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carel
Posts:
161
Registered:
12/12/04


Re: Root
Posted:
Jul 11, 1999 9:42 AM


The method of Newton is very elegant to solve for roots
Let f(x) = x^2  A where A is the number to get the square root from.
Then g(x) = f'(x) = 2x
Then xnew = xold  f(xold)/g(xold) = xold  [xold^2  A]/[2xold]
So guess a value for x and then obtain a new guess value.
You will soon obtain x = A^(.5)
In general you can obtain any root by letting f(x) = x^n  A and g(x) = nx^(n1)
Carel James Adelman wrote in message <378894F7.MD1.2.j.adelman@ukonline.co.uk>... >James Adelman wrote on 11 Jul 99 12:57:39 +0000 in sci.math: > >> Kaimbridge wrote on Sun, 11 Jul 1999 08:52:29 GMT in sci.math: >> >> > In article <k4Ph3.7360$Le1.136634@news2.randori.com>, >> > "User" <pohanl@aol.com> wrote: >> > > >> > > Does anyone know how to go about doing a square root or cube root >> > > calculation? >> > >> > There was a good post on "the hand method" done back in January: >> > >> > http://www.deja.com/=dnc/getdoc.xp?AN=433077696 >> > >> > ~Kaimbridge~ >> > >> An alternative might be to use the binomial expansion, although this >> will probably make for a lot more arithmetic than the one at the >> URL above; it _may_ need fewer terms for accuracy, but I don't think >> so, you's have to try it. >> >> Let's try 57: >> >> 57^.5=49^.5*(57/49)^.5 >> =7*(1+8/49) >> >Missed some brackets: should be > > =7*[(1+(1/2)(8/49)+(1/2)(1/2){(8/49)^2}/2+(3/2)(1/2)(1/2){(8/49)^3}/3!+.. .] >> >> It works, but without a calculator (or even with one) it requries lots >> of effort. I suppose the only good news is the factorial on the >> bottom. >> >> >  >> > Note that "news" in "@mydejanews.com" has been >> > dropped. >> > >> > >> > Sent via Deja.com http://www.deja.com/ >> > Share what you know. Learn what you don't. >> >>  >> >> James Adelman > > > >James Adelman



