My friend asked me what the roots of $y=x^3+x^2-2x-1$ was.

I didn't really know and when I graphed it, it had no integer solutions. So I asked him what the answer was, and he said that the $3$ roots were $2\cos\left(\frac {2\pi}{7}\right), 2\cos\left(\frac {4\pi}{7}\right)$ and $2\cos\left(\frac {8\pi}{7}\right)$.

Question: How would you get the roots without using a computer such as Mathematica? Can other equations have roots in Trigonometric forms?

Anything helps!

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    $\begingroup$ See the Wikipedia page.. There is a general method for producing these kinds of solutions. It's described in the section "General solution to the cubic with real coefficients" $\endgroup$ – Will R Jun 11 '16 at 17:59
  • $\begingroup$ Also related: math.stackexchange.com/questions/218441/… $\endgroup$ – Maximilian Gerhardt Jun 11 '16 at 17:59
  • $\begingroup$ OP, does the graph cross the x-axis at all? $\endgroup$ – moonman239 Jun 11 '16 at 20:17
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    $\begingroup$ @moonman239 can you think of any cubic equation whose graph doesn't cross the x-axis at all? $\endgroup$ – alephzero Jun 11 '16 at 23:22
  • $\begingroup$ @alephzero: Point taken. $\endgroup$ – moonman239 Jun 16 '16 at 7:06

Let $p(x) = x^3+x^2-2x-1$, we have $$p(t + t^{-1}) = t^3 + t^2 + t + 1 + t^{-1} + t^{-2} + t^{-3} = \frac{t^7-1}{t^3(t-1)}$$

The RHS has roots of the form $t = e^{\pm \frac{2k\pi}{7}i}$ ( coming from the $t^7 - 1$ factor in numerator ) for $k = 1,2,3$. So $p(x)$ has roots of the form $$e^{\frac{2k\pi}{7} i} + e^{-\frac{2k\pi}{7} i} = 2\cos\left(\frac{2 k\pi}{7}\right)$$ for $k = 1,2,3$.

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    $\begingroup$ Why (out of all the integers), did you choose $k$ to equal $1,2,3$? $\endgroup$ – Frank Jun 11 '16 at 22:01
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    $\begingroup$ And how did you come up with this method? Is great! :D $\endgroup$ – Ant Jun 11 '16 at 22:54
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    $\begingroup$ @Frank, For any integer $k$, $e^{\frac{2k \pi}{7}i}$ is a root of $t^7 - 1$. Since $e^{2\pi i} = 1$, integers differs by a multiple of $7$ give us the same root. For the six roots of $\frac{t^7 - 1}{t-1} =t^6 + t^5 + t^4 + t^3 + t^2 + t + 1$, the corresponding $k$ can be any $6$ integers as long as they are not divisible by $7$ and belong to different equivalent class under modulo $7$ arithmetic. We can take $k$ to be $1,2,3,4,5,6$, we can also take them to be $\pm 1,\pm 2,\pm 3$. I pick the later one because it is more convenient to combine the exponentials to cosines. $\endgroup$ – achille hui Jun 11 '16 at 23:49
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    $\begingroup$ @Ant, we are told the roots are $2\cos\left(\frac{2k\pi}{7}\right)$, the simplest way to get them are $2\cos\left(\frac{2k\pi}{7}\right) = e^{\frac{2k\pi}{7}i} + e^{-\frac{2k\pi}{7}i}$, the natural guess is substitute $x = t + t^{-1}$ in $p(x)$ and see what one get. As one expect, it is a polynomial for all the primitive $7$ roots of unity. $\endgroup$ – achille hui Jun 11 '16 at 23:54
  • $\begingroup$ Wait, so how did you know to substitute $x$ with $t+\frac {1}{t}$? It's either I'm not putting two and two together, or I forgot some important detail. :/ $\endgroup$ – Frank Jun 12 '16 at 23:52

Consider the equation $$\cos4\theta=\cos3\theta$$ whose roots are $$\theta=n\cdot\frac{2\pi}{7}$$

Representing this as a polynomial in $c=\cos\theta$, we have $$8c^4-4c^3-8c^2+3c+1=0$$ $$\Rightarrow (c-1)(8c^3+4c^2-4c-1)=0$$

Now write $x=2c$ and we see that the polynomial equation $$x^3+x^2-2x-1=0$$ has roots as stated in your question. Note that $$2\cos\frac{6\pi}{7}=2\cos\frac{8\pi}{7}$$

  • $\begingroup$ +1 for the $2\cos\frac{6\pi}{7} = 2\cos\frac{8\pi}{7}$ part. $\endgroup$ – achille hui Jun 11 '16 at 19:43
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    $\begingroup$ In the language of Chebyshev polynomials: $\frac{T_4(x/2)-T_3(x/2)}{x/2-1}$ is the polynomial under consideration. $\endgroup$ – J. M. ain't a mathematician Jun 11 '16 at 20:01

Set $x=t+t^{-1}$. Then the equation becomes $$ t^3+3t+3t^{-1}+t^{-3}+t^2+2+t^{-2}-2t-2t^{-1}-1=0 $$ and, multiplying by $t^3$, $$ t^6+t^5+t^4+t^3+t^2+t+1=0 $$ and it should be now clear what the solutions are. For each root there's another one giving the same solution in $x$.


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