Fifth degree polynomials cannot generally be solved analytically, but at least one solution always exists. Given the normal form $$ax^5+bx^4+cx^3+dx^2+ex+f=0,$$ is it possible to find sufficient conditions on $a,b,c,d,e,f$ that guarantee a unique solution?

One sufficient condition is obviously $a,b,c,d,e>0$, so that the LHS of the above equation is strictly increasing in $x$. In my case, the coefficients are functions of several parameter values, so that $a,c,e>0$ and $b,d<0$, while $f\lessgtr0$ is ambiguous (depends on parameter values).

My fifth degree polynomial is not always strictly increasing due to $b,d<0$, but still the solution is often unique, as the following numerical example shows: 5th degree polynomial with unique solution

I could "prove" for this example that the solution is unique, by taking the derivative (a fourth degree polynomial), finding the local maxima and minima, and verifying that the function crosses 0 only once between the extrema or between the extrema and $-\infty,\infty$. But I was hoping there was a simpler and more general way.

Moreover, my fifth degree polynomial is strictly increasing for some parameter values despite $b,d<0$. Is there a better way to figure out when this is the case than computing the derivative and determining whether $$5ax^4+4bx^3+3cx^2+2dx+e>0~\forall x\in\mathbb{R}?$$

  • 1
    $\begingroup$ Does a "triple root" count as one or as three roots in your problem? $\endgroup$ – Rory Daulton Aug 23 '15 at 21:16
  • $\begingroup$ Counts as one. I think of it as the number of solutions after removing duplicates. $\endgroup$ – Nameless Aug 23 '15 at 23:53
  • 1
    $\begingroup$ Find the critical values and determine their type. You should then be able to determine how many roots there are from the picture you can then sketch. $\endgroup$ – Mose Wintner Aug 24 '15 at 2:34
  • $\begingroup$ Yes that's what I thought. Doesn't even require the type of the critical points; just graphing them and connecting the dots will tell how many solutions there are. Still, I was hoping there was a simpler way, say a condition on the coefficients, since determining the solutions of a fourth degree polynomial is not a walk in the park either. $\endgroup$ – Nameless Aug 24 '15 at 11:43

You could use Sturm's theorem to count the number of roots inside any subinterval of the real line. In particular, if you choose the interval to be $(-\infty,\infty)$ (and evaluating the signs of the polynomials occuring in the Sturmian chain boils down to simply taking the sign of the leading coefficient), it will tell you the total number of real roots.

Note that this procedure is widely believed to be numerically unstable, so you should use arbitrary precision arithmetic in an implementation. On the other hand, it saves you from using the solution formulas for fourth-degree polynomials. Those are typically not designed for numerical evaluation as well, and overkill in the sense that they yield the roots of the derivative, which you are actually not interested in. Note that the instability is inherent to the problem: if you have multiple roots, an arbitrarily small perturbation of the input could transform a $k$-fold root to $k$ simple roots. Hence, short of exact calculation (or at least approximate arithmetic in roughly the same precision as the precision of the intermediate results in exact arithmetic), there's hardly anything you can do.

Besides, in some "lucky" cases Descartes' Rule of signs could help (if there is at most one simple positive and negative root, respectively, and the complex roots are sufficiently far away to ensure that the number of sign changes matches the number of real roots).

  • $\begingroup$ Thank you, I did not know this technique. Very elegant! It is possible to find sufficient conditions in terms of the coefficients using the Sturm sequences, although the polynomial long division makes it a lot of work and the resulting conditions are not particularly easy to work with. Then again, this is a 5th degree polynomial, so any uniqueness condition is likely not very simple. $\endgroup$ – Nameless Aug 25 '15 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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