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September 29th, 2017, 03:11 PM   #1
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The Mediant and new high-order root approximating methods.

In number theory the generalization of the operation called: Mediant
leads the way to new and extremely simple high-order root approximating methods which might be of interest to some number theorists, mainly because of its connection to true construction methods of real numbers:

https://domingogomezmorin.wordpress.com/

From the evidence at hand, these new high-order methods have no precedents in the math literature.


YOUTUBE CHANNEL:



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October 17th, 2017, 06:30 PM   #2
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Some new stuff have been included in the aforementioned webpage on new hig-order numerical methods, specifically on their matricial representation

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October 17th, 2017, 07:00 PM   #3
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Ignoring the fact that you are advertising on a free forum...

This method only works for taking the nth root of a number? Heck, you can do that with a simple logarithm table. I was hoping you could apply this to finding the zeros of polynomial functions. Is this possible?

-Dan
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October 17th, 2017, 08:00 PM   #4
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Yes, it can generate similar approximations for the general algebraic equation as shown in the book,
and it can be applied to complex roots.

However, a crucial important point here, is that one can generate --by using the SIMPLEST ARITHMETIC-- all those high-order root approximating methods that have been consecrated as an exclusive product of Infinitesimal Calculus.

Notice, that all this is new, there are no precedents on this ARITHMETICAL methods in the whole story of root approximating methods,
so there remains much more to be investigated in this area, mainly because there are an uncountable number of variants.

In reference to your comment on the logarithmic table, of couse, you must be joking, because we are talking here about
extremely simple High-Order iterating methods.

Last edited by arithmo; October 17th, 2017 at 08:09 PM.
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October 17th, 2017, 08:05 PM   #5
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The video seemed very unorganized but from what I gathered, in each case you are doing 1 of 2 things.

1. using Newton's method without the calculus point of view which I suppose is fine but certainly not groundbreaking by any means.

2. using something less precise than Newton's method in which case you should be using Newton's method.
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October 18th, 2017, 07:31 AM   #6
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Quote:
Originally Posted by SDK View Post
The video seemed very unorganized but from what I gathered, in each case you are doing 1 of 2 things.

There are two videos, not just once.

1. using Newton's method without the calculus point of view which I suppose is fine but certainly not groundbreaking by any means.

2. using something less precise than Newton's method in which case you should be using Newton's method.
I am not using Newton's method, I am using a New General and Unifying Arithmetical Principle that embraces,
at once, many celebrated methods (Newton's, Halley's, Householder's, Lucas', Daniel Bernoulli's, etc.
from which some of them have been consecrated as an exclusive achievement of Infinitesimal calculus)
as well as many other new high-order iterating functions for root approximating.

Thus, the new methods embrace not only quadratic convergence (as for Newton's) but any convergence rate.
No derivatives, no infinitesimal calculus, no Trial-&-Error checkings, but just the simplest and pure Arithmetic.


Such Natural Arithmetical Principle could had been easily implemented by ancient mathematicians because it only requires
the simplest arithmetic, unfortunately and to our surprise, they didn't.

The new principle also leads the way to the generalized continued fractions shown in the book for Pi, e, cube root, Golden Mean, ...
(one of them already published in Monthly, and shown at the front cover of the book).

Thus, for any educated person this not just a matter of 'algorithms' but about a very simple arithmetical principle
which has not been used in the whole history of math. This is a math history issue.

This is also a matter of a true Natural method for constructing irrational numbers.
No derivatives, no infinitesimal calculus, no Trial-&-Error checkings, but just the simplest and pure Arithmetic.

The two videos are just introductory, there is also a webpage and the book.
https://domingogomezmorin.wordpress.com/


Last edited by arithmo; October 18th, 2017 at 07:40 AM.
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October 18th, 2017, 09:21 AM   #7
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Quote:
Originally Posted by arithmo View Post
In reference to your comment on the logarithmic table, of couse, you must be joking, because we are talking here about
extremely simple High-Order iterating methods.
Actually I am not joking. Logarithms are typically introduced during the junior level (more or less) of American high schools. There is nothing terribly complicated about them though, yes, it is not an Algebraic method.

In any event the sound on my computer is acting up and I couldn't hear the narration in the video. I see no reason in the video as to why this method would do what you claim. Of course I suppose that is why I should buy the book. I would appreciate it if you would give me a quick run-down on how you would apply this method to the solution of equation $\displaystyle x^3 - x + 1 = 0$?

Thank you.

-Dan
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October 18th, 2017, 10:56 AM   #8
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$\displaystyle \dfrac{4}{3} \cdot \dfrac{5}{4} \cdot \dfrac{6}{5}$ has a product of 2.

Applying your algorithm we get

$\displaystyle \dfrac{9}{7} \cdot \dfrac{11}{9} \cdot \dfrac{14}{11}$ which also has a product of 2.

Doing it again we get

$\displaystyle \dfrac{20}{16} \cdot \dfrac{25}{20} \cdot \dfrac{32}{25}=\dfrac{5}{4} \cdot \dfrac{5}{4} \cdot \dfrac{32}{25}$ which also has a product of 2.

One more time yields

$\displaystyle \dfrac{10}{16} \cdot \dfrac{37}{29} \cdot \dfrac{42}{33}=\dfrac{5}{8} \cdot \dfrac{37}{29} \cdot \dfrac{14}{11}$ which has a product of $\displaystyle \dfrac{1295}{1276}$ which is not 2.

I'm confused.
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October 18th, 2017, 11:54 AM   #9
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Quote:
Originally Posted by mrtwhs View Post
$\displaystyle \dfrac{4}{3} \cdot \dfrac{5}{4} \cdot \dfrac{6}{5}$ has a product of 2.

Applying your algorithm we get

$\displaystyle \dfrac{9}{7} \cdot \dfrac{11}{9} \cdot \dfrac{14}{11}$ which also has a product of 2.

Doing it again we get

$\displaystyle \dfrac{20}{16} \cdot \dfrac{25}{20} \cdot \dfrac{32}{25}=\dfrac{5}{4} \cdot \dfrac{5}{4} \cdot \dfrac{32}{25}$ which also has a product of 2.

One more time yields

$\displaystyle \dfrac{10}{16} \cdot \dfrac{37}{29} \cdot \dfrac{42}{33}=\dfrac{5}{8} \cdot \dfrac{37}{29} \cdot \dfrac{14}{11}$ which has a product of $\displaystyle \dfrac{1295}{1276}$ which is not 2.

I'm confused.




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.

This is the "charm" of this method, notice that the numerators and denominators are canceling each other.
You don't need to reduce fractions nor make any other operations to check the product ¡¡¡

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)

Now, please notice that you are using the most basic method that holds
very slow convergence rate, linear convergence.

The special case of the Generalized Mediant called 'ARITHMONIC MEAN' allows to produce High-Order methods.



There is another lineal variant of the sample you have just worked on (specifically for the cube root)
which was excerpted from my book and kindly analyzed by Kevin Brown at:

http://www.mathpages.com/home/kmath055/kmath055.htm

But, notice that there are an uncountable number of variants at any convergence rate, many of them are shown in the book.
As said, this is an unexplored field. Just striking, indeed.
A true math-history issue.

Last edited by arithmo; October 18th, 2017 at 12:41 PM.
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October 18th, 2017, 12:06 PM   #10
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Quote:
Originally Posted by topsquark View Post
Actually I am not joking. Logarithms are typically introduced during the junior level (more or less) of American high schools. There is nothing terribly complicated about them though, yes, it is not an Algebraic method.

In any event the sound on my computer is acting up and I couldn't hear the narration in the video. I see no reason in the video as to why this method would do what you claim. Of course I suppose that is why I should buy the book. I would appreciate it if you would give me a quick run-down on how you would apply this method to the solution of equation $\displaystyle x^3 - x + 1 = 0$?

Thank you.

-Dan

Ok, that´s pretty simple, I will show here some hints for
the general case:
a x^3 + b x + c = 0


But firstly, let's see if there comes out any suggestion from some people in this audience.
I think it is not a big deal to wait some few hours, mainly considering that these extremely simple arithmetical
methods didn't show up in the math literature since ancient times up to now, so we are talking about FIVE thousands
years ago, at least.


In reference to using logarithms, it must be remarked that that is not just a matter of looking at a table, but the effort and sacrifices needed to develope such logarithmic tables. Such a headache cannot be compared with the trivial effort needed for constructing these extremely simple natural arithmetical methods, by any means, and that is certainly what I call: A huge math history issue.

The last point, about buying the book, I understand you don't need to buy the whole book, using Amazon Kindle method anyone can read the desired pages on any smartphone, tablet, or computer.
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Last edited by arithmo; October 18th, 2017 at 12:40 PM.
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