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0|0|0|||||Newton %26 Roots|John Ungar|1|10:22:32|06/25/2010|Hello everybody%21%0D%0AI found this one %3D%3E http%3A%2F%2Fwww.cut-the-knot.org%2Fwiki-math%2Findex.php%3Fn%3DCalculus.ApproximationOfRoots%0D%0Aon your Webproject and it seems to be a pretty hot hot piece of math. But one question is still open for me%3A Whilst the old Newton-mechanism in his approximation is correlating to the chosen N%2C the Reineke Algorhythm seems not to do anything like that. Seems this nice little formula has not any relation to the size of a chosen N%2C its always approxying in 5-9 steps. And the question therefor is%3A%0D%0AIs there a way to an analytic proof to show%2C that there is a relation between a chosen N and the number of approximations on the one hand and no relation between N and the approx count in the Reineke algorhythm on the other%3F%0D%0A%0D%0AThank U so far%0D%0AJohn Martin Ungar
1|1|0|||||RE%3A Newton %26 Roots|alexb||21:38:05|06/26/2010|As far as I understand%2C Chris did not mean to apply this formula iteratvely. But you are right. An estimate is still missing.
2|2|1|||||RE%3A Newton %26 Roots|C. Reineke|1|08:54:59|06/28/2010|Alex%2C%0D%0A%0D%0Ayou wrote%3A%0D%0A%0D%0A%93As far as I understand%2C Chris did not mean to apply this formula iteratively.%94%0D%0A%0D%0AYes and no.  Do you remember my example when I computed sqrt%2866%29 with two%0D%0Aiterations%3F With a %93good%94 initial guess you can use the formula to approximate a root%0D%0Ain one step%2C but with a %93poor%94 initial guess you need more iterations%2C of course.%0D%0A%0D%0AHowever%2C I would like to repeat John%92s observation%3A%0D%0A%0D%0AAssume we want to compute the nth root%281000%29%2C the initial guess is always%3D1%2C desired accuracy%3A EXCEL%92s 14 digits.  %0D%0A%0D%0ANow let n increase%2C say%2C 2%2C 3%2C 4%2C 5%2C 10%2C 100%2C 1000.  Then you will see that the number of iterations needed to reach the desired accuracy does not increase %96 contrary to Newton%92s method. To compute the 1000th root%281000%29 Newton%92s algorithm needs 692 iterations%2C the new algorithm needs %28always%29 7%21%0D%0A%0D%0ASo it seems that the number of iterations does not depend on n.%0D%0AI%92ve got absolutely no explanation for this property%2C but maybe it%92s possible to prove%0D%0Ait mathematically%2C as John wrote.%0D%0A%0D%0ABy the way%2C the algorithm also converges if n is negative. %0D%0A%0D%0AKind regards%0D%0A%0D%0AChris%0D%0A