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I need to conclude $n\xi_j^{n-1}=\prod_{i \ne j}(\xi_{i}-\xi_{j})$ from $x^n-1=\prod_{i}(x-\xi_{i})$ where $\xi_i$ is the $i$th root of the unit.

And another thing I do not quite understand is that why $\prod_{i=1}^n\xi_{i}^{-1}=(-1)^{n-1}$ where $\xi_i$ is the $i$th root of the unit.

I think maybe it can be done by differentiation but how can I ceal with the right hand side?

Could someone write a deduction explicitly? Thanks so much!

Alex M.
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Y.X.
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    One way to derive it is in my other answer here. To follow the letter of the book do what they say: take the derivative in $z$ of $\prod(z-\xi_{j})$ then evaluate it at $z=\xi_{k}$ (hint: there will be just one non-zero term left). You could try a simple case, maybe $n=3$, to see how it works step by step. – dxiv Feb 27 '17 at 20:43
  • @dxiv Thanks su much! Now I can understand the first identity that I asked. But how to deal with the second one mentioned above? – Y.X. Feb 27 '17 at 20:47
  • As answered already, that follows from Vieta's formula for the product of roots for the polynomial $x^n-1,$. – dxiv Feb 27 '17 at 20:52

4 Answers4

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Let $\Phi_n(x) = x^n - 1$. Then for each $1 \leq j \leq n$, we have $$\Phi_n(x) = \prod_{i=1}^{n}(x - \xi_n) = (x- \xi_j)\prod_{i \neq j}(x - \xi_i).$$ Using the product rule, we find that $$\Phi_n'(x) = (x - \xi_j)\frac{d}{dx}\prod_{i\neq j}(x - \xi_i) + \prod_{i\neq j}(x - \xi_i).$$ We also have $$\Phi_n'(x)= nx^{n-1}$$ from the power rule. Letting $x = \xi_j$ in both expressions gives $$n\xi_j^{n-1} = \prod_{i \neq j}(\xi_j - \xi_i).$$ As for the second question, in general the product of the roots of a degree $n$ polynomial $$f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$$ is given by $(-1)^n a_0$. In the case of $\Phi_n(x)$, $a_0 = -1$, thus $$\prod_{i=1}^{n}\xi_i^{-1} = \left(\prod_{i=1}^{n}\xi_i\right)^{-1} = (-(-1)^{n})^{-1} = (-1)^{n - 1}.$$

Ethan Alwaise
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First, you can factor $$x^n-1=(x-\xi_i)(x^{n-1}+x^{n-2}\xi_i+\cdots+x\xi_i^{n-2}+\xi_i^{n-1}).$$ Therefore, $$ (x^{n-1}+x^{n-2}\xi_i+\cdots+x\xi_i^{n-2}+\xi_i^{n-1}=\prod_{j\neq i}(x-\xi_j). $$ Evaluating this last expression at $x=\xi_i$ gives the first result.

I'm not quite sure how to interpret your second question.

David Hill
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  • I have edited the question. Does it seem to become clearer now? – Y.X. Feb 27 '17 at 20:45
  • @Y.X. for odd $n$, every term in the product $\prod_{i=1}^n\xi_i$ other than $1$ can be paired with its inverse, so the product is $1$. When $n$ is even, only the terms $(-1)$ and $1$ are their own inverses, so the product is $-1$. Putting that together you get $(-1)^{n-1}$. – David Hill Feb 28 '17 at 04:23
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First of all, we can expand $\prod_{i=1}^n(x-\xi_i)$ to get $$\prod_{i=1}^n(x-\xi_i) = x^n + x^{n-1}\sum_{i=1}^n(-\xi_i) + x^{n-2}\sum_{i=1}^n\sum_{j=1}^n(-\xi_i)(-\xi_j) + \cdots + x^0\prod_{k=1}^n(-\xi_k) = x^n-1$$ Two $n$-th polynomial of $x$ equal means each corresponding coefficient equal. This gives us a series of relations: $$\sum_{i=1}^n\xi_i = 0$$ $$\sum_{i=1}^n\sum_{j=1}^n\xi_i\xi_j = 0$$ $$\cdots$$ $$\prod_{k=1}^n(-\xi_k) = -1$$ The last relation leads to $\prod_{k=1}^n\xi_k^{-1} = (-1)^{n-1}$.

Now let's look at the term $\prod_{j \neq i}^n(\xi_i-\xi_j)$. WOLG, we can let $i = 1$. $$\prod_{j=2}^n(\xi_1-\xi_j) = \xi_1^{n-1}+\xi_1^{n-2}\sum_{j=2}^n(-\xi_j) + \xi_1^{n-3}\sum_{j=2}^n\sum_{k=2}^n(-\xi_j)(-\xi_k)+\cdots+\xi_1^0\prod_{j=2}^n(-\xi_j)$$ Using the relation we found before, we can see each term equals to $\xi_1^{n-1}$, and there $n$ terms. Thus $$\prod_{j=2}^n(\xi_1-\xi_j) = n\xi_1^{n-1}$$

Guangliang
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For the second identity: Note that by Vieta's formula, the constant term $-1$ here is either the elementary symmetric function $e_{n}$ or $-e_{n}$ and it depends on the parity of $n$. When $n$ is odd, $1$ should be $e_{n}$ and when $n$ is even, $-1$ should be $e_{n}$. So as $e_{n}$ has been defined to be $\prod_{i}\xi_{i}$, $\prod_{i}\xi_{i}=(-1)^{n-1}$. Thus $\prod_{i}\xi_{i}^{-1}=(\prod_{i}\xi_{i})^{1}=((-1)^{n-1})^{-1}=(-1)^{n-1}$.

I would like to know if I have got it right, so could someone please point out if there is any mistake.

Y.X.
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