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What is the fastest way to compute the value of the unique positive real root corresponding to the following polynom: :p(x) = a*x^5 + b*x^4 + c*x^3 + d*x^2 + e*x - f = 0 where a, b, c, d, e, f are all strictly positive. Please something faster or more subtle than newton...

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  • $\begingroup$ You could use the observation that the root is less than the min of $(f/a_i)^{1/i}$ and greater than the max of $(f/5a_i)^{1/i}$, where I have relabelled the coefficients. However, this is not subtle and at best I do not see how it will save much time or effort over using Newton-Raphson or even bisection. $\endgroup$ Feb 5, 2014 at 17:50
  • $\begingroup$ Sorry: in both cases one needs min, not min and max. Also, because the coefficients $a_i$ are all positive, you can restrict your attention to the appropriate roots of $ (f/5a_i)$ and use roots of $(f/a_i)$ only if it narrows the interval. $\endgroup$ Feb 5, 2014 at 18:24
  • $\begingroup$ Do you have some reason, brian, to think there is something faster than Newton's Method? $\endgroup$ Feb 6, 2014 at 0:16
  • $\begingroup$ there is the observation that the number exist and somehow easily bounded, hence I was thinking to some kind of magical number (expressed as a function of the coefficients exactly like Masked Avenger put forward) and then one or 2 round of newton... $\endgroup$
    – brian
    Feb 6, 2014 at 12:53
  • $\begingroup$ to Masked Avenger: thanks that is a good starting point for the Newton. I see however a potential computation tradeoff : we might have to pay a lot computationnaly to compute the cubic roots compared to an uneducated guess and several more rounds of analytical Newton... this is where a need a kind of closed form for the solution or a nice but simple to compute guess to start the newton. $\endgroup$
    – brian
    Feb 6, 2014 at 13:03

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