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How to find the value of $$1+4\omega+9\omega^{2}+\cdot\cdot\cdot +n ^{2}\omega^{n-1}$$ where $\omega$ is a primitive $n$th root of unity? I am trying to find the value using the fact that $$1+\omega+\omega^{2}+\cdot+\cdot\cdot+\omega^{n-1}=0$$ but did't get the answer. Please suggest me how to solve it. Thanks in advance.

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Hint: denote $S(x)=\sum_{k=1}^n x^k$ and calculate $(S'(x)x)'$.

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Following the idea suggested by A.G. you may denote $$C(x)=\sum_{k=1}^n k^2x^{k-1}$$Note that the desired term is $C(\omega)$. You somehow want to relate this to $$A(x)=\sum_{k=1}^n x^{k-1}$$ of which you know that $A(x)=\frac{x^n-1}{x-1}$ for $x \ne 1$. The intermediate step would be $$B(x)=\sum_{k=1}^n kx^{k-1}$$ Now, note that you have the relations $$(xA(x))'=B(x), (xB(x))'=C(x)$$ So it's easy to find a closed form for $C(x)$ by taking twice a derivative and then you find the desired value as $C(\omega)$.

Hint: The final result should be $$\frac{n^2(\omega-1)-2n}{(\omega-1)^2}$$

Edit: I will present a more detailed solution: From $A(x)=\frac{x^n-1}{x-1}$ and $B(x)=(xA(x))'$ you easily obtain $$B(x)=\frac{nx^n(x-1)-x^n+1}{(x-1)^2}$$ Now, using this and $C(x)=(xB(x))'$ you obtain $$C(x)=\frac{n^2x^n(x-1)^2-2nx^n(x-1)+(x+1)(x^n-1)}{(x-1)^3}$$ Now, plugging in $x=\omega$ and using $\omega^n=1$ we obtain $$C(\omega)=\frac{n^2(\omega-1)-2n}{(\omega-1)^2}$$

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  • $\begingroup$ You know the exact answer so please of possible write complete solution...thanks... $\endgroup$
    – neelkanth
    Oct 9 '15 at 13:21
  • $\begingroup$ @neela: I edited.. $\endgroup$
    – Tintarn
    Oct 9 '15 at 13:38
  • $\begingroup$ Thanks a lot for nice explanation. ... $\endgroup$
    – neelkanth
    Oct 9 '15 at 13:41
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If you multiply this by $1-w$, the square coefficients become 1,3,5,7.
If you multiply by $(1-w)^2$ instead, they become $1,2,2,2,....$
If you multiply by $(1-w)^3$, most of them disappear.
Some coefficients at the end remain (near $w^n$) remain.

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  • $\begingroup$ but last term is with negative $\endgroup$
    – neelkanth
    Oct 9 '15 at 11:19

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