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An apparently new method to compute the $n$th root of any complex number
I found a series of articles (in Portuguese) by a Brazilian mathematician named Ludenir Santos, where presents a series of iterative methods, he said new, to extract nth roots of any complex number...
View ArticleAnswer by hmakholm left over Monica for An apparently new method to compute...
If one allows successive approximation methods (which indeed should be fair game for numerical solutions), then Newton-Raphson has been known for a long time already.The iteration you quote seems to be...
View ArticleAnswer by Lutz Lehmann for An apparently new method to compute the $n$th root...
If $|x|<1$ then the binomial series for the n-th root reads as$$(1+x)^{\frac1n}=1+\frac1n x+\frac{\frac1n(\frac1n-1)}2x^2+…$$which shows that the given formula is an approximation to the truncation...
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