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Is there a general formulation for finding all real roots of a polynomial by using the Newton-Raphson Method?

I have seen the answer from https://math.stackexchange.com/a/998489/1205840. This method is great! But what if we regard both the function and the function's derivative as a black box and the inputs must be real numbers? Is there a similar way to find all real roots instead of all roots?

Thanks!

RobPratt
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Guanqun Ma
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  • Yes, if you prepare first using the Vincent/Akritas method to find transformations to polynomials with one positive root where then the Newton method works unconditionally (not an answer as not very practical). Reversing the transformation then finds the root of the original polynomial. Just using Newton with deflation might miss a pair of real roots for a conjugate pair of complex roots close to the real axis. Finding all roots and filtering out the real ones with extra attention to nearly real ones and generally tight clusters on the real axis is less complicated. – Lutz Lehmann Aug 02 '23 at 04:53
  • Hi Lutz, thanks for your comment. Sometimes the function is not represented in power basis, and I try to avoid deriving the explicit power basis from the original representation. Newton's Method only uses the function value and its first derivative. This is what I want. But the variables of the function are required to be real-number. This keeps me away from the above method. – Guanqun Ma Aug 02 '23 at 17:30
  • With the same ease you can compute also second derivatives or remainders (like for the Bairstow method). The real problem is that you do not have a deflation method without expanding the polynomial. So one would need a large systematic set of seeds/starting points. // This all requires complex numbers, which should not really be a restriction. Except if you program in Java. Or if you were using the "complex dimension" for the complex-step method of computing the derivative. – Lutz Lehmann Aug 02 '23 at 18:27

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