This quintic equation has $5$ real roots:

$$x^5-4x^4+2x^3+5x^2-2x-1=0 \tag{1}$$

The roots are, from left to right:

$$x_1=\frac{\cos \frac{19}{22} \pi}{\cos \frac{1}{22} \pi}$$

$$x_2=\frac{\cos \frac{9}{22} \pi}{\cos \frac{19}{22} \pi}$$

$$x_3=\frac{\cos \frac{7}{22} \pi}{\cos \frac{5}{22} \pi}$$

$$x_4=\frac{\cos \frac{1}{22} \pi}{\cos \frac{7}{22} \pi}$$

$$x_5=\frac{\cos \frac{5}{22} \pi}{\cos \frac{9}{22} \pi}$$

I found these roots numerically, using ISC. The equation was found on Wikipedia in a different form (for $x-4/5$), no solutions were provided.

I can derive this equation for each root individually. But I don't see how do all five roots 'fit' together.

Why ${1,5,7,9,19}$? Why there is no $3/22, 13/22$ or $17/22$ in any of the arguments?

In any case, I would be grateful for the explanation for how these roots all fit together.


Using the equality $\cos(\pi \pm x)=-\cos(x)$ you get

$$x_1=-\frac{\cos \frac{3 }{22} \pi}{\cos \frac{1}{22} \pi} \\ x_2=-\frac{\cos \frac{9}{22} \pi}{\cos \frac{3}{22} \pi} \\ x_3=-\frac{\cos \frac{15}{22} \pi}{\cos \frac{5}{22} \pi} \\ x_4=-\frac{\cos \frac{21}{22} \pi}{\cos \frac{7}{22} \pi} \\ x_5=-\frac{\cos \frac{27}{22} \pi}{\cos \frac{9}{22} \pi} $$

Note that the identity $$\cos(3x)= \cos(x) [2 \cos(2x)-1]$$ gives you a nicer form for the roots. With this form, after the proper substitution you should be able to reduce your polynomial to the minimal polynomial of $\cos(\frac{\pi}{11})$, which will explain where the roots are coming.


We just have to find the minimal polynomial of $$ \alpha = \frac{\cos\frac{19\pi}{22}}{\cos\frac{\pi}{22}}=-\frac{\cos\frac{3\pi}{22}}{\cos\frac{\pi}{22}}=3-4\cos^2\frac{\pi}{22}=1-2\cos\frac{\pi}{11} \tag{1}$$ then find the conjugate roots. But it is well-known that the minimal polynomial of $\cos\frac{2\pi}{m}$ has degree $\frac{\varphi(m)}{2}$ (i.e. $5$ in our case) and the algebraic conjugates of $\cos\frac{2\pi}{m}$ are $\cos\frac{2\pi k}{m}$ with $\gcd(k,m)=1$, so the claim is straightforward.


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.