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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

3 Answers3

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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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Your problem is a special case of this problem. Actually, you can look at all of my answers to get the approach in terms of the Lambert-Tsallis function, $W_r(z)$. In your case, after a few manipulations you find:

$$x^2=\frac{W_{2/3}\bigg[ -\frac{2}{3}\bigg(\frac{k-1}{k}\bigg)\bigg(\frac{r}{k}\bigg)^{2/3}\bigg]}{-\frac{2}{3}\bigg(\frac{k-1}{k}\bigg)}$$

You are interested in the case $0<x<1$ then $W_{2/3}(z) < 0$. Based on properties of $W_{2/3}(z)$ it has up to 3 real values when $z<0$ (which is the case).

However, when $-z_b < z < 0$ with $z_b=-(\frac{2}{3})^{5/3}$ and one of them is responsible to $x>1$ (ignored), $0<x<1$ (desired) and $x<0$ (ignored). In conclusion, you must have

$$r \le \frac{2}{5}\big(\frac{3}{5}\big)^{3/2}k \big(\frac{k}{k-1}\big)^{3/2}$$

If you look the right side of previous expression it is always $>1$, so I do believe that no matter the value you consider at $r,k$ respecting your initial conditions, you always will have one value for $0<x<1$.

In the equality, it happens that both positives solutions of $x$ will be unitary and it only happens in the case when $r=1, k=5/2$.

ZKZ
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If a polynomial equation cannot be solved in radicals, it can also not be solved in elementary functions.

Note that my answer isn't the usual answer.

Because the equation is a trinomial-like equation, we can solve it by the following method.

$$x^5-\frac{k}{k-1}x^3+\frac{r}{k-1}=0$$ $$-\frac{k-1}{r}x^5+\frac{k}{r}x^3-1=0$$ $x\to \frac{(trk^2)^\frac{1}{3}}{k}$: $$t-\frac{k^\frac{4}{3}r^\frac{2}{3}-k^\frac{1}{3}r^\frac{2}{3}}{k^2}t^\frac{5}{3}-1=0$$ $\frac{k^\frac{4}{3}r^\frac{2}{3}-k^\frac{1}{3}r^\frac{2}{3}}{k^2}\to y, \frac{5}{3}\to \alpha$: $$t-yt^\alpha-1=0$$

Now the equation is in the form of equation 8.1 of [Belkic 2019]. Solutions in terms of Bell polynomials, Pochhammer symbols or confluent Fox-Wright Function $\ _1\Psi_1$ can be obtained therefore.
$\ $

Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106

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