
Re: cube root of a given number
Posted:
Jul 25, 2007 11:11 AM


On 24 Jul, 22:07, gwh <ghug...@cei.net> wrote: > On Jul 24, 9:36 am, arithmonic <djes...@gmail.com> wrote: > > > On 24 jul, 06:44, "sttscitr...@tesco.net" <sttscitr...@tesco.net> > > wrote: > > > > Your historical claim may well be true, but at > > > least one poster states he has a reference predating your > > > claim. It would be interesting if he could post the method > > > he found in the Handbook > > > It is not just a problem on the false and unethical statements of your > > friend Grover Hughes who > > allegued that my methods were exactly the same than those from > > Eshbach's, 1945 > > "Handbook of Engineering Fundamentals". > > You endorsed his unethical claim and stated that he was right and my > > methods were not new, at all. > > > You and him are now facing nonethical, false and negligent statements > > before the sci.math audience. > > > Grover Hughes WILL NOT be able to prove his claim and you know that. > >  > > You and your friend Grover Hughes have not proved what you claimed > > about Hurtwitz and Eshbach, that's why I repeat to you: > > > NOTICE that your friend Grover Hughes have not shown any single proof > > of what he claimed, and > > left the discussion. what a cheek, indeed. That is totally unethical. > > My, my! Leave town for a few days, and look what happened while my > back was turned! I've never been so popular before, and all because I > remarked that an old text showed how to extract cube roots! Well, here > it is I'll do the best I can to type it in a form that I hope will > be readable. If anyone's interested, I'll be happy to scan the page > and email it directly to you just ask. Anyway, here's what page 204 > of Handbook of Engineering Fundamentals", edited by Ovid . Eshbach, > copyright 1936 , gives, to find the cube root of 158252.632929 : > > 158 252. 632 929  > 54.09 > 5^3 = 125 > _____ > 300 X 5^2 = 7500  33 252 > 30 X 5 X 4 = 600  > 4^2 = 16  > _____ > 8116  32 464 > 300 X 540^2=87480000  788 632 929 > 30 X 540 X 9= 145800  > 9^2 = 81  > _________  > 87625881  788 632 929 > > I know that it would have been better had Eshbach chosen a number > which was not a perfect cube, but the method works fine for that case, > anyway I used it sometimes to check my slide rule value, for > greater precision.... > > Sorry about the clumsy presentation, but that's the best I know how to > type this stuff the lines above which are blank except for an > underline are supposed to appear directly under the 4^2 = 16 > and 9^2 = 81 > but I can't underline and type numbers at the same time. Is there a > way? > > Eshbach of course gives a written explanation for each step, but to > save all of us time, I am assuming that everyone reading this is > perfectly capable of working that out for himself. If you do want the > entire text from Eshbach, lemme know and I'll either email it (as I > said earlier) or I'll type it out and post it here some day soon I > don't look at sci.math every day, just when I feel like it, so forgive > the time lapse. Write me directly if you wish, at ghug...@cei.net. > > BTW, does arithmonic always get so excited and upset? I never intended > to help his ulcer along...... Yes, he is quite mad. He thinks we have in some way lied and conspired to subvert his worldview  namely that the mathematical world since Babylonian times has schemed to prevent his "method" emerging and that even today we (for you are my secret friend) and the mathematical community are deliberately failing to understand his valuable insights and voluminous rantings.
Unfortunately, the method you describe is not his. But he would not have believed you, even if you had found a reference.
El Morono simply does not understand that his method is a trivial variant of Heron's method. Basically, to find say the square root of 7, you would start with a 1x7 rectangle. You want to preserve the area but make the sides more and more equal so you take the average of 1 and 7 (1+7)/2 = 4. This is one side of a new rectangle, the other side is 7/4 =1.75. The area of the rectangle is 7, but the sides are more equal. All Morin does is represent the sides as fractions (7/1, 1/1) and use the mediant as the averaging function. (7+1)/(1+1) = 4/1. The other side is 7/(4/1) =7/4. The area of the rectangle is preserved, the sides are more equal. (4 +7)/(4+1) = 11/5 and so on.
He will never understand what a best rational approximation is.

