My Math Forum The Mediant and new high-order root approximating methods.

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 October 25th, 2017, 04:26 PM #41 Senior Member   Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 Arithmo Are P, a, and b arbitrary? If not, how they are selected is a missing key. And I see why the mean of the mediants approximates the square root of P, but how does that help with the seventh root of P? So how the method is extended to other roots is another missing key. And here are the critical missing elements, which seem necessary if you wish to persuade others to buy your book or to give you accolades for a significant mathematical achievement. What is the proof that your method is usually successful in finding the roots of polynomials of any rational degree, and what is the proof that your method usually converges on a good enough approximation faster than other methods? Absent both proofs, or at least sketches of such proofs, I cannot see why anyone will buy your book. Thanks from topsquark and arithmo
October 25th, 2017, 06:20 PM   #42
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Quote:
 Originally Posted by JeffM1 Arithmo Are P, a, and b arbitrary? If not, how they are selected is a missing key. And I see why the mean of the mediants approximates the square root of P, but how does that help with the seventh root of P? So how the method is extended to other roots is another missing key. And here are the critical missing elements, which seem necessary if you wish to persuade others to buy your book or to give you accolades for a significant mathematical achievement. What is the proof that your method is usually successful in finding the roots of polynomials of any rational degree, and what is the proof that your method usually converges on a good enough approximation faster than other methods? Absent both proofs, or at least sketches of such proofs, I cannot see why anyone will buy your book.

The only condition is that the product is P, being P any positive number

You can find the complex root of P using complex numbers, in any of the next posts I will bring a particular example of complex cube roots of P, computed in exactly the same way, using the Generalized Mediant (The Rational Mean).

The methods applies for the nth-root and at any desired convergence rate, as will be explanined later in the next posts.
Also notice, that there are an uncountable ways of using this operation, I have shown many ways in the book, however,
I will give here examples on high-order convergence rate, just by using the generalized Rational Mean concept.

Since ancient times it has been already known that by using arithmetic and harmonic means you can get approximations
to the square root. The arithmetic mean produce the same values than when applying Newton's method to the equation x^2=P.
Edouard Lucas constructed a very cumbersome method (using combinatory) for applying the arithmetic and harmonic means to high order roots, a real headache.
And here is the essence of all this, the arithmetic and harmonic means are just particular cases of a more general and unifying concept: The Rational Mean (Generalized Mediant), which is much more than just a courious property of Number as Cauchy stated.

I will go, step by step

Last edited by arithmo; October 25th, 2017 at 06:46 PM.

 October 27th, 2017, 11:31 AM #43 Member   Joined: Mar 2017 From: venezuela Posts: 36 Thanks: 3 ... yes, indeed, the Cauchy's signs and the Freshman Trauma... Summarizing, in the previous example we computed a Mediant by previously multiplying by P both the numerator and denominator of one of the two fractions. ¿What happen if instead of multiplying by P, we multiply by x? that is, let's start with the following set of fractions: $\frac{x}{1}\text{= }\frac{P}{x}$ and compute the first Mediant by previously multiplying the numerator and denominator of the first fraction by x: $\frac{x*x+P}{x*1+x}$ and we know the another fraction will be its inverse multiplied by P. Let's see: ITERATIONS INITIAL SET $\frac{x}{1}\text{, }\frac{P}{x}$ STEP 1: $\frac{x^{2}+P}{2x}\text{, }\frac{2Px}{x^{2}+P}$ STEP 2: $\frac{x^{3}+3Px}{3x^{2}+P}\text{, }\frac{P\left( 3x^{2}+P\right) }{x^{3}+3Px}$ STEP 3: $\frac{P^{2}+6Px^{2}+x^{4}}{4x^{3}+4Px}\text{, }\frac{P\left( 4Px+4x^{3}\right) }{P^{2}+6Px^{2}+x^{4}}$ and so on... All those expressions can be independently used as Iterated-Functions for computing the square root of P. Now, take a look at their convergence rate Does any of them remind you Householder´s name? But, wait a minute¡. We have not used any derivatives here, nor even the superb Infinitesimal Calculus ¡ So, if it took them about 3 thousands years (or more) to develope those derivatives, then why they did not analyzed this extremely simple Operation and those trivial methods since antiquity? Now, this is the more trivial example that the new general and unifying arithmetical concept 'The Rational Mean' can bring you. Pretty simple, however, take a look at the math-literature and show me any book or any paper on this, and I mean from ancient times up to now. Notice that all we already know that in ancient times mathematicians used the particular cases: arithmetic and harmonic means, for computing the square root. Please remember, this operation tends to provoke so much anger and hate because of Freshman Trauma and Cauchy's signs, so take your time, think a lot. Don't be afraid. Don't worry, be happy. and notice that... I was once like you are now, and I know that it's not easy To be calm when you've found something going on But take your time, think a lot... Last edited by arithmo; October 27th, 2017 at 11:54 AM.
 October 27th, 2017, 04:29 PM #44 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,853 Thanks: 750 Math Focus: Wibbly wobbly timey-wimey stuff. Okay, I think I see where the polynomial expressions are coming from. Thanks. I'd like to ask you a question about the polynomials in the example of the cubic equation. Where do you get the expressions for the iteration functions? $\displaystyle a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0$ 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}}} \$ You are using the fractions in weird, but I presume Historical, ways. It's kind of hard to follow. I think you are doing something with $\displaystyle x = \frac{P_3}{a_3} = \frac{P_2}{a_2}$ but I'm not seeing the logic. -Dan Last edited by topsquark; October 27th, 2017 at 04:35 PM.
October 29th, 2017, 11:06 AM   #45
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Quote:
 Originally Posted by arithmo ... yes, indeed, the Cauchy's signs and the Freshman Trauma... ITERATIONS INITIAL SET $\frac{x}{1}\text{, }\frac{P}{x}$ STEP 1: $\frac{x^{2}+P}{2x}\text{, }\frac{2Px}{x^{2}+P}$ STEP 2: $\frac{x^{3}+3Px}{3x^{2}+P}\text{, }\frac{P\left( 3x^{2}+P\right) }{x^{3}+3Px}$ STEP 3: $\frac{P^{2}+6Px^{2}+x^{4}}{4x^{3}+4Px}\text{, }\frac{P\left( 4Px+4x^{3}\right) }{P^{2}+6Px^{2}+x^{4}}$ and so on... All those expressions can be independently used as Iterated-Functions for computing the square root of P. Now, take a look at their convergence rate Does any of them remind you Householder´s name? But, wait a minute¡. We have not used any derivatives here, nor even the superb Infinitesimal Calculus ¡ So, if it took them about 3 thousands years (or more) to develope those derivatives, then why they did not analyzed this extremely simple Operation and those trivial methods since antiquity? Pretty simple, however, take a look at the math-literature and show me any book or any paper on this, and I mean from ancient times up to now.

In the previous posts we were at primary school level,
but now we are at high school level ¡
So, it will be pretty easy to express those functions in matricial form:

isn't it? Let's see:

$% \begin{pmatrix} x & P \\ 1 & x% \end{pmatrix}% ^{2}=% \begin{pmatrix} x^{2}+P & 2Px \\ 2x & x^{2}+P% \end{pmatrix}%$

Cubic convergence:

$% \begin{pmatrix} x & P \\ 1 & x% \end{pmatrix}% ^{3}=% \begin{pmatrix} x^{3}+3Px & P\left( 3x^{2}+P\right) \\ 3x^{2}+P & x^{3}+3Px% \end{pmatrix}%$

Fourth order convergence:

$% \begin{pmatrix} x & P \\ 1 & x% \end{pmatrix}% ^{4}=% \begin{pmatrix} P^{2}+6Px^{2}+x^{4} & P\left( 4Px+4x^{3}\right) \\ 4x^{3}+4Px & P^{2}+6Px^{2}+x^{4}% \end{pmatrix}%$

Quintic convergence:

$% \begin{pmatrix} x & P \\ 1 & x% \end{pmatrix}% ^{5}=$ $% \begin{pmatrix} x\left( 5P^{2}+10Px^{2}+x^{4}\right) & P\left( P^{2}+10Px^{2}+5x^{4}\right) \\ P^{2}+10Px^{2}+5x^{4} & x\left( 5P^{2}+10Px^{2}+x^{4}\right) \end{pmatrix}%$

Dear Sirs, we are talking of a time period of approximately 3 thousands years (or
more), and no one can show any math textbook nor any math paper from any renown journal
showing something by far similar to this, and you already know what I am taling about: A math history issue.
This should be taken as a signal. But..., a signal from whom, from where?

Indeed, I hope nobody get offended by this fact as some math journalists certainly do.

And I must remark, there are an uncountable number of variants for all the methods based on the general
and unifying concept: The Rational Mean

https://domingogomezmorin.wordpress.com/

Last edited by arithmo; October 29th, 2017 at 11:21 AM.

October 29th, 2017, 11:29 AM   #46
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Quote:
 Originally Posted by arithmo Dear Sirs, we are talking of a time period of approximately 3 thousands years (or more), and no one can show any math textbook nor any math paper from any renown journal showing something by far similar to this, and you already know what I am taling about: A math history issue. This should be taken as a signal. But..., a signal from whom, from where?
It sure would be a lot easier to take you seriously if you stopped writing like some sort of nutty salesman. Reading this I expect you to start going on about the wonders of Johnson's Snake Oil Rejuvenating Tonic at any moment.

 October 29th, 2017, 01:25 PM #47 Member   Joined: Mar 2017 From: venezuela Posts: 36 Thanks: 3 THE CUBE ROOT Given the following set of three fractions whose product is trivial and equal to P. $A_{1}=\left\{ \frac{{a}_{1}}{b_{1}},\ \frac{{a}_{2}}{{a}_{1}},\ \frac{{Pb}% _{1}}{{a}_{2}}\right\}$ let's equalize the first two denominators and the last two numerators by keeping the values of the fractions the same, so only their forms get modified (see overlined and underlined expressions), as follows: $RM\left[ \frac{P{a}_{1}^{2}}{\underline{P{a}_{1}{b}_{1}}},\ \frac{% \overline{P{{a}_{2}{b}_{1}}}}{\underline{P{a}_{1}{ b}_{1}}},\ \frac{\overline{% {P{a}_{2}{b}_{1}}}}{{a}_{2}^{2}}\right]$ Example: ${{a}_{1}}=4,$ ${{a}_{2}}=5,$ ${{b}_{1}}=3$ and $P=2$: $A_{1}=\left\{ \frac{4}{3},\ \frac{5}{4},\ \frac{6}{5}\right\} =\left\{ \frac{32}{24},\ \frac{30}{24},\ \frac{30}{25}\right\}$ And let's compute their Rational Mean (Generalized Mediant): $RM\left[ \frac{32}{24},\ \frac{30}{24},\ \frac{30}{25}\right] =\frac{% 32+30+30}{24+24+25}= \frac{92}{73}$ That is called the Arithmonic Mean of the set $A_{1}$, and this is a particular case of the Rational Mean concept, as the Arithmetic and Harmonic means also are. As a matter of fact, in some cases the Arithmonic Mean become an Arithmetic or a Harmonic Mean. That is, all the means are just particular cases of the Rational Mean: The Arithmetic Mean is the Rational Mean of a set of fractions with equal denominators The Harmonic Mean is the Rational Mean of a set of fractions with equal numerators. The Arithmonic Mean is the Rational Mean of a set of fractions with equal numerators and equal denominators arranged according to a predefined chain-pattern. Notice the main difference between the Rational Mean concept and the well known Weigthed Mean concept: The Rational Mean deals with the form of the fractions by using factors for the numerators and denominators --keeping their values the same--, while the Weigthed Mean only uses factors for the decimal value of the fraction. You can make an equivalence for them both, however, they are pretty different, mainly because the Rational Mean wide scope certainly surpasses the Weigthed Mean. Notice that by changing the order of the fractions (not only their forms) then the Arithmonic Mean varies, and that's is the most important property for this operation, because we are not dealing only with the Forms of the fractions but also with their Location in the set. We are dealing with Relative Quantity, Math-Music, the harmonies of Number, The Simplest Arithmetic. Of course, some others probably would prefer just to say that the rational mean works with ordered pairs, while the weigthed mean works with rational numbers. Thus, changing the location in the set... ¿? $A_{1}=\left\{ \frac{{a}_{1}}{b_{1}},\ \frac{{a}_{2}}{{a}_{1}},\ \frac{{Pb}_{1}}{{a}_{2}}\right\}$ $A_{2}=\left\{ \frac{{{a}_{2}}}{{a}_{1}},\ \frac{{{a}_{1}}}{{b}_{1}},\ \frac{% {P{b}_{1}}}{{a}_{2}}\right\}$ $A_{3}=\left\{ \frac{{{a}_{2}}}{{a}_{1}},\ \frac{{P{b}_{1}}}{{a}_{2}},\ \frac{{{a}_{1}}}{{b}_{1}}\right\}$ and equalizing their numerators and denominators, respectively overlined and underlined as indicated, and computing the corresponding their Rational Means: $RM\left[ \frac{P{a}_{1}^{2}}{\underline{P{a}_{1}{b}_{1}}},\ \frac{\overline{% P{{a}_{2}{b}_{1}}}}{\underline{P{a}_{1}{b}_{1}}},\ \frac{\overline{{P{a}_{2}{% b}_{1}}}}{{a}_{2}^{2}}\right] =\dfrac{\left( {a}_{1}^{2}+2{{a}_{2}{b}_{1}}% \right) {P}}{{{a}_{2}^{2}+2P{a}_{1}{b}_{1}}}$ $RM\left[ \frac{\overline{{a}_{1}{{a}_{2}}}}{{a}_{1}^{2}},\ \frac{\overline{{% {a}_{1}{a}_{2}}}}{\underline{{a}_{2}{b}_{1}}},\ \frac{{P{b}_{1}^{2}}}{% \underline{{a}_{2}{b}_{1}}}\right] =\dfrac{{P{b}_{1}^{2}+2{{a}_{1}{a}_{2}}}}{% {a}_{1}^{2}+2{a}_{2}{b}_{1}}$ $RM\left[ \frac{{{a}_{2}^{2}}}{\underline{{a}_{1}{a}_{2}}},\ \frac{% \overline{{P{a}_{1}{b}_{1}}}}{\underline{{a}_{1}{a }_{2}}},\ \frac{\overline{P% {{a}_{1}{b}_{1}}}}{P{b}_{1}^{2}}\right] =\dfrac{{{a}_{2}^{2}}+2{P{a}_{1}{b}% _{1}}}{P{b}_{1}^{2}{+2a}_{1}{a}_{2}}$ We get a set of three new fractions whose product is aslo trivial and equal to P, you don't need to make any operation to check that their product is P, because numerators and denominators are canceling to each other: $\left\{ \dfrac{{P{b}_{1}^{2}+2{{a}_{1}{a}_{2}}}}{{a}_{1}^{ 2}+2{a}_{2}{b}_{1}% },\ \ \ \dfrac{{{a}_{2}^{2}}+2{P{a}_{1}{b}_{1}}}{P{b}_{1}^ {2}{+2a}_{1}{a}_{2}% },\ \ \dfrac{\left( {a}_{1}^{2}+2{{a}_{2}{b}_{1}}\right) {P}}{{{a}_{2}^{2}+2P% {a}_{1}{b}_{1}}}\right\} \ \$ The Arithmonic Mean... , o yes Indeed, the Freshman Trauma and those restrictive Cauchy's signs... I would like any guy from the audience to show me either any math textbook or any paper analyzing or even just mentioning this extremely simple operation, closely related to the ancient Arithmetic an Harmonic means. Remember this operation causes so much anger because of the aforementioned Freshman Trauma and Cauchy's signs https://domingogomezmorin.wordpress.com/ https://www.linkedin.com/in/domingo-...rin-500778127/ to be continued... Last edited by arithmo; October 29th, 2017 at 01:54 PM.
October 29th, 2017, 02:08 PM   #48
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Quote:
 Originally Posted by arithmo I would like any guy from the audience to show me either any math textbook or any paper analyzing or even just mentioning this extremely simple operation, closely related to the ancient Arithmetic an Harmonic means. Remember this operation causes so much anger because of the aforementioned Freshman Trauma and Cauchy's signs
I agree with romsek. Will you please stop trying impress us with your comments about how easy it is to use your methods and that they are all on the secondary school level. That is what I find most irritating in your posts, not because textbooks don't use it but because you are grandstanding. If your work has merit then that kind of acclaim might be useful. It is not useful here.

Thanks for the matrix comment. That makes it much easier to follow.

-Dan

 October 29th, 2017, 07:12 PM #49 Member   Joined: Mar 2017 From: venezuela Posts: 36 Thanks: 3 THE CUBE ROOT continues... So given a set of three fractions whose product is trivial and equal to P. $A_{1}=\left\{ \frac{{a}_{1}}{b_{1}},\ \frac{{a}_{2}}{{a}_{1}},\ \frac{{Pb}% _{1}}{{a}_{2}}\right\}$ by the agency of the Arithmonic Mean we got a new set of three new fractions whose product is also trivial and equal to P, you don't need to make any operation to check that their product is P, because numerators and denominators are canceling one to each other: $\left\{ \dfrac{{P{b}_{1}^{2}+2{{a}_{1}{a}_{2}}}}{{a}_{1}^{ 2}+2{a}_{2}{b}_{1}% },\ \ \ \dfrac{{{a}_{2}^{2}}+2{P{a}_{1}{b}_{1}}}{P{b}_{1}^ {2}{+2a}_{1}{a}_{2}% },\ \ \dfrac{\left( {a}_{1}^{2}+2{{a}_{2}{b}_{1}}\right) {P}}{{{a}_{2}^{2}+2P% {a}_{1}{b}_{1}}}\right\} \ \$ Now, if we make: $\frac{{a}_{1}}{b_{1}}=x,\text{ }\frac{{a}_{2}}{{a}_{1}}=x,\text{ }\frac{{Pb}_{1}}{{a}_{2}}=\dfrac{P}{x^{2}}$ then the new three fractions become: $\left\{ \dfrac{2x^{3}+P}{3x^{2}},\text{ }\dfrac{x^{4}+2Px}{2x^{3}+P},% \text{ }\dfrac{3Px^{2}}{x^{4}+2Px}\text{}\right\}$ and because this is at most high-school math it is straightforward to find its general matricial expression: $% \begin{pmatrix} x^{2} & x & P \\ P & x^{2} & Px \\ x & 1 & x^{2}% \end{pmatrix}% ^{2}=$ $% \begin{pmatrix} x^{4}+2Px & 2x^{3}+P & 3Px^{2} \\ 3Px^{2} & x^{4}+2Px & P^{2}+2Px^{3} \\ 2x^{3}+P & 3x^{2} & x^{4}+2Px% \end{pmatrix}%$ $% \begin{pmatrix} x^{2} & x & P \\ P & x^{2} & Px \\ x & 1 & x^{2}% \end{pmatrix}% ^{3}=% \begin{pmatrix} P^{2}+7Px^{3}+x^{6} & 3x^{2}\left( x^{3}+2P\right) & 3Px\left( 2x^{3}+P\right) \\ 3Px\left( 2x^{3}+P\right) & P^{2}+7Px^{3}+x^{6} & 3Px^{2}\left( x^{3}+2P\right) \\ 3x^{2}\left( x^{3}+2P\right) & 3x\left( 2x^{3}+P\right) & P^{2}+7Px^{3}+x^{6}% \end{pmatrix}%$ and so on... we get iterate functions with quadratically, cubic, quartic, quintic, ... convergence rate, for the cube root of P. n= nth convergence rate $% \begin{pmatrix} x^{2} & x & P \\ P & x^{2} & Px \\ x & 1 & x^{2}% \end{pmatrix}% ^{n}$ It is clear that nobody in this world could ever feel grandstanding (nor try to convince others to think such thing) for having developed such simple and trivial arithmetical methods, even though they trivially produce that which all math papers and textboks have consecrated as the exclusive product of the superb Infinitesimal Calculus and its fluxions. On the contrary, the fact that no ancient pictographic-logographic bamboo rod, nor bone script, nor clay tablet, nor papyrus, nor math-paper, nor math textbook contains any trace of these extremely trivial arithmetical methods certainly drive us to cogitate on what we have read and learned in there, and in reference to that, the last word we could think of is: "grandstanding", remind we are talking, at most, about high-school stuff before the whole history of math. This is the reason these methods, observations and cogitations are mainly dedicated to modern young students who truly feel the very esence of Number. The Arithmonic Mean a very particular case of the Rational Mean... , oh yes Indeed, the Freshman Trauma and those restrictive Cauchy's signs... https://domingogomezmorin.wordpress.com/ https://www.linkedin.com/in/domingo-...rin-500778127/ to be continued... Last edited by arithmo; October 29th, 2017 at 07:15 PM.
October 29th, 2017, 07:55 PM   #50
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 Originally Posted by arithmo Quadratic convergence: $% \begin{pmatrix} x & P \\ 1 & x% \end{pmatrix}% ^{2}=% \begin{pmatrix} x^{2}+P & 2Px \\ 2x & x^{2}+P% \end{pmatrix}%$ Cubic convergence: $% \begin{pmatrix} x & P \\ 1 & x% \end{pmatrix}% ^{3}=% \begin{pmatrix} x^{3}+3Px & P\left( 3x^{2}+P\right) \\ 3x^{2}+P & x^{3}+3Px% \end{pmatrix}%$ Fourth order convergence: $% \begin{pmatrix} x & P \\ 1 & x% \end{pmatrix}% ^{4}=% \begin{pmatrix} P^{2}+6Px^{2}+x^{4} & P\left( 4Px+4x^{3}\right) \\ 4x^{3}+4Px & P^{2}+6Px^{2}+x^{4}% \end{pmatrix}%$ Quintic convergence: $% \begin{pmatrix} x & P \\ 1 & x% \end{pmatrix}% ^{5}=$ $% \begin{pmatrix} x\left( 5P^{2}+10Px^{2}+x^{4}\right) & P\left( P^{2}+10Px^{2}+5x^{4}\right) \\ P^{2}+10Px^{2}+5x^{4} & x\left( 5P^{2}+10Px^{2}+x^{4}\right) \end{pmatrix}%$
Finally, you address the convergence of your methods and now I understand your problem. This is not what is meant by higher order convergence. Just because your iteration scheme becomes a higher order polynomial means nothing about convergence of a solution. Of course you get a higher degree polynomial....you are iterating a polynomial!

The claim that some iteration scheme, $N : \mathbb{R} \to \mathbb{R}$ converges quadratically to a solution of $f(x) = 0$ means
$f \circ N (x + \delta) = \mathcal{O}(\delta^2)$
For $k^{th}$ order convergence the definition follows as
$f \circ N (x + \delta) = \mathcal{O}(\delta^k)$.

These convergence properties are strongly tied to the fact that near a true solution, an iterative method is expected to be a strong contraction (i.e. Lipschitz with constant $L \ll 1$.

Since iterative methods are only really required when $f$ is nonlinear and the contraction mapping theorem guarantees a unique solution, it is clear that this kind of converge makes an implicit assumption that $\delta$ is "small". Here small must mean at least that if $x_0,x_1$ are both roots of $f$, then $N$ can not possibly be a contract if $x_0 \in [x_1 - \delta, x_1 + \delta]$.

Some sort of analysis of this sort is required before you can claim your methods have quadratic, cubic, or any other order of convergence. I have no idea how you can carry out this analysis without asymptotic analysis which requires a great deal of "the superb infinite calculus".

Moreover, I am very skeptical to believe that your method has higher order convergence, not because my feelings are hurt as you say, but because this methodology looks very similar to the secant method which has subquadratic convergence in general (mainly because it doesn't require any smoothness beyond continuity).

Best of luck

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