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Re: cube root of a given number
Posted:
Jul 22, 2007 8:38 PM
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On 22 Jul, 23:47, arithmonic <djes...@gmail.com> wrote:
> On 22 jul, 07:29, "sttscitr...@tesco.net" <sttscitr...@tesco.net>
> wrote:
>
> > On 22 Jul, 04:28, arithmonic <djes...@gmail.com> wrote:
>
> > > On 21 jul, 17:39, "sttscitr...@tesco.net" <sttscitr...@tesco.net>
> > > wrote:
>
> > > > On 21 Jul, 21:38, arithmonic <djes...@gmail.com> wrote:
>
> > > > > On 16 jul, 04:39, "sttscitr...@tesco.net" <sttscitr...@tesco.net>
>
> > > 1.- I challenge you to show such Eshbach's method in this thread,
> > > because both of you are trying to state that my methods --based on the
> > > Rational Mean-- are the same as the one you read in Eshbach's
> > > work ("Handbook of Engineering Fundamentals).
>
> > I'm sure the poster knows what he read.
>
> Why are you so sure? Are you his friend?
> I am sure that the poster called Grover Hughes does not know a single
> bit of all what he was talking about, tha's why I challenged him and
> be sure the method he mentioned is by far similar to the methods shown
> in my web pages. The sci.math audience is also waiting and observing
> the results of my challenges to both of you.
>
> Now to the very specific point on ROOTS-SOLVING METHODS:
>
>
>
> > > 2.- I challenged you in this posting to show to the sci.math audience
> > > a very simple numerical example on your alleged GENERAL Hurwitz's
> > > ROOT-SOLVING METHOD for computing, say, THE FIFHT ROOT OF 2.
>
> > There are 5 fifth roots of 2. Which one do you want ?
> > Are you saying you can find complex roots too ?
>
> > Finding the real root of x^5-2 =0 is simple.
> > s(x,y) is the sign of binary quntic x^5-2y^5
>
> > The root must lie between (0,1) = 0 and (1,0) = "inf"
> > calculate s(0,1) and s(1,0). Form the mediant (1,1)
> > At any stage in the process s(xn,yn) will be 1 or -1.
> > if s(xn,yn) = un The new new mediant is formed with
> > (xk,yk) where k is the largest index <n such that un*uk = -1
>
> > 0 1 -1
> > 1 0 1
> > 1 1 -1
> > 2 1 1
> > 3 2 1
> > 4 3 1
> > 5 4 1
> > 6 5 1
> > 7 6 1
> > 8 7 -1
> > 15 13 1
> > 23 20 1
> > 31 27 -1
> > 54 47 1
> > 85 74 -1
> > 139 121 1
> > 224 195 1
> > 309 269 1
> > 394 343 -1
> > 703 612 -1
> > 1012 881 -1
> > 1321 1150 -1
> > 1630 1419 -1
> > 1939 1688 -1
> > 2248 1957 -1
> > 2557 2226 -1
> > 2866 2495 -1
> > 3175 2764 -1
> > 3484 3033 -1
> > 3793 3302 -1
> > 4102 3571 -1
> > 4411 3840 -1
> > 4720 4109 -1
> > 5029 4378 -1
> > 5338 4647 -1
> > 5647 4916 -1
> > 5956 5185 -1
> > 6265 5454 -1
> > 6574 5723 -1
> > 6883 5992 -1
> > 7192 6261 -1
> > 7501 6530 -1
> > 7810 6799 -1
> > 8119 7068 1
> > 15929 13867 -1
> > 24048 20935 -1
> > 32167 28003 -1
> > 40286 35071 -1
> > 48405 42139 1
> > 88691 77210 1
> > 128977 112281 1
> > 169263 147352 -1
> > 298240 259633 -1
>
> > Can you solve x^3 -2x^2 -x+1 =0 ?
>
> > > Notice that I am challenging you and your friend Grover Hughes with two very simple inquires.
>
> > I don't know why you think Grover Hughes is a friend of mine.
>
> > You used to claim that you could find the best simultaneous
> > approximations to cubrt(2), cubrt(4).
> > As you methods are so revolutionary, I would have thought
> > this would have been possible too.
>
> > In fact, you can use your methods for simultaneous approximation of
> > cubrt(2), cubrt(4), but it would be
> > sheer luck if a best approximation was found.
> > Shall I give you a hint ?
>
> > Do you understand the diffference between fast convergence and best
> > rational approximation ?
>
> One step at the time, please.
>
> Let us focus on the main point of this thread: The extremely simple
> high-order arithmetical methods shown in my webpages, and your
> allegued Hurtwitz's Root-Solving Method as being the same thing that
> my methods, in such a way, that when talking about them you stated: "I
> don't think the claim that these methods are in any way new stands up
> to scrutiny."
>
> Only after clarifying this point to the sci.math audience I will
> procede to answer all the other
> remarks you have made in this posting.
>
> But... Please, If you don't mind, I want to ask you just two more
> questions specifically related to this table you have shown on the
> fifth root and your remark about the methods shown in my webpages: "I
> don't think the claim that these methods are in any way new stands up
> to scrutiny."
>
> ARE YOU JUST PLAYING A JOKE OR WHAT?
> HAVE YOU EVER READ A BOOK ON THE HISTORY OF MATHEMATICS?
What I find strabge is that you claim to have invented
a new method of root solving and yet seem to be incapable
of applying it to any problem I pose.
Can you solve x^3 -2x^2 -x+1 =0 ?
Why is it that I can apply your method to simultaneously
approximate cubrt(2), cubrt(4) but you can't. ?
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