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October 18th, 2017, 01:07 PM   #11
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Quote:
Originally Posted by arithmo View Post
The new set of fractions in your last step should be:
45/36
57/45
72/57

whose product is clearly 2.
I would strongly suggest you to use Excel.
I watched the youtube completely. No where do you say to not reduce the fractions.

Quote:
Do not reduce the fractions, at any step of the proccedure, the form
of the fractions are crucial when using the Rational Mean (Generalized mediant)
I guess it would have been nice to know this ahead of time.

Quote:
Now, please notice that you are using the most basic method ...
Well basic for me is to always write fractions in reduced form. On the other hand, 4th grade was the best three years of my life.
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October 18th, 2017, 01:46 PM   #12
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I watched the youtube completely. No where do you say to not reduce the fractions.

No where do I say you had to reduce fractions


I guess it would have been nice to know this ahead of time.


Many Number Theory students know that well ahead of time, that's the main property of this operation because it basically works with pairs of numbers.
Moreover, by constructing and using the general algebraic form for any of these methods (Generating the corresponding algebraic iteration functions), you will realize that you can also reduce the fractions at any step os the process if you wish to.
However, any person knows well ahead that you don´t have to reduce fractions at each step of Newton's method for approximating the cube root of 2.
On the other hand, if your personal task is to find all the best approximations for the cube root of 2, that's another topic.




Well basic for me is to always write fractions in reduced form. On the other hand, 4th grade was the best three years of my life.

Tell that also to Newton, Householder, and all the brilliant mathematicians who
produced those "superb" methods consecrated as the exclusive product of the superb infinitesimal calculus. It is really striking to realize that since ancient times they never devised such simple and powerful methods, worst when considering that this is the simplest arithmetic,
(though, greeks were very close when dealing with the MESOLABIUM).




Last edited by arithmo; October 18th, 2017 at 02:16 PM.
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October 18th, 2017, 06:27 PM   #13
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Well now I'm really confused.

At the 6 minute mark in your video you say:

"In Number Theory the name 'Mediant' is used exclusively when it operates with two reduced fractions."

You then talk about a new rational operator called a RATIONAL MEAN but the only difference you give between this and a mediant is that it "surpasses Cauchy's restriction on denominator signs."

At the 7 minute mark you say "given three fractions whose product is 2" but you don't say how the fractions were selected. OK I select

$\displaystyle \dfrac{5}{4} \cdot \dfrac{5}{4} \cdot \dfrac{32}{25}$ which has a product of 2.

One iteration yields $\displaystyle \dfrac{10}{16} \cdot \dfrac{37}{29} \cdot \dfrac{42}{33}$ (I promise not to reduce) which has a product of $\displaystyle \dfrac{15540}{15312}$ which is not 2.
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October 18th, 2017, 07:49 PM   #14
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The more I read about this the more it looks like absolute crankery. There is no disputing that your rational mean produces a ratio between the 2 you start with. When the product is fixed, the AM/GM inequality guarantees that the new value lies closer to the square root. This generalizes further for $n^{th}$ roots. However, you have quite a few noticeable problems.

1. You have not shown that your method has higher order convergence. Since your method for rational means is essentially equivalent to the secant method, I would guess that it has linear convergence only. Even worse, you haven't done any analysis to determine whether this is the case, much less proved higher order convergence.

2. The fundamental method behind Newton's method (and its higher order generalizations) amounts to "subtracting off" lower order terms and solving linear (or higher order) equations to correct the error. When used to find roots such as you are doing, these yield formulas which are rational much like yours. In fact, it appears to me you are obtaining at best the same exact formulas and more likely, less useful versions which subtract off non-optimal polynomial approximations for the root (which don't share derivatives as a Taylor polynomial does).

For example, Newton iteration for finding cube roots amounts to iteration of the function
$N(x) = x - \frac{x^3 - \lambda}{3x^2}$ which is a rational function.

3. Nothing in your method addresses the inherent issue which arises from multiple basins of attraction. For example, there are $n$-many $n^{th}$ roots and your initial guess in a typical iterative method determines to which root you limit under iteration. Your method seems immune to this which is not a virtue but rather a huge gap. This means there are difficulties which will arise when taking higher roots or trying to solve polynomials. For example, there are 2 real fourth roots of a positive real number. How does your method address the fact that appropriate guesses should be capable of converging to either.

4. Most importantly, I just want to repeat point 1. If you can't give any convergence estimates or analysis, then you don't have a "revolutionary" root finding method, you have a bunch of claims which nobody has any reason to believe.

Last edited by skipjack; October 24th, 2017 at 11:58 AM.
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October 18th, 2017, 08:47 PM   #15
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Quote:
Originally Posted by mrtwhs View Post
well now i'm really confused.

At the 6 minute mark in your video you say:

"in number theory the name 'mediant' is used exclusively when it operates with two reduced fractions."

yes, that's true


you then talk about a new rational operator called a rational mean but the only difference you give between this and a mediant is that it "surpasses cauchy's restriction on denominator signs."

no, that's not the only difference, the main difference is that no mathematician since ancient times neither used nor devised the huge scope and power of such arithmetical operator for constructing high order root approximating methods, among other things.
Worst when considering that since antiquity the mesolabium was shouting them out loud to do so. The very particular case called 'the arithmonic mean' which leads the way to high-order methods is a clear evidence of this




at the 7 minute mark you say "given three fractions whose product is 2" but you don't say how the fractions were selected. Ok i select

the phrase: "their product is trivial because their numerator and denominators are canceling each other", is all over the place,
the webpage, the videos, and of course the book.

Notice, that i didn't say, as for instance: "their product is trivial because the product of the numerators of the first two fractions cancels the denominator of the last fraction", or something like that.

The initial set of fractions whose product is p, is always denoted as:
$\displaystyle \dfrac{a1}{b1} \cdot \dfrac{a2}{a1} \cdot \dfrac{p* b1}{a2}$
as shown in the videos, the webpage and the book.



$\displaystyle \dfrac{5}{4} \cdot \dfrac{5}{4} \cdot \dfrac{32}{25}$ which has a product of 2.
One iteration yields $\displaystyle \dfrac{10}{16} \cdot \dfrac{37}{29} \cdot \dfrac{42}{33}$ (i promise not to reduce) which has a product of $\displaystyle \dfrac{15540}{15312}$ which is not 2.


for the particular rational method you are pointing out (the most basic with linear convergence), you should have used:

$\displaystyle \dfrac{a1}{b1} \cdot \dfrac{a2}{a1} \cdot \dfrac{p* b1}{a2}$

that is:

$\displaystyle \dfrac{20}{16} \cdot \dfrac{25}{20} \cdot \dfrac{32}{25}$


in the case you decide to use the very particular case of the rational mean called: 'arithmonic mean', you can use:
$\displaystyle \dfrac{5}{4} \cdot \dfrac{5}{4} \cdot \dfrac{32}{25}$

and in the first step you will get:

$\displaystyle \dfrac{160}{127} \cdot \dfrac{63}{50} \cdot \dfrac{635}{504}$

the error for the last fraction in the set is: 4.15 x 10-7

please see the webpage for more details on high-order methods and the arithmonic mean:
https://domingogomezmorin.wordpress.com/

as said, you must be aware that the general and unifying concept: 'rational mean' embraces an uncountable number of ways for approximating roots.

So, the videos and the webpage are just an introduction to some methods, the book shows more.
.
.
.
.

Last edited by skipjack; October 24th, 2017 at 12:03 PM.
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October 18th, 2017, 08:56 PM   #16
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Quote:
Originally Posted by SDK View Post
The more I read about this the more it looks like absolute crankery. There is no disputing that your rational mean produces a ratio between the 2 you start with. When the product is fixed the AM/GM inequality guarantees that the new value lies closer to the square root. This generalizes further for $n^{th}$ roots. However, you have quite a few noticeable problems.

1. You have not shown that your method has higher order convergence. Since your method for rational means is essentially equivalent to the secant method, I would guess that it has linear convergence only. Even worse, you haven't done any analysis to determine if this is the case, much less proved higher order convergence.

2. The fundamental method behind newton's method (and its higher order generalizations) amounts to "subtracting off" lower order terms and solving linear (or higher order) equations to correct the error. When used to find roots such as you are doing, these yield formulas which are rational much like yours. In fact, it appears to me you are obtaining at best the same exact formulas and more liikely, less useful versions which subtract off non-optimal polynomial approximations for the root (which don't share derivatives as a Taylor polynomial does).

For example, Newton iteration for finding cube roots amounts to iteration of the function
$N(x) = x - \frac{x^3 - \lambda}{3x^2}$ which is a rational function.

3. Nothing in your method addresses the inherent issue which arises from multiple basins of attraction. For example, there are $n$-many $n^{th}$ roots and your initial guess in a typical iterative method determines to which root you limit under iteration. Your method seems immune to this which is not a virtue but rather a huge gap. This means there are difficulties which will arise when taking higher roots or trying to solve polynomials. For example, there are 2 real fourth roots of a positive real number. How does your method address the fact that appropriate guesses should be capable of converging to either.

4. Most importantly, I just want to repeat point 1. If you can't give any convergence estimates or analysis, then you don't have a "revolutionary" root finding method, you have a bunch of claims which nobody has any reason to believe.


You say: "Your method", so it is clear that you have not read at all anything related to the UNCOUNTABLE METHODS pointed out in the videos and webpage. So please take aside all your crankery produced by a hurt Ego .
Take a look at the highorder methods shown at:

https://domingogomezmorin.wordpress.com/

And stop you cankery, because it is a real shame that these methods have not being taught nor written in the math literature
since ancient times up to now, That's a real and crude shame, and if your huge Ego get hurt because of that,
that's your problem, not mine. To your disgrace these methods will stay around here for ever, so take it easy, you cannot do nothing about that.

Last edited by arithmo; October 18th, 2017 at 09:12 PM.
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October 19th, 2017, 12:34 AM   #17
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Look, we went through this on MHF at one point. You are making a number of, pardon me for saying so, unverified comments. Of course we are going to be skeptical! And at no point have we seen anything like a proof of the method, just your comments that it works. That is why I asked for a concrete example of your method.

Extraordinary claims require extraordinary proof. You make comments and say how revolutionary this method is but at no point have you offered any proof! As I mentioned before you have to assume we are going to be skeptical.

We need some level of proof which is why I asked for the example. Otherwise you are just a person making a number of unjustifiable statements.

So prove that you aren't a troll... Why not work out that example I asked for? If you can show that the method works by a concrete example then that will reduce the increasingly growing skepticism.

-Dan
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October 20th, 2017, 04:46 PM   #18
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Here are some hints. This is just an introduction, for more details and convergence proofs, take a look at the book:

https://domingogomezmorin.wordpress.com/


Given the general cubic equation:

$\displaystyle {{a}_{3}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}+{{a}%
_{1}}x+{{a}_{0}}=0 $

or the same:

$\displaystyle {{a}_{3}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}+{{a}%
_{1}}x=-{{a}_{0}} $

You can express it as:

$\displaystyle {P}_{{1}}+{P}_{2}+P_{3}=-{{a}_{0}} $

with:

$\displaystyle {{a}_{3}}{{x}^{3}=P}_{{1}}
$

$\displaystyle {{a}_{2}}{{x}^{2}=P}_{2} $

$\displaystyle {{a}_{1}}x=P_{3} $

For the general cubic equation to be satisfied, $\displaystyle {P}_{{1}%
},{P}_{2},P_{3} $ are expressions where the x values
should converge to same solution while complying with:

$\displaystyle {P}_{{1}}+{P}_{2}+P_{3}=-{{a}_{0}} $

$\displaystyle

{{x}^{3}=}\frac{{P}_{{1}}}{{{a}_{3}}},\ \ \ \ {{x}^{2}=}\frac{{P}_{2}}{{{a}%
_{2}}},\ \ \ \ x=\frac{P_{3}}{{{a}_{1}}}
$

The general algebraic iteration functions for solving each of those
elemental cubic and quadratic expressions can be trivially generated by using ARITHMONIC MEANS, as
shown in the introductory videos and my webpage (and of course the book).

Because the x values from $\displaystyle {P}_{{1}},{P}_{2},P_{3} $ must converge at the same time to the same value (the solution of the equation),
their Rational Mean (Generalized Mediant) can be used to find the iteration function for approximating the solution,
always complying with the condition: $\displaystyle {P}_{{1}}+{P}_{2}+P_{3}=-{{a}_{0}} $

As a consequence, using two different arithmonic means, the new method
yields the following two iteration functions :

1st. Iteration function:

$\displaystyle \frac{2{{a}_{3}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}-{{a}_{0}}}{3{{a}_{3}}%
{{x}^{2}}+2{{a}_{2}}x+{{a}_{1}}} \ $

2nd. Iteration function:

$\displaystyle \frac{{{a}_{3}}{{x}^{4}}-2{{a}_{0}}x}{2{{a}_{3}}{{x}^{3}}+{{a}_{2}%
}{{x}^{2}}+{{a}_{1}}x-{{a}_{0}}} \ $


Thus, the Rational Mean method yielded two general iteration functions.

The first one coincidentally corresponds to the well known Newton's iteration function for the
general cubic equation, while the second one is kind of Newton's counterpart, mainly because many times it converges when Newton's fails and viceversa.



No derivatives, No Trial-&-Error checkings, No Infinitesimal Calculus, but just the Simplest Arithmetic.




************************************************** ****************

EXAMPLE 1:


Given the equation:

$\displaystyle x^{3}-3x+1=0 $

Whose solutions: x=0.34730, x=1.5321, x=-1.8794


1st. Iteration function:

$\displaystyle \frac{2{{a}_{3}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}-{{a}_{0}}}{3{{a}_{3}}%
{{x}^{2}}+2{{a}_{2}}x+{{a}_{1}}} = \frac{2x^{3}-1}{3x^{2}-3} \
$ ITERATION FAILS for x0=1.0

2nd. Iteration function:

$\displaystyle \frac{{{a}_{3}}{{x}^{4}}-2{{a}_{0}}x}{2{{a}_{3}}{{x}^{3}}+{{a}%
_{2}}{{x}^{2}}+{{a}_{1}}x-{{a}_{0}}}=\frac{2x-x^{4}}{-2x^{3}+3x+1%
} $

ITERATIONS:

1.0
0.5
0.41667
0.38151
0.36481
0.35643
0.35211
0.34984
0.34865




************************************************** *****************************

EXAMPLE 2:

$\displaystyle \ x^{3}+x^{2}-5x+1=0 $

Solution is: x=-2.8662, x=1.6554, x=0.21076

1st. Iteration function:

$\displaystyle \frac{2{{a}_{3}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}-{{a}_{0}}}{3{{a}_{3}}{{%
x}^{2}}+2{{a}_{2}}x+{{a}_{1}}}= \frac{1}{3x^{2}+2x-5}\left(
2x^{3}+x^{2}-1\right) $

ITERATIONS WITH SERIOUS TROUBLES:

0.9
-1.6468
45.659
30.354
20.162
13.386
8.8955
5.9394
4.0221
2.8174
2.1133

2nd. Iteration function:

$\displaystyle \frac{{{a}_{3}}{{x}^{4}}-2{{a}_{0}}x}{2{{a}_{3}}{{x}^{3}}+{{a}_{2}}{%
{x}^{2}}+{{a}_{1}}x-{{a}_{0}}}= -\frac{2x-x^{4}}{2x^{3}+x^{2}-5x-1%
} $

ITERATIONS:


0.9
0.35393
0.27083
0.23930
0.22507
0.21813
0.2146
0.21278
0.21182
0.21132
0.21105




and so on...



We can see how we can find not only newton's method for the general
algebraic equation but also others, just by the agency of the SIMPLEST
arithmetic.

Even worse: When computing the Rational Mean of the two general iteration expressions produced above,
then you get another new iteration function whose behavior is far better at any instance.

Indeed, this is not just a matter of a bunch of numerical methods, but a
huge math history issue.

As said, it is a real shame these methods have not been taught even at secondary level
nor displayed in the math literature since ancient times up to now.



If you want more details or more comments, or any other inquires then just take a look at the book, that's it.

https://domingogomezmorin.wordpress.com/

. .
. .
. .

Last edited by arithmo; October 20th, 2017 at 05:06 PM.
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October 20th, 2017, 05:34 PM   #19
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Thank you for working those out.

It strikes me that you are presenting no derivation of these rules. Does the book do this? I have a suspicion that the source of the fractions that are being used have a derivation that is far from what you might call High School Algebra. In other words I suspect that a derivation of these rules would be less than trivial.

For example, Newton's method for which you say that the arithmonic system replicates. I can derive Newton's method in about five lines if I want to be specific about each point. Yes, Calculus is involved in the derivation but it is otherwise very simple to understand and work with. I have no idea why your method works and I suspect a derivation would take quite a bit more effort.

Just my thoughts.

-Dan
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October 20th, 2017, 07:39 PM   #20
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Quote:
Originally Posted by topsquark View Post
Thank you for working those out.

It strikes me that you are presenting no derivation of these rules. Does the book do this?

Yes, of course it does. I will not show the whole book here.


Quote:
Originally Posted by topsquark View Post
I have a suspicion that the source of the fractions that are being used have a derivation that is far from what you might call High School Algebra. In other words I suspect that a derivation of these rules would be less than trivial.
Well, I would frankly say they are more precisely not at High School level,
but at primary school level (No pun intended)

They are just the same as that that you can see at my webpage
No more, no less.
WYSWYG



Quote:
Originally Posted by topsquark View Post
For example, Newton's method for which you say that the arithmonic system replicates. I can derive Newton's method in about five lines if I want to be specific about each point.
-Dan
I must remark, the Arithmonic Mean and in general terms the Rational Mean, not only replicates Newton's, but it also allows to generate (no replication) an uncountable number of linear and high-order methods. You can see the evidence (the iceberg tip) in the 2nd. Iteration Function shown in my last message (what I decided to call Newton's counterpart)


Quote:
Originally Posted by topsquark View Post
Yes, Calculus is involved in the derivation but it is otherwise very simple to understand and work with. I have no idea why your method works and I suspect a derivation would take quite a bit more effort.
Just my thoughts.
No, The iterations function shown in my last message took just two lines for being generated. They are just GENERALIZED MEDIANTS, just the sum of numerators and denominators (Primary school level).
The introductory explanations I gave here, are by far enough for constructing those iterations functions. Just arithmetic, sums of numerators and denominators.

Look at my webpage, you can find high-order expressions for approximating the nth-root just by suming numerators and denominators, that's all you need.
There is nothing hidden behind all this.
You know there are hundreds (may be Thousands) of papers about Nth-root approximating methods published in many Peer review journals.
Now, take a another look at my webpage. ¿Did you see something like that
in those journals?
Answer: No, you don't. Please cogitate on that.

That's what I call a huge math history issue.

The 'Rational-Mean' methods work fine, because they are simple, natural, and they do not use geometry, nor Cartesian System, nor Trial-and-error methods, nor derivatives, but NUMBER itself.

I must remark, that from a math-history perpective when you compare Newton's method to these Rational Methods, you certainly cannot put aside all the work he needed to find his method (Geometry, the Cartesian System , and Infinitesimal calculus).

However, even giving you the unfair advantage of puting aside all that, the derivations needed for constructing any of the Rational Methods shown in my book are FAR shorter, simpler and natural than using derivatives, when generating root-approximating methods.

We are just dealing with NUMBER itself.

It is far enough what I have explained here, as said, the book is at amazon
and there is the kindle version where enyone can see any desired number of pages in any tablet, smartphone of computer.


I am a structural engineer specialized in numerical methods for structural engineering.
I live in Venezuela where we all are having so many serious social and political problems by these days, so making the book has been a real issue for me and my family.

You can find me at Linkedin.com.
https://www.linkedin.com/in/domingo-...rin-500778127/

I think I am not a troll. Well... I will try not to become that.

That's it

WYSWYG


p.d. There are other issues here:

Apart for the aforementioned huge math-history issue,
These methods are True natural ways for constructing irrational numbers just by Number itself --by defect and excess--.
Everything is controlled by Quantity, by Number.
In the background, there always exists a simple Arithmetic Grid, as a solid structural frame,
that supports and controls all what you may construct or imagine in maths, including of course
the fluxions, Infinitesimal Calculus, etc.

Last edited by arithmo; October 20th, 2017 at 07:57 PM.
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