.com
Apr 14, 2017, 4:19:52 PM4/14/17
to
bird brain John Gabriel birdbrain, I don't disagree with 777777,
since he didn't yet fully explain his method. He says this and that.
As a bird brain John Gabriel birdbrain, you cannot distingzish
inquiry from opinion. Anyway this is a funny read, much more
funnier than the new fake calculoose:
An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf
since he didn't yet fully explain his method. He says this and that.
As a bird brain John Gabriel birdbrain, you cannot distingzish
inquiry from opinion. Anyway this is a funny read, much more
funnier than the new fake calculoose:
An Invitation to Smooth Infinitesimal Analysis John L. Bell
http://publish.uwo.ca/~jbell/invitation%20to%20SIA.pdf
burs...@gmail.com
Apr 14, 2017, 4:23:16 PM4/14/17
to
The big advantage of limes is that it doesn't break
trichotomy, since you still evaluate in the finite Quantities,
you can also compute limes of piecewise functions easily, etc..
trichotomy, since you still evaluate in the finite Quantities,
you can also compute limes of piecewise functions easily, etc..
Message has been deleted
burs...@gmail.com
Apr 14, 2017, 4:53:44 PM4/14/17
to
Am Freitag, 14. April 2017 22:32:29 UTC+2 schrieb Me:
> > it maybe be both 0 and non-zero at the same time.
> Not in the context of classical mathematics / classical logic.
> > it maybe be both 0 and non-zero at the same time.
Yes without further context, where it changes its
value, like an operator applied to it,
or two different equality signs, we usually
get a contradiction
e = 0 & ~ e = 0 /* is a contradiction */
e =_1 0 & ~ e =_2 0 /* is not per se a contradiction */
And of course (for the first sample, we need that
we are not in R, some other domain D, for the second
sample we assume n > 0):
e^2 = 0 & ~ e = 0 /* is not per se a contradiction */
lim 1/n = 0 & ~ 1/n = 0 /* is not per se a contradiction */
The later cases are case where an operator is involved.
The last case is very problematic for JG, since he
believes ~ 0.999... = 1.
But ~ 0.999... = 1 is possible when we are outside of R,
some other domain D. Its value could be 1 - e, with e^2 = 0
and ~ e = 0. Right?
Don't ask me how ~ 0.999... = 1 exactly works, just flunking
it. I tried duals once, but problem is to make it really
injective, i.e. from f : R -> R, to get an injective
f : D -> D, this was the obstactle. Basically when f'(x)=0,
it didn't work the extension. Bell mentions this when he
talks about stationary points I guess.
Me
Apr 14, 2017, 5:00:06 PM4/14/17
to
On Friday, April 14, 2017 at 6:48:51 PM UTC+2, 7777777 wrote:
> perjantai 14. huhtikuuta 2017 19.00.59 UTC+3 John Gabriel kirjoitti:
> >
> > Look you moron: It is either ZERO or it is NOT.
> >
> If it would be that simple ...
> perjantai 14. huhtikuuta 2017 19.00.59 UTC+3 John Gabriel kirjoitti:
> >
> > Look you moron: It is either ZERO or it is NOT.
> >
John is right. For every x: x is either ZERO or it is NOT ZERO, at least in the context of classical mathematics: x = 0 v ~(x = 0).
Moreover, if x e IR and |x| < 1/n for all n IN, then x = 0 (since the reals are archimedian).
> it maybe be both 0 and non-zero at the same time.
Btw.: One may call 0 e IR the only infitesimal in IR.
See:
https://math.stackexchange.com/questions/1086713/is-0-an-infinitesimal
7777777
Apr 14, 2017, 11:57:09 PM4/14/17
to
lauantai 15. huhtikuuta 2017 0.00.06 UTC+3 Me kirjoitti:
> On Friday, April 14, 2017 at 6:48:51 PM UTC+2, 7777777 wrote:
> > perjantai 14. huhtikuuta 2017 19.00.59 UTC+3 John Gabriel kirjoitti:
> > >
> > > Look you moron: It is either ZERO or it is NOT.
> > >
> > If it would be that simple ...
>
> John is right. For every x: x is either ZERO or it is NOT ZERO
Let's get this straight:
> On Friday, April 14, 2017 at 6:48:51 PM UTC+2, 7777777 wrote:
> > perjantai 14. huhtikuuta 2017 19.00.59 UTC+3 John Gabriel kirjoitti:
> > >
> > > Look you moron: It is either ZERO or it is NOT.
> > >
> > If it would be that simple ...
>
> John is right. For every x: x is either ZERO or it is NOT ZERO
is 0.000...001 either ZERO or is it NOT ZERO ?
William Elliot
Apr 14, 2017, 11:57:56 PM4/14/17
to
On Fri, 14 Apr 2017, Dan Christensen wrote:
> When it comes to calculus, JG is utterly clueless.
DC, who didn't take his promissed haitus, is clueles about sci.math
thinking it's a gossip forum. Good bye, you are blacklisted.
> When it comes to calculus, JG is utterly clueless.
DC, who didn't take his promissed haitus, is clueles about sci.math
thinking it's a gossip forum. Good bye, you are blacklisted.
Python
Apr 15, 2017, 4:50:50 AM4/15/17
to
0.1374 is a shortcut for a mapping from {1,2,3,4,...}
to {0,1,2,3,4,5,6,7,8,9} :
1 -> 1
2 -> 3
3 -> 7
4 -> 4
any n > 4 -> 0
similarly, 0.99999... is a shortcut for:
1 -> 9
2 -> 9
any n -> 9
being well defined, from this mapping the value it represents
is the sum for all n of the terms m(n)/10^n
On the contrary, when you write :
0.0000....001
there is NO WAY to consider it as a shortcut for such a mapping.
Ask yourself, Mr Seven, take 0.000...001, let n be ANY integer,
what is the nth digit of 0.000...001?
7777777
Apr 16, 2017, 12:51:56 AM4/16/17
to
your
f'(c) ={ f(c+n) - f(c-m) } / (m+n)
is the same as
f'(c) = {f(b) - f(a)}/(b-a)
because
c+n = b
c-m = a
meaning
b-a = c+n-(c-m) = c+n-c+m = m+n
7777777
Apr 16, 2017, 2:16:01 AM4/16/17
to
perjantai 14. huhtikuuta 2017 23.06.28 UTC+3 burs...@gmail.com kirjoitti:
> Thats actually not true. Can you please demonstrate
> step by step? I didn't use (0.000...001)^2 = 0
you did not use (0.000...001)^2 = 0 and therefore ended up with
> Thats actually not true. Can you please demonstrate
> step by step? I didn't use (0.000...001)^2 = 0
> x^2 + 2*x*dx + dx^2 - x^2
> ------------------------- = 2*x + dx
> dx
>
>
> If you use (0.000...001)^2 = 0, you don't compute
> (f(x+h)-f(x))/h. Instead you only compute f(x+h)-f(x)
no. you must be very confused to say so. If I use (0.000...001)^2 = 0 then
> (f(x+h)-f(x))/h. Instead you only compute f(x+h)-f(x)
I do compute (f(x+h)-f(x))/h as expected. I tried to understand your confusion,
to understand what you are doing. First of all I want to use dx instead of
h. So you are telling that I am computing only f(x+dx)-f(x). Also
f(x+dx)-f(x) = x^2 + 2xdx + (dx)^2 - x^2
>
> And for this computation you get:
>
> f(x+0.000...001)-f(x) = 2*x*0.000...001+0.000...001^2
>
> Now if you alow 0.000...001^2 = 0, you only get:
>
> g(x) = 2*x*0.000...001
>
> Which is still not f'(x).
It is very strange why do you keep telling this thing, and it is wrong, because that's not what I am doing. I am computing (f(x+dx)-f(x))/dx as expected.
Why did you omit division by dx?
>You are completely confused what f'(x) is and what it is not.
by dx claiming that it is me who is doing so when I use (0.000...001)^2 = 0,
but that's not something that I do. It is something that you do, and I don't
know your reasons for doing so except that you just made an error.
The correct way is, and this is what I do:
f'(x) = (f(x+dx)-f(x))/dx = (x^2 + 2xdx + (dx)^2 - x^2)/dx =
(2xdx + (dx)^2)/dx
now I use (dx)^2 = (0.000...001)^2 = 0
so that
(2xdx + (dx)^2)/dx = 2xdx/dx = 2x(0.000...001/0.000...001) = 2x
which is f'(x)
there is no division by 0 because 0.000...001 is nonzero
John Gabriel
Apr 16, 2017, 3:17:15 AM4/16/17
to
Yes, the mean value theorem is about a special kind of finite difference - one called a "natural arithmetic mean". You are too stupid to waste my time.
John Gabriel
Apr 16, 2017, 3:18:51 AM4/16/17
to
On Sunday, 16 April 2017 01:16:01 UTC-5, 7777777 wrote:
> Of course it is not f'(x) but that's not what I am doing.
99.99% of the time, you have no clue what you are saying. Pure idiot is all you are.
> Of course it is not f'(x) but that's not what I am doing.
7777777
Apr 16, 2017, 3:37:25 AM4/16/17
to
f'(c) = {Delta f}/{Delta x} <---------- finite difference quotient
is the definition of the derivative?
What is m ? What is n? Why do you want to look for them? Why do you want to look
for c? Why you never tell what is the derivative at point x, the point at which
we want to know the value of the derivative?
burs...@gmail.com
Apr 16, 2017, 4:52:48 AM4/16/17
to
Am Sonntag, 16. April 2017 08:16:01 UTC+2 schrieb 7777777:
> (2xdx + (dx)^2)/dx = 2xdx/dx = 2x(0.000...001/0.000...001) = 2x
But it is:
> (2xdx + (dx)^2)/dx = 2xdx/dx = 2x(0.000...001/0.000...001) = 2x
(2*x*dx+dx^2)/dx=2*x+dx
And you say:
(2*x*dx+dx^2)/dx=2*x
Therefore dx=0
But you say:
> there is no division by 0 because 0.000...001 is nonzero
But the above gives 0.000...001=0.
7777777
Apr 16, 2017, 3:53:59 PM4/16/17
to
Do you want to deny the infinitesimal difference ratio f'(x) = dy/dx ?
And if so, what's your suggestion what should be used instead of it?
I don't necessarily have answers to all of your questions, and you
should also yourself struggle finding them yourself, just like everyone else
here. Do you expect that I am some kind of answering machine?
I have only shown that the finite difference quotient is not enough to
define the derivative, and I have told why, and I have also told that
the infinitesimal difference ratio is needed. Now, if you want to deny
also this, what do you have got left?
John Gabriel
Apr 16, 2017, 4:00:10 PM4/16/17
to
On Sunday, 16 April 2017 14:53:59 UTC-5, 7777777 wrote:
> I have only shown that the finite difference quotient is not enough to
> define the derivative,
In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
> I have only shown that the finite difference quotient is not enough to
> define the derivative,
No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
> and I have told why, and I have also told that
> the infinitesimal difference ratio is needed.
7777777
Apr 16, 2017, 4:28:58 PM4/16/17
to
sunnuntai 16. huhtikuuta 2017 23.00.10 UTC+3 John Gabriel kirjoitti:
> On Sunday, 16 April 2017 14:53:59 UTC-5, 7777777 wrote:
>
>
> > I have only shown that the finite difference quotient is not enough to
> > define the derivative,
>
> In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
>
> No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
>
you can't seem to understand that the problem is, what you have is
> On Sunday, 16 April 2017 14:53:59 UTC-5, 7777777 wrote:
>
>
> > I have only shown that the finite difference quotient is not enough to
> > define the derivative,
>
> In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
>
> No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
>
f'(c) = { f(c+n)-f(c-m) } / (m+n)
and it is not f'(x)
You are misleading by telling that x = c, two different points are not the one
and the same point. Furthermore, you are unable to deal with the infinitesimal
range, I seems that just because you deny them so you can't work with them.
Dan Christensen
Apr 16, 2017, 6:03:05 PM4/16/17
to
On Friday, April 14, 2017 at 11:57:56 PM UTC-4, William Elliot wrote:
> On Fri, 14 Apr 2017, Dan Christensen wrote:
>
> > When it comes to calculus, JG is utterly clueless.
>
> DC, who didn't take his promissed haitus,
Bags are packed. One delay after another. Hurry up and wait. Sigh...
> On Fri, 14 Apr 2017, Dan Christensen wrote:
>
> > When it comes to calculus, JG is utterly clueless.
>
> DC, who didn't take his promissed haitus,
> thinking it's a gossip forum. Good bye, you are blacklisted.
Toodles.
Dan
John Gabriel
Apr 17, 2017, 8:41:18 AM4/17/17
to
On Sunday, 16 April 2017 15:28:58 UTC-5, 7777777 wrote:
> sunnuntai 16. huhtikuuta 2017 23.00.10 UTC+3 John Gabriel kirjoitti:
> > On Sunday, 16 April 2017 14:53:59 UTC-5, 7777777 wrote:
> >
> >
> > > I have only shown that the finite difference quotient is not enough to
> > > define the derivative,
> >
> > In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
> >
> > No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
> >
>
> you can't seem to understand that the problem is, what you have is
>
> f'(c) = { f(c+n)-f(c-m) } / (m+n)
>
> and it is not f'(x)
No moron. It is you who do not understand because you are fixated on x being located at the beginning of an interval. Your obtuseness knows no limits.
> sunnuntai 16. huhtikuuta 2017 23.00.10 UTC+3 John Gabriel kirjoitti:
> > On Sunday, 16 April 2017 14:53:59 UTC-5, 7777777 wrote:
> >
> >
> > > I have only shown that the finite difference quotient is not enough to
> > > define the derivative,
> >
> > In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
> >
> > No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
> >
>
> you can't seem to understand that the problem is, what you have is
>
> f'(c) = { f(c+n)-f(c-m) } / (m+n)
>
> and it is not f'(x)
f'(c) = { f(c+n)-f(c-m) } / (m+n) is the same as f'(x) = { f(x+n)-f(x-m) } / (m+n) and in this case x=c but x is in (x-m,x+n).
burs...@gmail.com
Apr 17, 2017, 8:56:12 AM4/17/17
to
Thats not the mean value theorem, and its not true
for arbitrary m,n. Trivially if you only consider
m,n so that f'(x) = { f(x+n)-f(x-m) } / (m+n) is
true, then also f'(c) = { f(c+n)-f(c-m) } / (m+n)
with x = c is true.
for arbitrary m,n. Trivially if you only consider
m,n so that f'(x) = { f(x+n)-f(x-m) } / (m+n) is
true, then also f'(c) = { f(c+n)-f(c-m) } / (m+n)
with x = c is true.
7777777
Apr 17, 2017, 12:44:07 PM4/17/17
to
maanantai 17. huhtikuuta 2017 15.41.18 UTC+3 John Gabriel kirjoitti:
> On Sunday, 16 April 2017 15:28:58 UTC-5, 7777777 wrote:
> > sunnuntai 16. huhtikuuta 2017 23.00.10 UTC+3 John Gabriel kirjoitti:
> > > On Sunday, 16 April 2017 14:53:59 UTC-5, 7777777 wrote:
> > >
> > >
> > > > I have only shown that the finite difference quotient is not enough to
> > > > define the derivative,
> > >
> > > In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
> > >
> > > No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
> > >
> >
> > you can't seem to understand that the problem is, what you have is
> >
> > f'(c) = { f(c+n)-f(c-m) } / (m+n)
> >
> > and it is not f'(x)
>
> No moron. It is you who do not understand because you are fixated on x being located at the beginning of an interval. Your obtuseness knows no limits.
>
> f'(c) = { f(c+n)-f(c-m) } / (m+n) is the same as f'(x) = { f(x+n)-f(x-m) } / (m+n) and in this case x=c but x is in (x-m,x+n)
What is f'(x) for f(x) = sin(x)?
> On Sunday, 16 April 2017 15:28:58 UTC-5, 7777777 wrote:
> > sunnuntai 16. huhtikuuta 2017 23.00.10 UTC+3 John Gabriel kirjoitti:
> > > On Sunday, 16 April 2017 14:53:59 UTC-5, 7777777 wrote:
> > >
> > >
> > > > I have only shown that the finite difference quotient is not enough to
> > > > define the derivative,
> > >
> > > In the mainstream calculus it is not only NOT enough, it is bullshit. In the New Calculus, all that is required is the finite difference quotient.
> > >
> > > No one can rubbish my definition: f'(x) = { f(x+n)-f(x-m) } / (m+n)
> > >
> >
> > you can't seem to understand that the problem is, what you have is
> >
> > f'(c) = { f(c+n)-f(c-m) } / (m+n)
> >
> > and it is not f'(x)
>
> No moron. It is you who do not understand because you are fixated on x being located at the beginning of an interval. Your obtuseness knows no limits.
>
> f'(c) = { f(c+n)-f(c-m) } / (m+n) is the same as f'(x) = { f(x+n)-f(x-m) } / (m+n) and in this case x=c but x is in (x-m,x+n)
John Gabriel
Apr 17, 2017, 5:07:57 PM4/17/17
to
7777777
Apr 18, 2017, 1:25:56 AM4/18/17
to
dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
dear genius, am I right to assume that f'(x) = dy/dx is not equal to
{Delta f}/ {Delta x} ?
dear genius, what is c?
dear genius, am I right to assume that f'(c) = {Delta f}/ {Delta x} ?
dear genius, what is m, what is n?
dear genius, why do you want to go looking for c?
dear genius, why do you want to go looking for m and n?
dear genius, am I right to assume that the point x is not the same point as the point c if they are two different points?
Dear genius, why do you keep insisting that two different points x and c
are one and the same point contrary to the facts?
Dear genius, would you nevertheless like to write that x=c?
Dear genius, would this not lead to
dy/dx = {Delta f}/ {Delta x} = (f(x+h)-f(x))/h ?
Dear genius, what do you tell to people who say that it is correct to tell that dy/dx = Lim_Delta x ->0 {Delta f}/ {Delta x}= Lim_h->0(f(x+h)-f(x))/h and not
dy/dx = {Delta f}/ {Delta x} = (f(x+h)-f(x))/h?
Dear genius, do you think that
f'(x) = {Delta f}/ {Delta x} = (f(x+h)-f(x))/h
makes sense?
Dear genius, don't you think that we would need to write:
f'(c) = (f(x+h)-f(x))/h
And say there is c in (x,x+h)?
John Gabriel
Apr 18, 2017, 10:39:36 AM4/18/17
to
On Tuesday, 18 April 2017 00:25:56 UTC-5, 7777777 wrote:
> >
> > You've been shown many times you moron.
>
> dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
> dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
> dear genius, am I right to assume that f'(x) = dy/dx is not equal to
> {Delta f}/ {Delta x} ?
Yes.
> >
> > You've been shown many times you moron.
>
> dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
> dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
> dear genius, am I right to assume that f'(x) = dy/dx is not equal to
> {Delta f}/ {Delta x} ?
> dear genius, what is c?
> dear genius, am I right to assume that f'(c) = {Delta f}/ {Delta x} ?
> dear genius, what is m, what is n?
> dear genius, why do you want to go looking for c?
> dear genius, why do you want to go looking for m and n?
> dear genius, am I right to assume that the point x is not the same point as the point c if they are two different points?
> Dear genius, why do you keep insisting that two different points x and c
> are one and the same point contrary to the facts?
> f'(x) = {Delta f}/ {Delta x} = (f(x+h)-f(x))/h makes sense?
> Dear genius, don't you think that we would need to write:
>
> f'(c) = (f(x+h)-f(x))/h
>
> And say there is c in (x,x+h)?
f'(c) = { f(c+n)-f(c-m) } / (m+n)
burs...@gmail.com
Apr 18, 2017, 11:28:50 AM4/18/17
to
But the below is not correct, m,n are not independent.
Take f(x)=sin(x) at x=pi/6 and n=pi/30 and m=pi/30, does
the equality hold? No.
sin(pi/6+pi/30) = 0.58778525229247312916870595463907
sin(pi/6-pi/30) = 0.41887902047863909846168578443727
pi/30+pi/30 = 0.20943951023931954923084289221863
cos(pi/6) = 0.86602540378443864676372317075294
f(x+m)-f(x-m)
------------- = 0.80646785136591709963875719845893 <> f'(x)
m+n
But choosing m,n so that the equation holds doesn't
make any sense, how do you have f'(x) from the beginning?
By Geogebra auto diff CAS? Not very helpful...
Take f(x)=sin(x) at x=pi/6 and n=pi/30 and m=pi/30, does
the equality hold? No.
sin(pi/6+pi/30) = 0.58778525229247312916870595463907
sin(pi/6-pi/30) = 0.41887902047863909846168578443727
pi/30+pi/30 = 0.20943951023931954923084289221863
cos(pi/6) = 0.86602540378443864676372317075294
f(x+m)-f(x-m)
------------- = 0.80646785136591709963875719845893 <> f'(x)
m+n
But choosing m,n so that the equation holds doesn't
make any sense, how do you have f'(x) from the beginning?
By Geogebra auto diff CAS? Not very helpful...
7777777
Apr 19, 2017, 1:36:52 AM4/19/17
to
tiistai 18. huhtikuuta 2017 17.39.36 UTC+3 John Gabriel kirjoitti:
> On Tuesday, 18 April 2017 00:25:56 UTC-5, 7777777 wrote:
> >
> > dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
> > dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
> > dear genius, am I right to assume that f'(x) = dy/dx is not equal to
> > {Delta f}/ {Delta x} ?
>
> Yes.
So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
> On Tuesday, 18 April 2017 00:25:56 UTC-5, 7777777 wrote:
> >
> > dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
> > dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
> > dear genius, am I right to assume that f'(x) = dy/dx is not equal to
> > {Delta f}/ {Delta x} ?
>
> Yes.
should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
But you used to tell that f'(0) does not exist, because according to you inflection points have no derivative, so did you change your mind?
Or would you again like to invent an exception to the rules, in order to deny the undeniable?
John Gabriel
Apr 19, 2017, 8:27:15 AM4/19/17
to
On Wednesday, 19 April 2017 00:36:52 UTC-5, 7777777 wrote:
> tiistai 18. huhtikuuta 2017 17.39.36 UTC+3 John Gabriel kirjoitti:
> > On Tuesday, 18 April 2017 00:25:56 UTC-5, 7777777 wrote:
> > >
> > > dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
> > > dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
> > > dear genius, am I right to assume that f'(x) = dy/dx is not equal to
> > > {Delta f}/ {Delta x} ?
> >
> > Yes.
>
> So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
> should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
No stupid. I don't agree. There are no derivatives possible at points of inflection.
> tiistai 18. huhtikuuta 2017 17.39.36 UTC+3 John Gabriel kirjoitti:
> > On Tuesday, 18 April 2017 00:25:56 UTC-5, 7777777 wrote:
> > >
> > > dear genius, am I right to assume that f'(x) = cos(x) for f(x) = sin(x) ?
> > > dear genius, am I right to assume that f'(x) = dy/dx for f(x) = sin(x) ?
> > > dear genius, am I right to assume that f'(x) = dy/dx is not equal to
> > > {Delta f}/ {Delta x} ?
> >
> > Yes.
>
> So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
> should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
> But you used to tell that f'(0) does not exist, because according to you inflection points have no derivative, so did you change your mind?
> Or would you again like to invent an exception to the rules, in order to deny the undeniable?
burs...@gmail.com
Apr 19, 2017, 10:34:10 AM4/19/17
to
But teachel, you said all men have too many holes
in head, and that this cancelation and equality is
korrekt:
P = P * (x-k) / (x-k) /* JGs rule */
Evem when x - k = 0. Now I take x=m and k=-n, and
compute your secant limes for an inflection point.
Take the case f(x)=x^3. We have:
f(x+m)-f(x-n) x^3+3*m*x^2+3*m^2*x+m^3-x^3+3*n*x^2-3*n^2*x-n^3
--------------- = -----------------------------------------------
m+n m+n
3*(m+n)*x^2-3*(m+n)*(m-n)*x+(m+n)*(m^2+m*n+n^2)
= -----------------------------------------------
m+n
Now we apply JGs cancellation rule:
= 3*x^2 - 3*(m-n)*x + (m^2+m*n+n^2)
Now if we evaluate this for m=0,n=0 which is allowed
by JGs cancellation rule, we get:
= 3*x^2 - 3*0*x + 0
= 3*x^2
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
P.S.: The stuff works for any f that has a taylor expansion.
But JG didn't show us yet how f(x)=e^(-1/x^2) works.
in head, and that this cancelation and equality is
korrekt:
P = P * (x-k) / (x-k) /* JGs rule */
Evem when x - k = 0. Now I take x=m and k=-n, and
compute your secant limes for an inflection point.
Take the case f(x)=x^3. We have:
f(x+m)-f(x-n) x^3+3*m*x^2+3*m^2*x+m^3-x^3+3*n*x^2-3*n^2*x-n^3
--------------- = -----------------------------------------------
m+n m+n
3*(m+n)*x^2-3*(m+n)*(m-n)*x+(m+n)*(m^2+m*n+n^2)
= -----------------------------------------------
m+n
Now we apply JGs cancellation rule:
= 3*x^2 - 3*(m-n)*x + (m^2+m*n+n^2)
Now if we evaluate this for m=0,n=0 which is allowed
by JGs cancellation rule, we get:
= 3*x^2 - 3*0*x + 0
= 3*x^2
JGs cancellation rule didn't care whether x=0, according
to him it holds for any P, and we anyway apply JGs
canncellation rule for the case where x=m.
Therefore we get f'(x)=3*x^2 even for inflection points.
P.S.: The stuff works for any f that has a taylor expansion.
But JG didn't show us yet how f(x)=e^(-1/x^2) works.
burs...@gmail.com
Apr 19, 2017, 10:42:47 AM4/19/17
to
Corr.:
Sorry, I made two little sign errors:
f(x+m)-f(x-n) x^3+3*m*x^2+3*m^2*x+m^3-x^3+3*n*x^2-3*n^2*x+n^3
Sorry, I made two little sign errors:
f(x+m)-f(x-n) x^3+3*m*x^2+3*m^2*x+m^3-x^3+3*n*x^2-3*n^2*x+n^3
--------------- = -----------------------------------------------
m+n m+n
3*(m+n)*x^2-3*(m+n)*(m-n)*x+(m+n)*(m^2-m*n+n^2)
m+n m+n
= -----------------------------------------------
m+n
Now we apply JGs cancellation rule:
= 3*x^2 - 3*(m-n)*x + (m^2-m*n+n^2)
m+n
Now we apply JGs cancellation rule:
7777777
Apr 19, 2017, 3:09:15 PM4/19/17
to
keskiviikko 19. huhtikuuta 2017 17.34.10 UTC+3 burs...@gmail.com kirjoitti:
>
> JGs cancellation rule didn't care whether x=0, according
> to him it holds for any P, and we anyway apply JGs
> canncellation rule for the case where x=m.
>
> Therefore we get f'(x)=3*x^2 even for inflection points.
JG does not agree: There are no derivatives possible at points of inflection.
>
> JGs cancellation rule didn't care whether x=0, according
> to him it holds for any P, and we anyway apply JGs
> canncellation rule for the case where x=m.
>
> Therefore we get f'(x)=3*x^2 even for inflection points.
According to him you are an idiot and the one who must change his mind because
he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
>
> P.S.: The stuff works for any f that has a taylor expansion.
> But JG didn't show us yet how f(x)=e^(-1/x^2) works.
>
> Am Mittwoch, 19. April 2017 14:27:15 UTC+2 schrieb John Gabriel:
> > > So you seem to agree that f'(x) = cos(x) for f(x) = sin(x), good. Then you
> > > should agree also that f'(0) = 1 because f'(0) = cos(0) = 1, right?
> >
> > No stupid. I don't agree. There are no derivatives possible at points of inflection.
John Gabriel
Apr 20, 2017, 11:50:02 AM4/20/17
to
On Wednesday, 19 April 2017 14:09:15 UTC-5, 7777777 wrote:
> keskiviikko 19. huhtikuuta 2017 17.34.10 UTC+3 burs...@gmail.com kirjoitti:
> >
> > JGs cancellation rule didn't care whether x=0, according
> > to him it holds for any P, and we anyway apply JGs
> > canncellation rule for the case where x=m.
> >
> > Therefore we get f'(x)=3*x^2 even for inflection points.
>
> JG does not agree: There are no derivatives possible at points of inflection.
> According to him you are an idiot and the one who must change his mind because
> he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
>
> Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
> keskiviikko 19. huhtikuuta 2017 17.34.10 UTC+3 burs...@gmail.com kirjoitti:
> >
> > JGs cancellation rule didn't care whether x=0, according
> > to him it holds for any P, and we anyway apply JGs
> > canncellation rule for the case where x=m.
> >
> > Therefore we get f'(x)=3*x^2 even for inflection points.
>
> JG does not agree: There are no derivatives possible at points of inflection.
> According to him you are an idiot and the one who must change his mind because
> he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
>
> Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
Hey stupid, what is the derivative of f(x)=sqrt(1-x^2) ?
It is this: f'(x)= -x/sqrt(1-x^2)
Now tell me you big baboon, what is the derivative at f'(1) and f'(-1)?
Do you see there is a difference between the general derivative f'(x)= -x/sqrt(1-x^2) and the derivative at a particular point say c, that is, f'(c) ?
Of course not, you are a nincompoop!!
7777777
Apr 20, 2017, 12:02:19 PM4/20/17
to
torstai 20. huhtikuuta 2017 18.50.02 UTC+3 John Gabriel kirjoitti:
> On Wednesday, 19 April 2017 14:09:15 UTC-5, 7777777 wrote:
> > keskiviikko 19. huhtikuuta 2017 17.34.10 UTC+3 burs...@gmail.com kirjoitti:
> > >
> > > JGs cancellation rule didn't care whether x=0, according
> > > to him it holds for any P, and we anyway apply JGs
> > > canncellation rule for the case where x=m.
> > >
> > > Therefore we get f'(x)=3*x^2 even for inflection points.
> >
> > JG does not agree: There are no derivatives possible at points of inflection.
> > According to him you are an idiot and the one who must change his mind because
> > he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
> >
> > Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
>
>
> I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
>
> Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
hey apeman, did you think that no-one would notice your lie:
> On Wednesday, 19 April 2017 14:09:15 UTC-5, 7777777 wrote:
> > keskiviikko 19. huhtikuuta 2017 17.34.10 UTC+3 burs...@gmail.com kirjoitti:
> > >
> > > JGs cancellation rule didn't care whether x=0, according
> > > to him it holds for any P, and we anyway apply JGs
> > > canncellation rule for the case where x=m.
> > >
> > > Therefore we get f'(x)=3*x^2 even for inflection points.
> >
> > JG does not agree: There are no derivatives possible at points of inflection.
> > According to him you are an idiot and the one who must change his mind because
> > he wrote that "the New Calculus does not change you moron, it is the first and only rigorous formulation of calculus in history."
> >
> > Nevertheless, I don't know how he can both agree that the derivative of sin(x) is cos(x) and deny it at the same time.
>
>
> I have never denied it. But Jack Huizenga and David Joyce and Anders Kaesorg do not believe that the derivative of sin x is always cos x.
>
> Where I am concerned, the derivative of sin x is always cos x and you are just an ape attributing lies to me.
How can the derivative of sin(x) be always cos(x) if the derivative of
sin(x) is not cos(x) at the point x = 0 ?
John Gabriel
Apr 20, 2017, 3:11:33 PM4/20/17
to
7777777
Apr 20, 2017, 3:46:41 PM4/20/17
to
x = 0 ?
Do you think that the derivative of sin(x) at
x = 0 is undefined, just like the derivative of f(x)=sqrt(1-x^2) at x = 1
and x = -1 ?
John Gabriel
Apr 20, 2017, 7:43:59 PM4/20/17
to
If Newton thought the way idiots of today think, there wouldn't even be the finite difference quotient f(x+h)-f(x) / h.
That "limit" of that difference quotient which is a secant line that most resembles a tangent line, is what is called the "derivative".
7777777
Apr 21, 2017, 12:12:58 AM4/21/17
to
You think that there can't be a tangent line at an inflection point.
The truth is only that there is no secant at an inflection point, therefore your
"method" does not work there. It does not mean that there is no tangent line at an inflection point. You have incorrectly assumed that because
your "method" does not work at an inflection point, you need to re-define what is the tangent line, without considering that it is your "method" which is wrong. You don't have a correct definition for the derivative, as I have said
over and over again, you only have the mean value theorem, that is the finite
difference quotient, but you need the infinitesimal difference ratio dy/dx.
Your mistake derives from your rejection of infinitesimals.
You have completely ignored the possibility that you can be wrong.
Instead, you have chosen to attack and ridicule people who tell the truth.
>
> If Newton thought the way idiots of today think, there wouldn't even be the finite difference quotient f(x+h)-f(x) / h.
>
> That "limit" of that difference quotient which is a secant line that most resembles a tangent line, is what is called the "derivative".
dy/dx = Lim h-> (f(x+h)-f(x))/h means that the finite difference quotient
f(x+h)-f(x)/h is never equal to dy/dx.
In other words, dy/dx is never finite, it is never equal to a finite difference quotient.
Because you have refused to deal with dy/dx, and instead work only with
the finite difference quotient f(x+h)-f(x)/h, you have introduced an error
into your calculation. That error manifests in your attempt to calculate
the derivative of sin(x) at x=0. You can never get the correct answer,
instead you start looking for the point c. The error disappears only
when the range becomes infinitesimal, also with the introduction of the infinitesimal dx.
John Gabriel
Apr 21, 2017, 10:23:14 AM4/21/17
to
Grabiner writes:
"How then was the tangent to be defined at the point (0,O) for a curve like y =x^3 (FIGUREl), or to a point on a curve with many turning points (FIGURE 2)? "
Page 2 of
https://www.maa.org/sites/default/files/0025570x04690.di021131.02p02223.pdf
Grabiner is part of the BIG STUPID and in her September 1983 publication, she admitted that it was not possible to have a tangent line at x=0 when f(x)=x^3. If it were, then there wouldn't be a need to "define" a tangent.
See, there's the first problem with morons of the BIG STUPID: if the facts don't fit the theory, change the facts.
By definition, a tangent line intersects a curve in one point, extends finitely to both sides and crosses the curve nowhere.
Ironically, in their redefinition, the BIG STUPID used the difference quotient to define the tangent line slope, not noticing the circularity of their actions. Till that point, a derivative was derived from a tangent line, but now the tangent line was derived from the derivative. Circular? Of course.
Grabiner claims the Greeks didn't study many curves. Evidently she never read the Works of Archimedes - just like most of you morons haven't.
Pull up your pants and give your mouth a chance you moron. You do not have even a tiny bit of my intelligence.
The rest of your comment is shit.
Eram semper recta
Mar 19, 2023, 10:36:45 AM3/19/23
to
On Tuesday, 28 March 2017 at 21:32:50 UTC-4, John Gabriel wrote:
> 9/21/2016 A Harvard alumnus comments! | The New Calculus
>
> Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.
>
> The question was:
>
> “What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.
>
> Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.
>
> One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.
>
> In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!
>
> There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga
>
> The great mathematics historian Carl Boyer had this to say:
> "Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.
>
> That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer
>
> I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.
>
> Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga
>
> One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
> don’t exist.
>
> Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga
>
> The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.
>
> Needless to say, if calculus were already extremely well understood, we would not have other mathematics PhDs making statements as follows:
>
> Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
> it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)
>
> Knisley goes on to say:
>
> However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)
>
> As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.
>
> There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga
>
> It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga
>
> Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.
>
> In the following comment I debunk the concept of Dedekind cuts and Cauchy sequences:
>
> https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU
>
> Our buffoon continues:
>
> The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga
>
> It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.
>
> In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.
>
> His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga
>
> Of course we know the slope of the tangent line, provided we know the slope of a parallel secant line. This is grade 8 mathematics! What he states in parenthesis is even more amusing. It demonstrates clearly that he, like many of his colleagues never understood calculus:
>
> If a given function is not continuous and smooth, then any of the methods of calculus are null and void.
>
> One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
> tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.
>
> He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga
>
> I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
> to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.
>
> It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.
>
> That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article
>
> https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E
>
> The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!
>
> Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga
>
> Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.
>
> The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
> ceased to continue, when he could no longer understand what he was reading.
>
> The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.
>
> (*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.
>
> And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.
>
> This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.
>
>
> Comments are unwelcome and will be ignored.
>
> Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
>
> gils...@gmail.com (MIT)
> huiz...@psu.edu (HARVARD)
> and...@mit.edu (MIT)
> david....@math.okstate.edu (David Ullrich)
> djo...@clarku.edu
> mar...@gmail.com
I wonder what these fucking morons might have to say about my historic geometric theorem now? Chuckle.
https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020
Still think that calculus requires "limits", you dumb fucks?!
> 9/21/2016 A Harvard alumnus comments! | The New Calculus
>
> Just a few weeks ago, some ignoramus posted a question on that ridiculous site Quora.
>
> The question was:
>
> “What do people think of John Gabriel’s New Calculus, which claims it is the only rigorous formulation of calculus”.
>
> Quora is a question and answer site, that is run by a group of mainstream academics and their sock puppets. They control both the questions and answers. Those who post questions or comments with opposing views, have both their questions and comments mercilessly edited, and deleted if need be. In a sense, it’s Wikipedia-esque, in that the questions and answers are the views of mainstream academics. Nothing that is different or new is tolerated.
>
> One of the fools who sits atop the trash heap, is one Prof. Jack Huizenga, a Harvard alumnus and mathematics teacher at Illinois University. The following link contains a snapshot of Huizenga’s comment: The ridiculous comment at the ridiculous site Quora.
>
> In this very first post, I will address the comments made by this ignoramus (Huizenga) and horrify those, who will realise that this dimwit actually teaches mathematics!
>
> There has been a completely rigorous notion of limit for well over a hundred years. – Huizenga
>
> The great mathematics historian Carl Boyer had this to say:
> "Cauchy had stated in his Cours d'analyse that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since a limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted.
>
> That is, one cannot define the number 'square root of 2', as the limit of the sequence 1, 1.4, 1.41,1.414, ... because to prove that this sequence has a limit one must assume, in view of the definitions of limits and convergence, the existence of this number as previously demonstrated or defined. Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit." - The History of Calculus and its Conceptual Development' (Page. 281) Carl B. Boyer
>
> I doubt that Huizenga ever read this book, and if he did, there is no doubt he did not understand it. Not only has the notion of limit never been rigorous, but it was never challenged by any intelligent mathematician.
>
> Rigorous treatments of infinitesimals are a bit more tricky, but have also been made. – Huizenga
>
> One can only assume the moron is referring to the abortion of non-standard analysis, by a failed Jewish mathematician called Abraham Robinson. Even today, there are many academics (including PhDs) who do not accept non-standard analysis. In my opinion, it is pure rot because infinitesimals
> don’t exist.
>
> Reformulations of calculus as attempted in New Calculus, are essentially no interest to mathematicians, as the field of calculus is already extremely well understood and rigorous. – Huizenga
>
> The buffoon wrongly assumes that the New Calculus is only a reformulation, which is evidently false. He goes on to talk about what interests mathematicians, but how could he know? Huizenga is not a mathematician, he is a teacher with no great works behind his name, unless of course one calls his juvenile papers on algebraic geometry a work of any kind.
>
> Needless to say, if calculus were already extremely well understood, we would not have other mathematics PhDs making statements as follows:
>
> Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don’t get it the first time around, our colleagues don’t get it, and our students are still not getting
> it. It’s no wonder that one of the most common occurrences in higher education is that of a non mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction. – Knisley (Crisis in calculus education)
>
> Knisley goes on to say:
>
> However, in the calculus curriculum, many of the associations are circular. All too often a given concept is associated with a concept that is defined in terms of the original concept. Such connections increase the complexity of a concept without shedding any insight on the concept itself. Not surprisingly, concepts motivated with circular associations are the ones most often memorized with little or no comprehension. – Knisley (Facing the crisis in calculus education)
>
> As for calculus being rigorous, we will hold off on that one until you have read and studied my New Calculus.
>
> There is no point in trying to remove limits or infinitesimals from a discussion of calculus. – Huizenga
>
> It’s easy to see that our moron academic knows nothing about Newton’s calculus and how he (Newton) arrived at the knowledge of it. The author’s main objection to the standard treatment of calculus via real analysis seems to be that he does not understand it. – Huizenga
>
> Well, that’s certainly news to me. I have never made such a claim on the internet or anywhere else. I reject real analysis, not because I don’t understand it (few can ever understand it as well as I), but because it deals with a topic about a non-existent concept – the real number. Real numbers do not exist, because irrational numbers do not exist. In fact, until I arrived on the scene, no one before me and after Euclid, understood what is a number.
>
> In the following comment I debunk the concept of Dedekind cuts and Cauchy sequences:
>
> https://drive.google.com/open?id=0B-mOEooW03iLSTROakNyVXlQUEU
>
> Our buffoon continues:
>
> The errors in New Calculus are too numerous for it, to be worth going through the whole text. – Huizenga
>
> It’s rather funny that he should say this, because he goes on to show that there is not a single error. Unsurprisingly, the baboon dismissed the rest of the text because he could not understand it up till that point. Huizenga mentions that one has to consider an (m,n) pair which is outright false.
>
> In fact, if he had only continued to study and reread the text carefully, he would have soon realised that the values of m and n play no role in that of the gradient. Tsk, tsk.
>
> His definition therefore presupposes that we know the slope of the tangent line (and that we have a notion of a tangent line to a graph of an arbitrary function) and merely computes the slope of a parallel secant line. This is incredibly circular. – Huizenga
>
> Of course we know the slope of the tangent line, provided we know the slope of a parallel secant line. This is grade 8 mathematics! What he states in parenthesis is even more amusing. It demonstrates clearly that he, like many of his colleagues never understood calculus:
>
> If a given function is not continuous and smooth, then any of the methods of calculus are null and void.
>
> One, and only one tangent line exists at every point, provided the function is continuous and smooth, and a given point is not a point of inflection, that is, only half-tangent lines are possible at points of inflection. Our moron then continues to quote the example of the cubic, where there is no
> tangent line at x=0 (because an inflection point exists at x=0). Unlike Newton's flawed formulation, the New Calculus handles this correctly.
>
> He further goes to great length to try and convince the reader that he can actually divide the numerator in his difference quotient by m+n. – Huizenga
>
> I don’t go to any great lengths. The proof that every term in the numerator has a factor of m+n can be shown by a high school student. It requires no special knowledge. The first page of the New Calculus website http://thenewcalculus.weebly.com has many examples on this, and includes links
> to dynamic Geogebra applets which prove conclusively that it is based on sound analytic geometry.
>
> It is legal to do this in the New Calculus, but illegal in Cauchy's flawed formulation.
>
> That m+n is a factor of every term in the numerator finite difference f(x+n)-f(x-m) can be seen by anyone with a modicum of intelligence, but our chump Huizenga lacks even this. The chump way to see this fact, is to investigate actual functions using the finite difference, and it is soon realised, that this fact is true for all functions. It gets slightly more complicated in the case of terms containing only m and/or n, but this is easy to prove by mathematical induction and constructive proofs, as I have shown in my article
>
> https://drive.google.com/open?id=0B-mOEooW03iLWldTU1ZkTDVQR0E
>
> The easiest proof is given with the equation of a straight line, say f(x)=kx+p. We have from the New Calculus derivative definition: f ' (x) = {k(x+n)+p - [k(x-m)+p] } / (m+n) =k(m+n) / (m+n) = k. Observe that m and n play no role in the value of k which is the gradient. What amuses me, is that so many grade 8 students understand this, and most PhD chumps just don't get it!
>
> Of course the New Calculus is likely to be revised over time to address concerns brought up by people. As this happens, I am confident it will look more and more like standard calculus (or the theory of the symmetric derivative which is closely related). – Huizenga
>
> Another presumptuous claim by our buffoon Huizenga, because the New Calculus has never been revised and there are no plans to revise it whatsoever. There is no need to revise theory that is based on well-defined concepts. It will stand the test of time, just as Euclid’s Elements did.
>
> The statement in parenthesis is quite amusing because it once again demonstrates the lack of understanding displayed by Huizenga. The New Calculus is not just about a new derivative definition, but also about a new integral definition, and much more (*). He might have gotten to that information had he continued studying the text. But ‘open-minded’ academic that he is not, he
> ceased to continue, when he could no longer understand what he was reading.
>
> The new calculus derivative definition has nothing in common with the symmetric derivative which Huizenga clearly does not understand. The symmetric derivative requires no special relationship exist between m and n. In fact, the symmetric derivative is used mostly in numeric differentiation.
>
> (*) There are many new methods and theorems in the New Calculus that are not possible using Newton's flawed formulation.
>
> And that covers his comment at Quora. In fact, he proves by all his statements, that there are no errors in the New Calculus, only serious issues in his ability to comprehend.
>
> This same moron thinks that the derivative of sin(x) is not always cos(x). He also fancies that the sine function can take degrees as input, which is outright false. The trigonometric ratios operate only on radian input.
>
>
> Comments are unwelcome and will be ignored.
>
> Posted on this newsgroup in the interests of public education and to eradicate ignorance and stupidity from mainstream mythmatics.
>
> gils...@gmail.com (MIT)
> huiz...@psu.edu (HARVARD)
> and...@mit.edu (MIT)
> david....@math.okstate.edu (David Ullrich)
> djo...@clarku.edu
> mar...@gmail.com
I wonder what these fucking morons might have to say about my historic geometric theorem now? Chuckle.
https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020
Still think that calculus requires "limits", you dumb fucks?!
Eram semper recta
Mar 19, 2023, 10:38:00 AM3/19/23
to
https://www.academia.edu/98758410/Was_Newtons_and_Leibnizs_method_of_setting_h_0_valid_The_answer_is_YES
Mostowski Collapse
Mar 19, 2023, 10:51:45 AM3/19/23
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I though you invented New Calculus 40-50 years ago,
when you were a tiny wanker?
when you were a tiny wanker?
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