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Topic: cube root of a given number
Replies: 112   Last Post: Jan 10, 2013 1:39 PM

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 arithmetic Posts: 106 From: venezuela Registered: 1/23/06
Re: cube root of a given number
Posted: Jul 21, 2007 11:51 PM

On 16 jul, 18:24, gwh <ghug...@cei.net> wrote:
> On Jul 14, 10:30 pm, arithmeticae <djes...@gmail.com> wrote:
>

> > If you really like to analyze the most simple high-order root-solving algorithms then you should take a look at:
>
> >http://mipagina.cantv.net/arithmetic/rmdef.htm
>
> > It is striking to realize that these new extremely simple artihmetical algorithms do not appear in any text on numbers since Babylonian times up to now.
>
> Maybe not in "any text on numbers", but back in 1945 I purchased a
> copy of "Handbook of Engineering Fundamentals", by Eshbach, and the
> cube root extraction scheme described there was precisely the same as
> the scheme described on one of the links given on the above website. I
> used that method lots of times in my engineering career when I needed
> more precision than my trusty log-log duplex decitrig slide rule was
> able to give me.
>
> Regards,
>
> Grover Hughes

Both you and I certainly know that you will not answer to my simple
challenge, i mean, to show any
simple numerical example on Eschbach's method, so sci.math audience
could be able to realize that you are telling the truth, mainly
because you certainly know that the new simple arithmetical methods
shown in my webpages and book have no precedents, at all.

However, I will do a favor to you and your friend
sttscitrans@tesco.net. Yes I will
show to you a very simple example on computing square roots by agency
of the RATIONAL MEAN (notice
that i am not talking about NEITHER FAREY FRACTIONS NOR PELL'S
EQUATION)

THE VERY SIMPLE EXAMPLE FOLLOWS. I am sure you will also be able to
what follows in so many books on numbers (PUN INTENDED, OF COURSE),
worst when considering
that we are talking at this moment about SIMPLE SQUARE ROOTS:

All this comments comes from the contents of the webpage:
http://mipagina.cantv.net/arithmetic/rmdef.htm
and the book: LA QUINTA OPERACI?N ARITM?TICA. Arithmonic Mean. ?
-----------------------------------PRELIMINARY
NOTE-----------------------------------
The Rational Mean of the fractions: f1=a1/b1 and f2=a2/b2 is:

Rm[f1, f2] = (a1+a2)/(b1+b2)

By agency of such a simple arithmetical operation you can produce all
the Householder expressions for the Nth root of any number P.
Notice that if you change the form of the fraction a1/b1= (x/x)*(a1/
b1)= (x*a1)/(x*b1) then you will get another result (provided that x
is not equal to 1):

For example: Rm[(x/x)*f1, f2]= (x*a1 + a2) / (x*b1 + b2)

--------------------------------END OF
NOTE----------------------------------------

Higher-order rational process based on the Rational Mean:

FUNDAMENTAL PRINCIPLE:
Any two fractions whose product is P represent two rational
approximations -by defect and excess-- to the square root of P.

If departing from those two fractions you can compute two mean values
whose product is also P then you have another two closer
approximations to the root. By continuing this process you will get a
root-solving algorithm for the square root of P, moreover, by using
the Rational Mean you will get a higher-order root-solving algorithm.

Starting with a set of two fractions f1, f2 whose product is f1*f2 =
P.
For example:
f1 =x/1 f2=P/x

Compute the following two rational means:

Rm[(x/x)*f1, f2] = (P+x^2) / (2x) (Newton)

Rm [(P/P)*f1, (x/x)*f2]= (2Px) / (P+x^2)

It yields, two expressions whose product is trivial and equal to P
and are closer to the square root of P.
You can use each of those new functions as independent iterating
functions both of them converging quadratically.

If you don't like quadratic convergence then compute another two
similar rational means by previously assigning those new functions to
f1 and f2, as follows:

f1 = (P+x^2) / (2x)
f2 = (2Px) / (P+x^2)

The two new rational means yields:
Rm [(x/x)*f1, f2] = (x^3 + 3Px) / (P
+3x^2) (Halley)
Rm [(P/P)*f1, (x/x)*f2]= (P^2 + 3Px^2) / (x^3 + 3Px)

two expressions whose product is trivial and equal to P, both of them
multiply by THREE the number of exact digits in each iteration.

If you prefer more convergence speed, then make:

f1 = (x^3 + 3Px) / (P+3x^2)

f2 = (P^2 + 3Px^2) / (x^3 + 3Px)

and compute other two rational means:

Rm [(x/x)*f1, f2] = (x^4 + 6Px^2 + P^2) / (4x^3 + 4Px)
(Householder)
Rm [(P/P)*f1, (x/x)*f2]= (4Px^3 + 4xP^2)/ (x^3 + 3Px)

Two expressions whose product is trivial and equal to P, both of them
multiply by FOUR the number of exact digits in each iteration.

By continuing this process, in the next step you will get two
functions which multiply by five the number of exact digits in each
iteration.

And so on.

Believe it or Not!.
Based on the evidence at hand, this so na?ve, trivial, natural and
simple rational process has no precedents since Sumerians times up to
now. We have not used neither any Cartesian-decimal system, nor any
derivatives, nor infinitesimal calculus, at all
I think, many experts on the history of mathematics should cogitate on
the very long story on root-solving. Indeed, it is disturbing to
realize these so simple rational processes based on the rational mean
do not appear in any book on numbers since ancient times up to now.

All this is fully explained in my book:
LA QUINTA OPERACI?N ARITM?TICA. Arithmonic Mean ? Domingo Gomez Morin.

and its webpage:
http://mipagina.cantv.net/arithmetic

Date Subject Author
4/20/04 vsvasan
4/20/04 A N Niel
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