4
$\begingroup$

The task is to prove the following non-equality by hand:

$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$

Wolframalpha shows this, but I can't prove it.

http://www.wolframalpha.com/input/?i=1%2Bcos(2pi%2F7)-4cos%5E2(2pi%2F7)-8cos%5E3(2pi%2F7)%3D0

$\endgroup$
  • $\begingroup$ You don't need to prove sth you can just calculate it (as wolfram did) and see that it's false. $\endgroup$ – MorphhproM Mar 30 '16 at 17:22
  • 2
    $\begingroup$ @MorphhproM Sure..Why not stop doing Math alltogether? $\endgroup$ – MathematicianByMistake Mar 30 '16 at 17:23
  • 1
    $\begingroup$ @MorphhproM He's asking how to see it without a calculator. $\endgroup$ – S.C.B. Mar 30 '16 at 17:24
  • 2
    $\begingroup$ There could be a typo. The "minimal polynomial" for $\cos(2\pi/7)$ is $1 + 4 x - 4 x^2 - 8 x^3$. So, if there were a $4$ on the $\cos(2\pi/7)$ term, you'd have an identity. $\endgroup$ – Blue Mar 30 '16 at 17:27
  • 1
    $\begingroup$ @Mongol-genius: The typo might not be yours. Books make mistakes, too. $\endgroup$ – Blue Mar 30 '16 at 17:28
5
$\begingroup$

We prove a closely related result, which in particular shows there is a typo in the given equation.

The number $e^{2\pi i/7}$ is a root of $x^7=1$, and therefore of $x^6+x^5+\cdots+x+1=0$, or equivalently of $$(x^3+x^{-3})+(x^2+x^{-2})+(x+x^{-1})+1=0.\tag{1}$$ (We divided through by $x^3$.) Let $w=\frac{1}{2}(x+x^{-1})$.

Note that $x^3 +x^{-3}=8w^3-6w$ and $x^2+x^{-2}=4w^2-2$ and $x+x^{-1}=2w$ So our equation can be rewritten as $$8w^3+4w^2-4w-1=0.\tag{2}$$ Since $e^{2\pi i/7}$ is a root of (1), it follows that $\cos(2\pi/7)$ is a root of (2).

$\endgroup$
  • $\begingroup$ As always whole different approach $+1$ $\endgroup$ – Archis Welankar Mar 30 '16 at 17:36
  • $\begingroup$ A whole different approach to what, @ArchisWelankar(I thought they used this often)? Anyway, +1. $\endgroup$ – S.C.B. Mar 30 '16 at 17:39
  • 1
    $\begingroup$ @K.Power That is not true in general. Notice the argument used shifts from talking about $x$ to talking about $w = \frac{x + x^{-1}}{2} = \frac{e^{it}+e^{-it}}{2} \equiv \cos(t)$. The last equation is in terms of $w$ not $x$. $\endgroup$ – Winther Mar 30 '16 at 17:59
  • 2
    $\begingroup$ Let $\theta=2\pi/7$. Then $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$. I think this comes quite early, it is a direct consequence of the familiar identity $e^{i\theta}=\cos\theta+i\sin\theta$. Since $e^{i\theta}$ is a root of (1), and $w=\frac{1}{2}(x+x^{-1})$, it follows that $\cos\theta$ is a root of (2). $\endgroup$ – André Nicolas Mar 30 '16 at 18:00
  • 1
    $\begingroup$ @Piquito: Compare the equation of my answer above with the equation of the question. If both were true, we would have $3\cos(2\pi/7)=0$. $\endgroup$ – André Nicolas Mar 30 '16 at 18:14
3
$\begingroup$

Here, we will prove the equation by @AndreNicolas and @Blue in a more elementary manner.

$$ 1+4\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) = 0 $$

Note that since $\cos 3x=4\cos^3 x-3\cos x$, and since $\cos 2x=2\cos^2 x-1$, our equation simplifies to $$ \cos \left(\frac{\pi}{7}\right) -\cos \left(\frac{2\pi}{7}\right) + \cos \left(\frac{3\pi}{7}\right) = \frac{1}{2}$$

Let $x=\frac{\pi}{7}$.

$$\cos x - \cos 2x + \cos 3x = \frac{1}{2}$$

$$\cos x + \cos 3x + \cos 5x = \frac{1}{2}$$ $$\cos x + \cos 3x + \cos 5x + ... = \frac{{\sin 2nx}}{{2\sin x}}$$ Since $n = 3$ $$\cos x + \cos 3x + \cos 5x = \frac{{\sin 6x}}{{2\sin x}}$$ $$\frac{{\sin 6x}}{{2\sin x}} = \frac{1}{2} \Leftrightarrow \sin 6x = \sin x$$ Which is true since $\sin (\pi-x)=\sin x$.

Or,similarly if $$K=\cos x - \cos 2x + \cos 3x $$ then $$K\sin\frac{\pi}{7}=\frac{\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2}=\frac{\sin\frac{6\pi}{7}}{2}\implies K=\frac{1}{2}$$

Since $ 2\sin A\cos B=\sin(A+B)+\sin(A-B)$

$\endgroup$
  • $\begingroup$ MXYMXY Please correct the identities cos(3x) and cos(2x) and your answer :) $\endgroup$ – Mongol-genius Mar 30 '16 at 21:17
  • $\begingroup$ @Mongol-genius Does is work now? $\endgroup$ – S.C.B. Mar 30 '16 at 22:35
  • $\begingroup$ MXYMXY --I think after you have changed the identities your answer must change: your equation simplifies to cos(2pi/7)+cos(4pi/7)+cos(6pi/7)= -1/2 $\endgroup$ – Mongol-genius Mar 30 '16 at 22:44
  • $\begingroup$ MXYMXY - However Thanks for your essential solution :) $\endgroup$ – Mongol-genius Mar 30 '16 at 22:50
  • $\begingroup$ @Mongol-genius That simplifies to the equation above, using $\cos (\pi -x)=-\cos x$. $\endgroup$ – S.C.B. Mar 30 '16 at 23:00
2
$\begingroup$

Putting $x=\cos(\frac{2\pi}{7})$ you have the polynomial $1+x-4x^2-8x^3\not= 0$ then $8x^3+4x^2-x-1\not=0$ from that $$8x^3+4x^2-x-1=4x^2(2x+1)-x-1-x+x=4x^2(2x+1)-(2x+1)+x=(2x+1)(4x^2-1)+x=(2x+1)^2(2x-1)+x$$ Since $(2x+1)^2>0$ and $x>\frac{1}{2}$ since $\frac{2\pi}{7}<\frac{\pi}{3}$ we have that $(2x-1)>0$ and a sum of positive numbers isn't zero.

$\endgroup$
2
$\begingroup$

I will address the problem as it has been stated not as to whether there is a conjectured or assumed typo.

First off the problem is equivalent to showing

$$\frac{(1+\zeta)}{\zeta^{2}(1+2\zeta)}\neq4$$ where $\zeta=cos(2\pi/7)$.

Since $1/2\lt\zeta\lt1$, the left hand side of the inequality is less than $3$ and therefore we are done.

$\endgroup$
0
$\begingroup$

Consider $p(x)=1+x-4x^2-8x^3$, that have only one real root, $x_0\approx0,4$. Now, calculating $x=\cos(2\pi/7)\approx0,6$ we find that $x\neq x_0$

$\endgroup$
  • $\begingroup$ You are asked to prove the non-equality by hand. If you meant for these approximations to be computed by hand, then please indicate this together with (a sketch of) a method of finding this approximation by hand. $\endgroup$ – Wojowu Mar 30 '16 at 17:40
  • $\begingroup$ Because is continuous, I apply Bolzano+ bisection $\endgroup$ – Martín Vacas Vignolo Mar 30 '16 at 17:42
  • $\begingroup$ You also have to compute the cosine. Either way, this is supposed to be in your answer, not a comment. $\endgroup$ – Wojowu Mar 30 '16 at 17:43
0
$\begingroup$

We consider $$f(x) = x - 4x^2 - 8x^3 $$

Notice that for $x \ge \frac{1}{2}, \>\> f(x) < -1$ siince $f(\frac{1}{2}) = -\frac{3}{2} $ and $f'(\frac{1}{2}) < 0$ and $f''(x) < 0$ for all $x > -1$

Now it's only important to show that $\cos \frac{2 \pi}{7} \ge \frac{1}{2}$

It should be relatively easy to show $\cos \frac{2 \pi}{7} \ge \cos \frac{\pi}{3}$

$\endgroup$
  • 2
    $\begingroup$ Please Include a proof of why $x \ge \frac{1}{2}, \>\> f(x) < -1$, and In your answer, not a comment. $\endgroup$ – S.C.B. Mar 30 '16 at 17:44

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.