Value of trigonometric sum

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Value of trigonometric sum

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#1 h Aug 27, 2010, 11:53 AM • 2 Y

Y by Adventure10, Mango247

Find the value of the sum
$\sqrt[3]{\cos{\frac{2\pi}{7}}} + \sqrt[3]{\cos\frac{4\pi}{7}} + \sqrt[3]{\cos\frac{6\pi}{7}}$

the solution I know is a bit ugly. Well, if you take to long to make it I post it.

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#2 h Aug 27, 2010, 12:07 PM • 1 Y

Y by Adventure10

Are you sure that there is a neat solution and answer ?
I just know that
${\cos{\frac{2\pi}{7}}} + {\cos\frac{4\pi}{7}} + {\cos\frac{6\pi}{7}}=-\dfrac 12.$

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pco

#3 h Aug 27, 2010, 12:37 PM • 2 Y

Y by Adventure10, Mango247

jonasrla wrote:

Find the value of the sum
$\sqrt[3]{\cos{\frac{2\pi}{7}}} + \sqrt[3]{\cos\frac{4\pi}{7}} + \sqrt[3]{\cos\frac{6\pi}{7}}$

the solution I know is a bit ugly. Well, if you take to long to make it I post it.

Numeric calculus gives $-0.71751508...$

Could you just give us the closed form of the result you obtained thru your ugly method, so we can concentrate on finding a clever method ?

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#4 h Aug 27, 2010, 1:00 PM • 2 Y

Y by Adventure10, Mango247

The problem may be Ramanujan's problem, I think.

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#5 h Aug 27, 2010, 4:15 PM • 2 Y

Y by Adventure10, Mango247

Well, when I said that I wanted the value, I'm meant the exact value, like $\sqrt2$ instead of $1,41...$

And yes, it's a neat problem, as I said I have the solution, but I just think it's ugly.

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#6 h Aug 27, 2010, 4:17 PM • 2 Y

Y by Adventure10, Mango247

jonasrla wrote:

Well, when I said that I wanted the value, I'm meant the exact value, like $\sqrt2$ instead of $1,41...$

I have quite well understood. That is what I called a "closed form" value.

So, since you have it, I asked you to give us this value so that we can try to find a clever method.

Thanks for giving us this result.

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Rust

#7 h Aug 27, 2010, 5:01 PM • 2 Y

Y by Adventure10 and 1 other user

Let $x_k=\sqrt[3]{2\cos\frac{2\pi k}{7}},y_k=x_k^3.$
Then $y_k$ are roots of $y^3+y^2-2y-1$ , because $\sum{}_y{}_i=-1,y{}_1y{}_2+y{}_2y{}_3+y{}_3y{}_2=-2,y{}_1y{}_2y{}_3=1.$
Therefore $x_k.x_kr,x_kr^2,r=exp(\frac{2\pi i}{3})$ are roots of $x^9+x^6-2x^3-1=0.$
If $(x{}_x{}_1)(x-x_2)(x-x_3)=x^3+ax^2+bx+c$ , then $(x-x_1r^l)(x-x_2r^l)(x-x_3r^l)=x^3+a*r{}^lx{}^2+b*r^{2l}x+c, l=1,2$
It give $\prod_k(x-x_kr)(x-x_kr^2)=x^6-ax^5+(a^2-b)x^4+x^3(2c-ab)+(b^2-ac)x^2-bcx+c^2.$
Therefore $c^3=-1,3c+a^3-3ab=1,3c^2+b^3-3abc=-2.$
It give $c=-1,b^3=-1-b^3$ and $a^3$ is real root of $x^3+15x^2+75x-64=(x+5)^3-189=0$ .
Therefore $S=\sum_k\sqrt[3]{\cos\frac{2\pi k}{7}}=\sqrt[3]{\frac{-x}{2}}=\sqrt[3]{\frac{5-\sqrt[3]{189}}{2}}.$

This post has been edited 2 times. Last edited by Rust, Aug 27, 2010, 5:07 PM

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#8 h Aug 27, 2010, 5:02 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$\sqrt[3]{{\frac{5-3\sqrt[3]{7}}{2}}}$
Let $p=-(a+b+c), q=(ab+bc+ca), r=-abc$ and $p\sqrt[3]{r}+3\sqrt[3]{r^2}+q=0$ ,
than $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{3\sqrt[3]{p^2\sqrt[3]{r}-3q\sqrt[3]{r}}-p-6\sqrt[3]{r}}$

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#9 h Aug 28, 2010, 12:55 AM • 4 Y

Y by luiseduardo, Adventure10, Mango247, and 1 other user

Looks like that the only way is doing it using polynomial, it may be not very fast, but it's practical.
The solution that I know is from my teacher. I can't surely tell that he did it for himself, but he showed me so the credit's goes to him. Some parts of this solution looks like the rust's solution. I just wanted to say wow.

$z=\cos\frac{2\pi}{7}+i\sin \frac{2\pi}{7} \Rightarrow z^6+z^5+z^4+z^3+z^2+z+1=0 \\ \Rightarrow z^3+z^{-3}+z^2+z^{-2}+z^1+z^{-1}=0$
If $y=z+z^{-1}$ then $(y^{3}-3y)+(y^{2}-2)+y=0 \Rightarrow y^3+y^2-2y+1=0$
And the roots are $y_1= 2\cos\frac{2\pi}{7}; y_2=2\cos\frac{4\pi}{7};y_3=2\cos\frac{6\pi}{7}$

Let $P(x)=x^3-Ax^2+Bx-C$ be a polynomial such that $x_1=\sqrt[3]{y_1}; x_2=\sqrt[3]{y_2}; x_3=\sqrt[3]{y_3}$
$A=x_1+x_2+x_3;\ B=x_1x_2+x_2x_3+x_3x_1;\ C=x_1x_2x_3=\sqrt[3]{1}=1$

Identity: $(a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc$

$A^3=-1+3AB-3C \Rightarrow A^3=3AB-4 \\ B^3=-2+3AB-3C \Rightarrow B^3=3AB-5$

Multiply the equations: $(AB)^3=(3AB-4)(3AB-5)$ . Making $t=AB$ and opening the product
$t^3-9t^2+27t-20=0 \Rightarrow t^3-9t^2+27t-27=-7 \Rightarrow (t-3)^3=-7$
We want just the real root: $t=3-\sqrt[3]{7}$

$A^3=3t-4=9-3\sqrt[3]{7}-4=5-3\sqrt[3]{7} \\ \Rightarrow \frac{A}{2}=\sqrt[3]{\cos{\frac{2\pi}{7}}}+\sqrt[3]{\cos\frac{4\pi}{7}}+\sqrt[3]{\cos\frac{6\pi}{7}} =\sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}} = \sqrt[3]{\frac{5-\sqrt[3]{189}}{2}}$

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