
Re: cube root of a given number
Posted:
Jul 22, 2007 8:38 PM


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 ROOTSSOLVING 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 > > > ROOTSOLVING 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^52 =0 is simple. > > s(x,y) is the sign of binary quntic x^52y^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 > highorder arithmetical methods shown in my webpages, and your > allegued Hurtwitz's RootSolving 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. ?

