$$ax^{2}+bx+c=0$$ $$x=-\frac{b}{a}+\frac{1}{\frac{b}{c}+\frac{1}{-\frac{b}{a}+\frac{1}{\frac{b}{c}+\frac{1}{...}}}}$$ $$ax^{3}+bx^{2}+cx+d=0$$ $$x=?$$ I know it can't recur like the quadratic continued fraction. Must be arithmetic operations. Got ideas?
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1What is exactly the questions here? – PC1 Jul 01 '22 at 00:35
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1Does this answer your question? Compare five ways of solving cubic equation by iterations (nested expressions). See also Finding a substitution that eliminates the squared term from a cubic equation for how to reduce it to a depressed cubic in the first place. – dxiv Jul 01 '22 at 00:41
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1So far as I know, apart from the quadratic case, there is no simple description of continued-fraction expansions of ... anything. But/and, also, your question title, etc. is misleading... – paul garrett Jul 01 '22 at 00:41
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Second question: since we are using infinitely many operations, could these types of formulas actually solve the quintic or higher degree polynomials? 1 of the restrictions of the unsolvability of the quintic (Ruffini-Abel theorem) is we only use finitely many operations, but here we use infinitely many. – Baby Hearty Bear Jul 01 '22 at 01:14
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1@BabyHeartyBear Any of the numerical methods for finding polynomial roots with arbitrary precision is iterative at core, though the recurrence relations become complicated. – dxiv Jul 01 '22 at 02:56