1
$\begingroup$

I'm interesting in finding analytical solutions for the equation $$\alpha K x^\alpha + x -N = 0,$$ where $\alpha$ is a positive integer and both $K$ and $N$ are positive real constants.

Based on the meaning of the equation (derived from a problem in chemistry), there must be one (and only one) real solution between zero and $N$. Actually, I'm interested in only such solution, not any other outside the $(0, \, N)$ range.

I know that I can use numerical methods to solve for specific $\alpha$, $N$ and $K$. I'm seeking a more elegant method.

I also know that no general solution for a general polynomial of order 5 or more can exist, but maybe this one falls in some special case I'm not aware of.

I'm a chemist, so I apology if the question is malformed. Let me know if you need more details.

$\endgroup$
2
  • 2
    $\begingroup$ There are closed-form power series (obtained by change of variable from the power series of the inverse function of $y (1-y)^t$, for which see e.g. math.harvard.edu/~elkies/Misc/catalan.pdf or math.harvard.edu/~elkies/Misc/catalan2.pdf); if memory serves, these can be expressed as hypergeometric functions when $\alpha$ is an integer. But indeed no solution in radicals for $\alpha>4$ and general $N,K$. $\endgroup$ Apr 14, 2017 at 16:30
  • 2
    $\begingroup$ The power series in powers of $K$ is $$x = N - \alpha N^\alpha K + \sum_{j=2}^\infty \frac{\alpha^{1+j} N^{j\alpha - j+1} (-K)^j}{(j-1)!} \prod_{i=1}^{j-2} (j\alpha-i)$$ $\endgroup$ Apr 14, 2017 at 18:23

1 Answer 1

1
$\begingroup$

Dividing by $\alpha K$ we obtain a polynomial equation of the form $$x^{\alpha} + d_1x + d_0=0.$$

Already for $\alpha=5$ it is known that such an equation (called in Bring-Jerrard normal form) is not solvable by radicals for general $d_0$, $d_1$. A solution of the Bring-Jerrard quintic in terms of hypergeometric functions can be found in the corresponding Wikipedia article.

$\endgroup$
2
  • $\begingroup$ Ok, many thanks. So I can safely say there is no easy solution for general $\alpha$. $\endgroup$ Apr 14, 2017 at 22:12
  • $\begingroup$ @Simone Conti: Everything depends on what you mean by "analytic". If this word has a usual meaning, then an analytic solution is given in the comment of Robert Israel. $\endgroup$ Apr 15, 2017 at 1:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.