4

According to this:

If $\omega^7 =1$ and $\omega \neq 1$ then find value of

$\displaystyle\frac{1}{(\omega+1)^2} + \frac{1}{(\omega^2+1)^2} + \frac{1}{(\omega^3+1)^2} + ... + \frac{1}{(\omega^6+1)^2}=?$

First I try like $\displaystyle\frac{1}{\omega+1} + \frac{1}{\omega^2+1} + \frac{1}{\omega^3+1} + ... + \frac{1}{\omega^6+1} = 3 $

I have done distribution them and finally got the solution $\dfrac{5}{3}$

However, this is, without a doubt, a time-consuming way.

Can someone please suggest easier way to solve this one.

Aryaman Maithani
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insidenothingisnothing
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  • If $\xi_i$ are all seven roots of unity $\xi_i^7 = 1$, then your sum is $$-\frac 14 + \sum_i \frac{1}{(\xi_i+1)^2}.$$ – Paweł Czyż Jun 05 '20 at 12:14

3 Answers3

2

Note that $\omega, \ldots, \omega^6$ are precisely the roots of the sixth degree polynomial: $$p(x) = x^6 + \cdots + 1 = \dfrac{x^7-1}{x-1}.$$ Thus, we can write $$p(x) = (x-\omega)\cdots(x-\omega^6).$$ Taking (natural) $\log$ on both sides and differentiating gives us $$\dfrac{p'(x)}{p(x)} = \dfrac{1}{x-\omega}+\cdots+\dfrac{1}{x-\omega^6}.$$ Note that \begin{align} \log p(x) &= \log(x^7 - 1) - \log(x-1)\\ \implies \dfrac{p'(x)}{p(x)} &= \dfrac{7x^6}{x^7-1} - \dfrac{1}{x-1}. \end{align}

This gives us that $$\dfrac{7x^6}{x^7-1} - \dfrac{1}{x-1} = \dfrac{1}{x-\omega}+\cdots+\dfrac{1}{x-\omega^6}.$$

Differentating both sides again gives us $$\dfrac{(x^7-1)(42x^5) - (7x^6)(7x^6)}{(x^7-1)^2} + \left(\dfrac{1}{x-1}\right)^2 = -\left(\dfrac{1}{x-\omega}\right)^2-\cdots-\left(\dfrac{1}{x-\omega^6}\right)^2.$$

Now, we simply substitute $x = -1$ both sides. It is clear that the RHS transforms to the negative of what we want, whereas the LHS becomes \begin{align} \dfrac{(-2)(-42) - (7)(7)}{(-2)^2} + \left(\dfrac{1}{-2}\right)^2 &= \dfrac{84-49}{4} + \dfrac{1}{4}\\ &= \dfrac{36}{4} = 9 \end{align}

This gives us the answer as $-9$.

Aryaman Maithani
  • 15,771
1

Let $\dfrac1{w+1}=x\implies w=\dfrac{1-x}x$

$$\implies\left(\dfrac{1-x}x\right)^7=1$$

As $x\ne0,$ $$x^6-3x^5+5x^4-\cdots=0$$

We need $$\sum_{r=1}^6x_r^2=\left(\sum_{r=1}^6x_r\right)^2-2\sum_{1\le i< j\le6}x_ix_j=\left(\dfrac31\right)^2-2\cdot\dfrac51$$

lab bhattacharjee
  • 274,582
  • See https://math.stackexchange.com/questions/1909362/product-of-one-minus-the-tenth-roots-of-unity and https://math.stackexchange.com/questions/2043534/if-alpha-i-i-0-1-2-n-1-be-the-nth-roots-of-unity-the-sum-i-0n-1 and https://math.stackexchange.com/questions/1685279/alternating-sum-of-roots-of-unity-sum-k-0n-1-1k-omegak and https://math.stackexchange.com/questions/3649434/evaluate-fracq1q2-fracq21q4-fracq31q6-where-q7-1-an/3649503#3649503 – lab bhattacharjee Jun 05 '20 at 13:52
0

Using $\omega^7=1$ the second sum computes to $$ \frac{3(\omega^6 + 2\omega^4 + \omega^3 + \omega^2 + \omega + 1)}{\omega^6 + 2\omega^4 + \omega^3 + \omega^2 + \omega + 1}=3 $$ For the first sum I do not obtain $5/3$. I obtain $$ 3\cdot\frac{5\omega^6+ 2\omega^5 + 11\omega^4 - 4\omega^3 + 11\omega^2 + 2\omega + 5}{9\omega^6 + 10\omega^5 + 7\omega^4 + 12\omega^3 + 7\omega^2 + 10\omega + 9}=-9, $$ because $\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1=0$.

Dietrich Burde
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