$\prod_{k=1}^{n-1}(1-e^{2kπi/n})=n(-1)^{n-1}$
#$\prod_{k=1}^{n-1}(1−e^2kπi/n)=n(-1)^{n-1}
languid zealot
thin wolfBOT
naokye
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languid zealot
i have a proof but
its got a singularity
\begin{align*}
x^n-1 &= (x-1)(x-e^{\frac{2\pi}{n}i})(x-e^{\frac{4\pi}{n}i})(x-e^{\frac{6\pi}{n}i})\cdots(x-e^{\frac{(n-1)\pi}{n}i})\
(x-1)(x^{n-1}+x^{n-2}+\cdots+1) &= (x-1)\prod_{\ell=1}^{n-1}(x-e^{\left(\frac{2\ell\pi}{n}\right)i})\
x^{n-1}+x^{n-2}+\cdots+1 &= (-1)^{n-1}\prod_{\ell=1}^{n-1}(e^{\left(\frac{2\ell\pi}{n}\right)i}-x)\
\text{setting $x=1$ will give:} \
n &= (-1)^{n-1}\prod_{\ell=1}^{n-1}(e^{\left(\frac{2\ell\pi}{n}\right)i}-1)\
\prod_{\ell=1}^{n-1}(e^{\left(\frac{2\ell\pi}{n}\right)i}-1) &= n(-1)^{1-n}
\end{align*}
thin wolfBOT
naokye
weary sand
\begin{align*}
x^n-1 &= (x-1)(x-e^{\frac{2\pi}{n}i})(x-e^{\frac{4\pi}{n}i})(...
Yup, seems good.
languid zealot
we cancelled out the $x=1$
thin wolfBOT
naokye
languid zealot
but we set x=1
so theres the singularity there
and i dont think my proof is solid
weary sand
Both sides are still defined at x = 1, there's no problem.
languid zealot
my lecturer said theres the singularity so he doesnt accept it tho
weary sand
Well, you can do it like this:
(x^n - 1)/(x - 1) = Π(x - e^(2πik/n), k = 1 to n - 1)
The find the limit of both sides as x -> 1. Though, I think the given answer is wrong - it should just be n. (-1)^(n - 1) would be present if we had e^(2πik/n) - 1 under the product.
fervent nexus
my lecturer said theres the singularity so he doesnt accept it tho
take a limit ig