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Consider roots of $f(u)=u^{-1/4}-2/15 u-1/6$ which can be written as $$ u^{-1/4}e^{-1/4 (2n \pi i)}-2/15 u-1/6=0 $$ where $u^{-1/4}$ is taken as principal-value. Or as a GroebenerBasis:

$$ f(u)=-810000 + 625 u + 2000 u^2 + 2400 u^3 + 1280 u^4 + 256 u^5=0 $$ which has five solutions. I was wondering if someone could explain to me why when I use NSolve to solve $u^{-1/4}e^{-1/4 (2n \pi i)}-2/15 u-1/6=0$, it returns two solutions for n=2 but only one solution for n=0,1,3?

theK[n_] := Exp[-1/4 2 n Pi I];
theUVals = (u /. 
     NSolve[u^(-1/4) theK[#] - 2/15 u - 1/6 == 0, u] & /@ {0, 1, 2, 3})

{{4.04009},
{0.554451 -4.74207 I},
{-5.0745-2.93375 I,-5.0745+2.93375 I},
{0.554451 +4.74207 I}}

Update as per comment below:

I can back-substitute the five solutions into the original equation but must use the branch form $u^{-1/4}=u^{-1/4}e^{-1/4(2 n\pi i)}$ to do so as Mathematica I believe, will interpret the naked $u^{-1/4}$ expression as principal-value:

MapThread[(#1^(-1/4) theK[#2] - 2/15 #1 - 
    1/6) &, {theUVals, {0, 1, 2, 3}}]

    {{8.32667*10^-17},
{2.77556*10^-17+2.22045*10^-16 I},
{-6.93889*10^-16+1.55431*10^-15 I,-6.93889*10^-16-1.55431*10^-15 I}, 
 {2.77556*10^-17-2.22045*10^-16 I}}
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  • $\begingroup$ Have you tried substituting your five solutions back into your original equation? $\endgroup$
    – John Doty
    Commented Feb 8, 2021 at 17:16
  • $\begingroup$ Yes. Updated post above. $\endgroup$
    – Dominic
    Commented Feb 8, 2021 at 17:36
  • $\begingroup$ Mathematica cannot read your mind. If you want to solve an equation involving a multivalued function, you must ask for the solution for each value. Or, in some special cases like this one, use VerifySolutions->False. $\endgroup$
    – John Doty
    Commented Feb 8, 2021 at 17:43
  • $\begingroup$ Ok. My main question though is why is NSolve returning two solutions for a single branch $u^{-1/4(4 \pi i)}$ and single solutions for the other branches? $\endgroup$
    – Dominic
    Commented Feb 8, 2021 at 17:45
  • $\begingroup$ Why do both solutions satisfy the equation in that case? $\endgroup$
    – John Doty
    Commented Feb 8, 2021 at 17:50

1 Answer 1

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Suppressing solution verification, using Solve and RootReduce to get true zeros, and substituting the results back into the equation, I get:

u^(-1/4) theK[2] - 2/15 u - 1/6 /. Solve[u^(-1/4) theK[2] - 2/15 u - 1/6 == 0, u, 
VerifySolutions -> False] // RootReduce // N
(* {-1.41069, 0., 0., -0.87287 + 0.391683 I, -0.87287 - 0.391683 I} *)

for case 2. So, in that case, only two of the five roots you get from using the inverse of the fourth root to make a polynomial actually solve the original equation. For other cases, only one solution works. The default for VerifySolutions is rejecting the others.

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  • $\begingroup$ But my back-substitution for all five solutions above satisfies the original branch expression (numerically) using the factor $e^{-1/4(2n\pi i)}$. Bit confusing to me. $\endgroup$
    – Dominic
    Commented Feb 8, 2021 at 17:39

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