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I have a following polynomial. (See WolframAlpha ):

$$x^9-6x^8+14x^7-16x^6+36x^5-56x^4+ 24x^3-320x+\frac{640}{9}=0 \tag{1}$$

Wolfram says that $(1)$ has three real roots and three pairs of complex conjugate roots.

I know one of the real roots of this polynomial is equal to:

$$x_1=\tan \left(\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right) =0.22267663636945$$

I know this is a root, because I derived the polynomial from this expression.

Edit

By @IvanNeretin's comment we also have two other real roots:

$$x_2=\tan \left(\frac{\pi}{3} +\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$$

$$x_3=\tan \left(\frac{2\pi}{3} +\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$$


A very surprising (for me) discovery: all six complex roots have the same absolute value of the imaginary part:

$$b=Im(x)= \pm \frac{\sqrt[3]{2}}{\sqrt{3}} (1+\sqrt[3]{2})= \pm 1.643902181980216,~~~~\text{for all } x \notin \mathbb{R}$$

I found it using ISC, but hadn't found closed forms for the real parts.


How can we find explicitly the real parts of the complex roots?


My thoughts - if we substitute in $(1)$:

$$x=a \pm i b$$

then we obtain two equations for one real varibale $a$ (for each three pairs of complex roots of course), and by adding and subtracting these equations we can lower their order.

Since there should be three distinct real parts $a$, we could probably bring it down to a cubic equation.

But I need a CAS for that and I'm not sure it will lead to a solution.

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  • $\begingroup$ @IvanNeretin, could you please explain this? My algebra probably stayed at school level. This seems connected to the three cubic roots of unity $\endgroup$ – Yuriy S Jun 23 '16 at 11:45
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The other two real roots, quite expectedly, are $\tan \left({\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$ and $\tan \left({2\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$. The complex roots are produced in a similar manner, remembering that $\tanh$ is a periodic function with imaginary period.

Long story short, all nine roots are $$\tan \left({\pi\over3}\cdot m+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left({\pi i\over3}\cdot n+\frac{1}{3} \text{arctanh} \frac{1}{3} \right),\;n,m\in\{0,1,2\}$$ A rigorous explanation is beyond my expertise. Informally speaking, inverse trigonometric and inverse hyperbolic functions are multivalued, but the polynomial "knows nothing" of this; to it, they are all the same, so if one value is a root, so must be the others.

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  • $\begingroup$ This is great, thank you. $\endgroup$ – Yuriy S Jun 23 '16 at 11:54
  • $\begingroup$ I figured out how to get the real parts explicitly, so I'm accepting your answer, thanks again $\endgroup$ – Yuriy S Jun 23 '16 at 11:58

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