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November 9th, 2017, 05:08 PM   #61
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Quote:
Originally Posted by romsek View Post
It does however provide relief from ... and halitosis!
It doesn't.

-Dan
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November 9th, 2017, 05:19 PM   #62
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Originally Posted by arithmo View Post
Summarizing, you both have not seen such an elemental and powerful arithmetical operation in any math text. Just a comment,
you are not obligued to show how your feelings and nerve have been touched by that, and your feelings are not relevant to this case, at all.

There are many others reading this who surely knows more on the history and philosophy of maths.
I've been looking over things and I have to agree with JeffM1 and romsek, ignoring the rudeness anyway. You have posted all of this on a public forum. And you are still refusing to come out with statements of how to use this stuff in a general way and you haven't provided any proof of your claims except in a few specific cases. I've always had this problem with your posts and I haven't seen anything to indicate that you are going to change that. So why are you posting here if you aren't going to share how you have proven your comments about, at least, convergence?

We don't really care about when this method was first expounded upon. That is a trivial thing. What we are interested in is the method and how and why it works. I'll say it again: You haven't shown us that you can prove your claims. That is what we are looking for. And until you do you will likely be peppered with sarcasm from members who think you are nothing but a troll.

-Dan
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November 12th, 2017, 02:29 PM   #63
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Quote:
Originally Posted by topsquark View Post
... So why are you posting here if you aren't going to share how you have proven your comments about, at least, convergence?
We don't really care about when this method was first expounded upon. That is a trivial thing. What we are interested in is the method and how and why it works. I'll say it again: You haven't shown us that you can prove your claims. That is what we are looking for. And until you do you will likely be peppered with sarcasm from members who think you are nothing but a troll.
-Dan
etc. etc. etc.
Ok, Summarizing another one who can not show any math text --since antiquity up to now -- that deals with this extremely simple, trivial and powerful operation:
THE ARITHMONIC MEAN which produces all those high-order iterated functions that have been unjustifiably consecrated as the exclusive Superior creation of the Superb Infinitesimal Calculus (as unjustifiably asserted in so many math booktexts and peer-review journals).

Now, Who is the troll? I think that those who only ask for answers, never responding a very simple question.

So, Neeext...

It is good for young students to see this discussion because this is the customary grandstanding behavior of many math-Egos,
firstly they make jokes, secondly they just deny and state that the thing does not work, suddenly,
they realize the thing certainly "might" work so they start asking you to write a book just for them -- "apparently"
because they do not understand such an "extremely complicated" thing and they cannot imagine how the convergence could be ever proved !!! --
(Very funny, indeed),
Finally and because they certainly know the thing is certainly a trivial and an extremely powerful arithmetical operation,
then they start calling you a Troll, this way they manage to kick you off the forum without having to recognize that they never see
such a trivial and powerful thing in any math text since ancient times up to now.

Now, based on the number of views in this thread, It is fairly clear that only two or three swollen egos are NOT the whole audience,
and because I know there are many other humble people reading this, most of them being young students, I repeat again:

Quote:
Originally Posted by arithmo View Post

Ok, fine, I will talk about the corresponding convergence proofs (which you certainly can infer are truly
straightforward), however, it is not fair to talk about that without previously
talking about the Arithmonic Mean and the history of maths.

$\dfrac{{{w_{3}(w_{1}+2w_{2})}}}{{{w_{2}+2w_{3}}}} ,\ \ \dfrac{{{w_{2}(w_{3}+{%
2w_{1}})}}}{{{w_{1}+2w_{2}}}},\dfrac{{{w_{1}(w_{2} +2w_{3})}}}{{{w_{3}+2w_{1}}%
}} $

Indeed, the Arithmonic Mean is quite an extremely simple Arithmetical Operation which
does not appear in any Arithmetic treatise since antiquity up to now, so we may wonder how
many other elementary and useful things do not appear in those arithmetic treatises.

I mean, we are talking about an elemental arithmetic operation which generates
all those iterating functions that have been consecrated as an exclusive achievement
of the glorious and superb Infinitesimal Calculus and that allows to produce
a true definition of the Algebraic Irrationals and their arithmetical operations.

Thus, I think that young students deserve an explanation, mainly because
teachers tend to overwhelm them specifically with the subject of the completeness,
excellence and extreme rigor of mathematics.

So, what happened here?

We are talking, at least, about 3000 thousand of years, and tens of thousands of Arithmetic treatises.

Let's talk about that, in the same way you all, day by day, talk about completeness, excellence
and extreme rigor of mathematics.

Dear math teacher, please show young students any reference on the Arithmonic Mean from any
arithmetic treatise, since ancient times up to now.


Let young students read though this discussion because history of mathematics is as
important as mathematics itself. Mathematics is what it is just because of its history,
and this issue is frankly relevant for the number theory field.


https://domingogomezmorin.wordpress.com/

https://www.linkedin.com/in/domingo-...rin-500778127/



Quote:
Originally Posted by arithmo View Post
THE CUBE ROOT continues...

So, in general terms given any set of values whose product is $P={w}%
_{1}{w}_{2}{w}_{3}$:

${{w_{1},w_{2},w_{3}}}$

The aforementioned three arithmonic means are:

$\dfrac{{{w_{3}(w_{1}+2w_{2})}}}{{{w_{2}+2w_{3}}}} ,\ \ \dfrac{{{w_{2}(w_{3}+{%
2w_{1}})}}}{{{w_{1}+2w_{2}}}},\dfrac{{{w_{1}(w_{2} +2w_{3})}}}{{{w_{3}+2w_{1}}%
}}= $

whose product is also trivial and equal to $P={w}_{1}{w}_{2}{w}_{3}$

as an example for computing the cube root of 2, given the set:

${w}_{1}=1,$ $\ {{w_{2}}}=1,$ $\ {{w_{3}}}=2$

By computing the three arithmonic means:

it yields:

STEP 1

$\frac{6}{5} \frac{4}{3}\frac{5}{4}$ $\ error=9.92\ast
10^{-3}$

STEP 2
Using:

${w}_{1}=\frac{5}{4},$ $\ {{w_{2}}}=\frac{5}{4},$ $\ {{w_{3}}}=\frac{P}{%
\frac{5}{4}^{2}}$

it yields:

$ \frac{160}{127},\frac{63}{50},\frac{635}{504}$ $%
error=4.15\ast 10^{-7}$

STEP 3

${w}_{1}=\frac{635}{504},$ $\ {{w_{2}}}=\frac{635}{504},$ $\ {{%
w_{3}}}=\frac{P}{\frac{635}{504}^{2}}$

it yields:

$\frac{967800960}{768144131},\frac{384071939}{%
304838100},\frac{487771523185}{387144514512}$ $error=3.00\ast 10^{-20}$

and so on...


$ $Summarizing:

$\frac{5}{4}$ $error=9.92\ast 10^{-3}$

$\frac{635}{504}$ $error=4.15\ast 10^{-7}$

$\frac{487771523185}{387144514512}$ $%
error=3.00\ast 10^{-20}$

It triples the number of exact digits in each iteration, the convergence
rate is cubic.


Now...

Do we really need to carry out the whole set of fractions in each iteration
for approximating the root?

Well, firstly, that's pretty fine because you are always getting
approximations by defect and excess,

however,

notice that in each iteration this Rational Mean Method is basically using
the general expression (shown in my last post):

${w}_{1}=x,$ $\ {{w_{2}}}=x,$ $\ {{w_{3}}}=$ $\frac{P}{x^{2}}$


Thus, in the numerical example above we are just computing:

$\dfrac{2x^{3}+P}{3x^{2}},$ $\dfrac{3Px^{2}}{x^{4}+2Px},$ $\dfrac{x^{4}+2Px%
}{2x^{3}+P}$

And as a consequence, in the numerical example above we were actually
iterating the function (starting with x=1):


$\dfrac{x^{4}+2Px}{2x^{3}+P}$
which yields the values: $\frac{5}{4},\frac{635}{504},\frac{%
487771523185}{387144514512},...$ and their corresponding errors: $9.92\ast 10^{-3},$ $\ 4.15\ast 10^{-7},$ $\
3.00\ast 10^{-20}$


Summarizing, computing those three arithmonic means and using one of their
values as the seed for the next computation,
is exactly the same than iterating the corresponding algebraic iterated
expression, that is: the corresponding polynomial we found using
the general expression:


${w}_{1}=x,$ $\ {{w_{2}}}=x,$ $\ {{w_{3}}}=$ $\frac{P}{x^{2}}$

that is, iterating any of these polynomials:

$\dfrac{2x^{3}+P}{3x^{2}},$ $\dfrac{3Px^{2}}{x^{4}+2Px},$ $\dfrac{x^{4}+2Px%
}{2x^{3}+P}$

and this observation of course applies to ALL the other polynomial
expressions (high order iteration functions).

So the answer to the question:

Do we need to show the whole set of fractions in each iteration for
approximating the root?

is NO you don't,

you can just use any of those polynomials as independent iteration
functions, that's just the same thing

Thus, the Rational Mean Concept certainly yields high-order iterating
methods, and notice that not only those consecrated as the
exclusive achievement of the superb Infinitesimal Calculus but also many
other iterated functions.

The Rational Mean Concept also embraces the well known Daniel
Bernoulli's method for approximating roots,

as well as many others, some of them might have linear convergence while
others high-order convergence rate

The Arithmonic Mean is a very particular case of the Rational Mean.
The scope of the Rational Mean Concept is so wide, indeed.
The Rational Mean Concept also embraces many other variants,
as the linear one kindly referenced and analyzed by Kevin Brown at:
http://www.mathpages.com/home/kmath055/kmath055.htm

there are an uncountable number of variants, many of them already shown on the book.


This is at most, just high school stuff, just the Simplest Arithmetic,
so there is nothing here to make anyone grandstanding, on the contrary,
it seems to make the whole trivial arithmetical-thing much more irritating,
even worse when adding the plus: The Freshman Trauma and those restrictive Cauchy's signs.

It might just be, there are other ways to see the world than exclusively using the superb Cartesian-Infinitesimal lens.


https://domingogomezmorin.wordpress.com/

https://www.linkedin.com/in/domingo-...rin-500778127/


Last edited by skipjack; November 12th, 2017 at 08:55 PM.
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November 12th, 2017, 05:36 PM   #64
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I tried. I really tried but your monomania is too much for me.

As you said, etc. etc. I'm done.

-Dan
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November 19th, 2017, 09:06 AM   #65
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Quote:
Originally Posted by topsquark View Post
I tried. I really tried but your monomania is too much for me.
As you said, etc. etc. I'm done.
-Dan

The true is that you and the others have never seen such an extremely simple arithmetical operation that produces an uncountable number high-order root-approximating algorithms.
Worst, you have never seen a numerical example like the one shown here: A trivial arithmetical operation that produces cubic convergence rate at the first step, worst, that produces any desired rate of convergence.

That's the only true, as you said, etc, etc.

Of course you are done, you have nothing to say, these extremely simple and trivial high order arithmetical algorithms does not appear in any math text since antiquity up to now. that's why you are totally done ¡¡¡


However, that you and other two guys are done it really does not matter at all.

What really does matter here is what young people can learn from this:

If these methods would have been analyzed 3000 years ago then most likely the course of mathematics had not been finally oriented towards the infinitesimal calculus, and as a consequence nowadays we were all dealing with a very different scheme of quantity.
Young student, just imagine, they spent more than 3000 years until they could finally find high-order artificial methods --using geometry, cartesian system, infinitesimals--, however, they certainly had at hand a trivial arithmetical tool for producing those methods, just imagine what we would have got at present if they instead had devoted all their efforts to the analysis of this general and unifying arithmetical concept: The Rational mean.

You might say: probably nothing. Ok that's fine, however, here you have an open door, a new way to go.

You have two options now, the red pill or the blue one, your choice



All the other discussion relating huge egos from some guys in any math forums is totally irelevant here.
They will never admit they have missed something.
Well, young student, to their disgrace, here you have something huge they have totally missed.

Again, it follows the extremely simple numerical example that does not appear in any math text since antiquity uo to now:


Quote:
Originally Posted by arithmo View Post
THE CUBE ROOT continues...

So, in general terms given any set of values whose product is $P={w}%
_{1}{w}_{2}{w}_{3}$:

${{w_{1},w_{2},w_{3}}}$

The aforementioned three arithmonic means are:

$\dfrac{{{w_{3}(w_{1}+2w_{2})}}}{{{w_{2}+2w_{3}}}} ,\ \ \dfrac{{{w_{2}(w_{3}+{%
2w_{1}})}}}{{{w_{1}+2w_{2}}}},\dfrac{{{w_{1}(w_{2} +2w_{3})}}}{{{w_{3}+2w_{1}}%
}}= $

whose product is also trivial and equal to $P={w}_{1}{w}_{2}{w}_{3}$

as an example for computing the cube root of 2, given the set:

${w}_{1}=1,$ $\ {{w_{2}}}=1,$ $\ {{w_{3}}}=2$

By computing the three arithmonic means:

it yields:

STEP 1

$\frac{6}{5} \frac{4}{3}\frac{5}{4}$ $\ error=9.92\ast
10^{-3}$

STEP 2
Using:

${w}_{1}=\frac{5}{4},$ $\ {{w_{2}}}=\frac{5}{4},$ $\ {{w_{3}}}=\frac{P}{%
\frac{5}{4}^{2}}$

it yields:

$ \frac{160}{127},\frac{63}{50},\frac{635}{504}$ $%
error=4.15\ast 10^{-7}$

STEP 3

${w}_{1}=\frac{635}{504},$ $\ {{w_{2}}}=\frac{635}{504},$ $\ {{%
w_{3}}}=\frac{P}{\frac{635}{504}^{2}}$

it yields:

$\frac{967800960}{768144131},\frac{384071939}{%
304838100},\frac{487771523185}{387144514512}$ $error=3.00\ast 10^{-20}$

and so on...


$ $Summarizing:

$\frac{5}{4}$ $error=9.92\ast 10^{-3}$

$\frac{635}{504}$ $error=4.15\ast 10^{-7}$

$\frac{487771523185}{387144514512}$ $%
error=3.00\ast 10^{-20}$

It triples the number of exact digits in each iteration, the convergence
rate is cubic.


Now...

Do we really need to carry out the whole set of fractions in each iteration
for approximating the root?

Well, firstly, that's pretty fine because you are always getting
approximations by defect and excess,

however,

notice that in each iteration this Rational Mean Method is basically using
the general expression (shown in my last post):

${w}_{1}=x,$ $\ {{w_{2}}}=x,$ $\ {{w_{3}}}=$ $\frac{P}{x^{2}}$


Thus, in the numerical example above we are just computing:

$\dfrac{2x^{3}+P}{3x^{2}},$ $\dfrac{3Px^{2}}{x^{4}+2Px},$ $\dfrac{x^{4}+2Px%
}{2x^{3}+P}$

And as a consequence, in the numerical example above we were actually
iterating the function (starting with x=1):


$\dfrac{x^{4}+2Px}{2x^{3}+P}$
which yields the values: $\frac{5}{4},\frac{635}{504},\frac{%
487771523185}{387144514512},...$ and their corresponding errors: $9.92\ast 10^{-3},$ $\ 4.15\ast 10^{-7},$ $\
3.00\ast 10^{-20}$


Summarizing, computing those three arithmonic means and using one of their
values as the seed for the next computation,
is exactly the same than iterating the corresponding algebraic iterated
expression, that is: the corresponding polynomial we found using
the general expression:


${w}_{1}=x,$ $\ {{w_{2}}}=x,$ $\ {{w_{3}}}=$ $\frac{P}{x^{2}}$

that is, iterating any of these polynomials:

$\dfrac{2x^{3}+P}{3x^{2}},$ $\dfrac{3Px^{2}}{x^{4}+2Px},$ $\dfrac{x^{4}+2Px%
}{2x^{3}+P}$

and this observation of course applies to ALL the other polynomial
expressions (high order iteration functions).

So the answer to the question:

Do we need to show the whole set of fractions in each iteration for
approximating the root?

is NO you don't,

you can just use any of those polynomials as independent iteration
functions, that's just the same thing

Thus, the Rational Mean Concept certainly yields high-order iterating
methods, and notice that not only those consecrated as the
exclusive achievement of the superb Infinitesimal Calculus but also many
other iterated functions.

The Rational Mean Concept also embraces the well known Daniel
Bernoulli's method for approximating roots,

as well as many others, some of them might have linear convergence while
others high-order convegence rate

The Arithmonic Mean is a very particular case of the Rational Mean.
The scope of the Rational Mean Concept is so wide, indeed.
The Rational Mean Concept also ambraces many other variants,
as the linear one kindly referenced and analyzed by Kevin Brown at:
http://www.mathpages.com/home/kmath055/kmath055.htm

there are an uncountable number of variants, many of them already shown on the book.


This is at most, just high school stuff, just the Simplest Arithmetic,
so there is nothing here to make anyone grandstanding, on the contrary,
it seems to make the whole trivial arithmetical-thing much more irritating,
even worse when adding the plus: The Freshman Trauma and those restrictive Cauchy's signs.

It might just be, there are other ways to see the world than exclusively
using the superb Cartesian-Infinitesimal lens.


https://domingogomezmorin.wordpress.com/

https://www.linkedin.com/in/domingo-...rin-500778127/

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November 19th, 2017, 12:03 PM   #66
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Quote:
Originally Posted by arithmo View Post
You have two options now, the red pill or the blue one, your choice
which one makes you go away?
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