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I've met this quintic equation in my research: $$x^5 - \frac{k}{k-1} \cdot x^3 +\frac{r}{k-1}=0$$ with the additional conditions: $$k>1; \quad 0<r<1; \quad 0<x<1$$ My background in math is limited. I know that it cannot be solved in radicals, but can be using the elliptic functions. However, elliptic function method requires transforming the equation to the Bring-Jerrard form. I just wondered if it is possible to find a analytic solution easier way. Basically, I need just one function x(r,k) in real positive number which satisfies the equation. Is it possible? How to get one?

DuzaBF
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  • There may be up to five real solutions. Have you tried using power series? – Somos Mar 17 '20 at 11:52
  • When using numerical solutions for the equation with $k$ and $r$ values in sensible range, as they are measured physically, I've got 3 real solutions and 2 of them positive, and only one of them satisfies condition $0<x<1$

    I've tried using method from the paper "Polynomial Transformations of Tschirnhaus, Bring and Jerrard" by Victor S. Adamchik and David J. Jeffrey and remove the $x^3$ term. However it caused the $x^2$ and $x$ terms to appear and solution became quite clumsy. I hope there is a simpler and less laborious way.

     – DuzaBF Mar 17 '20 at 12:51
  • If this helps, by inverting the unknown, $x^5+ax^3+b$ can be rewritten $y^5+py^2+q$. –  Mar 17 '20 at 13:01
  • Can't you do with $r(x,k)$ or $k(x,r)$ ? –  Mar 17 '20 at 13:07
  • Unfortunatly, no. I need $x(r,k)$ because $r$ and $k$ are measurable values and $x$ can be only calculated from model and in this case from the equation. $x$ value is important for my research as it used in decision making. – DuzaBF Mar 17 '20 at 13:12
  • By Descartes' rule of signs, this has 2 or 4 positive zeros (if the coefficients written are positive). – vonbrand Mar 17 '20 at 13:25
  • For trinomial with one parameter (as in answer below) work perfect hypergeometric formulas Mellin/Lagrange (see Belardinelli, p.56, formulas (4) and (5)), and formula by Лахтинъ with definite integration – Dmitry Ezhov Mar 17 '20 at 14:26

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Writing $x = (r/(k-1))^{1/5} y$, the equation becomes $$ y^5 - \frac{k}{(k-1)^{3/5} r^{2/5}} y^3+ 1 = 0$$ Let $c = k/((k-1)^{3/5} r^{2/5}$, so this is $y^5 - c y^3 + 1 = 0$.

Now, if you want a solution with $y > 0$, $c$ must not be too small. If $c > 0$, the minimum value of $y^5 - c y^3 + 1$ for $y \ge 0$ is $1 - 6 \sqrt{15} c^{5/2}/125$, and you want that $\le 0$, so you need $$c \ge \frac{5 \cdot 2^{3/5} \cdot 3^{2/5}}{6}$$ If $c$ is greater than this, there will be two positive real solutions for $y$ (and thus for $x$), if $c$ is less there will be none.

Robert Israel
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