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Let's say you have the polynomial with fractional power: $$P(z)=z^{9/2}-\frac{3}{2} z^4 +\frac{1}{2} z^3=0, \quad z \in \mathbb{C} \tag{1}$$ I have found two real roots of Eq. (1): $$z_1=0,$$ and $$z_2=1.$$

I'm looking for a theorem that define the number of roots and their multiplicity in the interval $[-1, 1]$.

I have found the Sturm's theorem that expresses the number of distinct real roots of $P(x)$ located in an interval. Also I have read the Budan's theorem, is the theorem for computing an upper bound on the number of real roots a polynomial has inside an open interval.

The Sturm's theorem and the Budan's theorem are both for an integer power of real polynomial equation $P(x)$, $x \in \mathbb{R}$. Later I found bounds on (complex) polynomial roots based on the Rouche's theorem that is usually used to simplify the problem of locating zeros.

In my case the highest power is fractional, $\frac{9}{2} \in \mathbb{R}$.

Question.

Does the rule exist to find or estimate the
lower/upper bound on the number of roots (real or comlex) a polynomial in an interval when polynomial's power is not integer?

Edit. If I wrote: solve x^(9/2)-3/2*x^4 +x^3/2 then Wolframalpha gives two roots only: 0 and 1.

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    $\begingroup$ say you have the polynomial equation That's not a polynomial since it has fractional powers of the variable. However, this particular expression is easy to factor, and you are half-way there already. $\endgroup$ – dxiv Jun 29 '18 at 3:38
  • $\begingroup$ @dxiv, thank for the comment, I have edited the post. I need to find the eatimation for common case, but not for the particular expression only. $\endgroup$ – Nick Jun 29 '18 at 3:48
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    $\begingroup$ You keep calling it a polynomial, but it's not. If the non-integer exponent is always a rational, you could technically reduce it to a polynomial by substitution (though may not be optimal), otherwise if you mean to cover general cases like $x^{\pi} - 3 x^e + 2 x^2 +1 = 0$ then the methods for polynomials will not help much. $\endgroup$ – dxiv Jun 29 '18 at 3:57
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    $\begingroup$ By Laguerre's generalization of Descartes' rule of signs, the Dirichlet polynomial (yes, this stuff do have a name) you have has 2 sign changes, so the number of positive roots counting multiplicity is either $2$ or $0$. The root at $z = 1$ is a double root, this exhaust all positive real roots. It is clear your Dirichlet polynomial doesn't have any negative real roots, so $z = 0$ and $1$ is all the real roots you have. $\endgroup$ – achille hui Jun 29 '18 at 4:08
  • $\begingroup$ @achillehui, how to be with the root $z_3=-1/2$? It is real too. $\endgroup$ – Nick Jun 29 '18 at 6:59
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In this specific way, we can make a substitution $u = z^{1/2}$ to end up with $$ \begin{split} 0 &= u^9-\frac32u^8+\frac12u^6\\ &= \frac{u^6}{2}\left(2u^3-3u^2+1\right)\\ &= \frac{u^6}{2} (u-1) \left(2u^2-u-1\right)\\ &= \frac{u^6}{2} (u-1)^2 \left(2u+1\right).\\ \end{split} $$ So there are $9$ roots: $6$ multiplicities of $u_1=0$, $2$ multiplicities of $u_2=1$, and $u_3=-1/2$. So there must be $18$ roots of the original equation.

This suggests the approach for the rational powers -- substitute $u=z^{1/n}$ where $n$ is what makes all remaining powers integers in the new equation in $u$, and apply the Fundamental Theorem of Algebra to get your upper bound.

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  • $\begingroup$ Thank you for the upper bound. How to be with the lower bound? I hope, in my case it should be $3$. $\endgroup$ – Nick Jun 29 '18 at 7:07
  • $\begingroup$ @Nick i think the lower and upper will be the same here, since the Fundamental Theorem of Algebra is exact and you only need to choose branches of square roots correctly when you impose the solutions in $z$ instead of the ones in $u$... $\endgroup$ – gt6989b Jun 29 '18 at 12:38

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