jamesvnguyen

DH, here is my appreciation for your advice! How to find square root of a number Sqrt(b) ? find (a) such that (a*a) is close to (b) For example: if b=178, (a) can be 6, 7, 8, 9, 10, 11, 12, or 13…… Let V = a + b/a Let Z = (V+4*b/V+16*b*V/(V*V+4*b)+64*b*V*(V*V+4*b)/(POWER((V*V+4*b),2)+16*b*V*V)) You will see Sqrt(b) = Z/32 + 8*b/Z [with either (a) that you chose] Is it powerful? DH, there is no body here but you and me. Why are you here, just curiosity? I think that I have something that I would like to contribute. But I don't know how to start. I need to talk to some one who is an expert about this. And you seem like the one that I am looking for. Will you willing to discuss about this subject. If you do, here is my email: dog92707@yahoo.com

I'm not an expert about this. But this is very straight forward, no guessing, no looping, involves exact number of steps, small (can do by hand). I stop short to call it a formular (two big formular). Forget about it? DH, I would like to learn from you. I'm wondering, is there any technique can find the roots of the polynomial (for example, hexic) which does not have a real root (only complex roots)?

I appreciate you to check me out. You can give me any set of coefficients of quintic, hexic, and septic; and I will give you the answers. (I stop at autic, since I don't have time to do it- Is it worth to continue?) My goal is finding the roots of the polynomial of fifth order and higher. The accuracy is up to the floating point of the system allowed. Is it worth to continue? (It is not a big program or anything likes that. Everyone can have it in their calculator)

the solutions of X^5 + 2X + 1 = 0 are: -0.70187357 + -0.8796972 i = 1st Solution -0.70187357 + 0.8796972 i = 2nd 0.94506809 + 0.8545175 i = 3rd 0.94506809 + -0.8545175 i = 4th -0.48638904 = 5th Only take me a few nano second to get them!

I understand why you said that. It is hard to believe. But how hard is it to prove it? If you give me a RANDOM set of coefficients of the polynomial (fifth order +), I will give you ALL the roots in nano second. How hard it is to know if they are the correct answers! All you need to do is plug them in to see if they yield to Zeros. Any way, If it is true what I just said, is it new and useful? I appriciate very much for your response! (Yes, for the fifth order and lower, I can do it by hand.)

If I say that I can find the roots of the general form of quintic, hexic, septic, autic, nonic ... PRECISELY without guessing the initial values (involved exactly few iteration, of course); I'm wondering how useful it is? Don't think that I am crazy, just tell me if it is new, and worth to publicize. Thanks for your advice!