17

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!

Martin Sleziak
  • 56,060
Frank
  • 6,594

3 Answers3

26

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 $k = 1,2,3$.

achille hui
  • 125,323
  • 1
    Why (out of all the integers), did you choose $k$ to equal $1,2,3$? – Frank Jun 11 '16 at 22:01
  • 1
    And how did you come up with this method? Is great! :D – Ant Jun 11 '16 at 22:54
  • 1
    @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. – achille hui Jun 11 '16 at 23:49
  • 1
    @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. – achille hui Jun 11 '16 at 23:54
  • 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. :/ – Frank Jun 12 '16 at 23:52
  • @Frank by experience, there are not too many ways to construct $\cos z$ from something even simpler, $\cos z = e^{iz} + e^{-iz}$ is one. In general, if you are given a real polynomial whose roots are very simple trigonometry functions, you should try to see whether you can express them in terms of a single root of unity. – achille hui Jun 13 '16 at 00:00
  • Okay! Thanks for the help! – Frank Jun 13 '16 at 00:02
  • By the way, do you know any other polynomial equations whose roots are trigonometric functions? – Frank Jun 13 '16 at 00:03
  • As a start, lookup Chebyshev polynomials on wiki. Most equations like this can be build from one or two Chebyshev polynomials. e.g. the current polynomial is $\frac{T_4(x/2)-T_3(x/2)}{x/2-1}$ (first pointed out by J.M in comment in David's answer). – achille hui Jun 13 '16 at 00:14
  • Is it possible to let $x=a\cdot\cos(b)$ and solve it from there? – Frank Jun 14 '16 at 02:24
  • @Frank, It is possible but dealing with trigonometic identities is usually harder, you basically don't know which identities to use. If you do it through polynomials, you have a small list (Chebyshev polynomails) to check and see whether there is something that can be used to simplify the expression. – achille hui Jun 14 '16 at 02:34
  • Using your substitution method, I solved $y=x^3-3x-1$ with roots $2\cos\left(\frac {\pi}{9}\right), 2\cos\left(\frac {5\pi}{9}\right)$ and $2\cos\left(\frac {7\pi}{9}\right)$ – Frank Jun 16 '16 at 13:33
9

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 $$2\cos\frac{6\pi}{7}=2\cos\frac{8\pi}{7}$$

David Quinn
  • 35,087
7

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 $x$.

egreg
  • 244,946