Let n be a positive integer and let z be a complex number such that z^{2^n-1}-1=0. Evaluate
\prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right).
Proposed by Titu Andreescu.
Solution:
It's clear that if z=1, then \displaystyle \prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right)=1. Assume that z \neq 1. We have z^{2^n-1}-1=(z-1)(z^{2^n-2}+z^{2^n-3}+\ldots+z+1)=0, so z^{2^n-2}+z^{2^n-3}+\ldots+z+1=0. Then,
\begin{array}{lll} \displaystyle \prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right) & = & \displaystyle \prod_{k=0}^{n-1}\left(z^{2^{k+1}}-z^{2^k}+1\right) \\ & = & \displaystyle \prod_{k=0}^{n-1} \dfrac{z^{3\cdot2^k}+1}{z^{2^k}+1}.\end{array} Clearly, \displaystyle \prod_{k=0}^{n-1} (z^{2^k}+1) = \sum_{k=0}^{2^n-1} z^k = z^{2^n-1} + \sum_{k=0}^{2^n-2} z^k = 1. Let n be odd. Since \gcd(2^n-1,3)=1, then z^{3i} \neq z^{3j} \quad \forall i \not \equiv j \pmod{2^n-1}. So, \displaystyle \prod_{k=0}^{n-1} (z^{3\cdot2^k}+1) = \sum_{k=0}^{2^n-1} z^{3k} = z^{3 (2^n-1)} + \sum_{k=0}^{2^n-2} z^{3k} = 1 + \sum_{k=0}^{2^n-2} z^k = 1, which gives \displaystyle \prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right)=1. Now, let n be even. Then z^{2^n-1}-1=(z^3-1)(z^{2^n-4}+z^{2^n-7}+\ldots+z^3+1)=0. If z is a third root of unity, then z^3=1 and
\prod_{k=0}^{n-1} (z^{3\cdot2^k}+1) = \sum_{k=0}^{2^n-1} z^{3k} = 2^n, which gives \displaystyle \prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right)=2^n. If z is not a third root of unity, since z^{2^n-(3k+1)}=z^{3(2^n-1-k)}, then
\prod_{k=0}^{n-1} (z^{3\cdot2^k}+1) = \sum_{k=0}^{2^n-1} z^{3k} = z^{3k(2^n-1)} + \sum_{k=1}^{2^n-1} z^{3(2^n-1-k)} = 1 + 3 \sum_{k=1}^{\frac{2^n-1}{3}} z^{2^n-(3k+1)} = 1.
In summary,
\prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right)=\left\{ \begin{array}{lll} 2^n & \textrm{if } n \textrm{ is even and } z = e^{\frac{2i\pi}{3}},e^{\frac{4i\pi}{3}} \\ 1 & \textrm{otherwise.} \end{array} \right.
\begin{array}{lll} \displaystyle \prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right) & = & \displaystyle \prod_{k=0}^{n-1}\left(z^{2^{k+1}}-z^{2^k}+1\right) \\ & = & \displaystyle \prod_{k=0}^{n-1} \dfrac{z^{3\cdot2^k}+1}{z^{2^k}+1}.\end{array} Clearly, \displaystyle \prod_{k=0}^{n-1} (z^{2^k}+1) = \sum_{k=0}^{2^n-1} z^k = z^{2^n-1} + \sum_{k=0}^{2^n-2} z^k = 1. Let n be odd. Since \gcd(2^n-1,3)=1, then z^{3i} \neq z^{3j} \quad \forall i \not \equiv j \pmod{2^n-1}. So, \displaystyle \prod_{k=0}^{n-1} (z^{3\cdot2^k}+1) = \sum_{k=0}^{2^n-1} z^{3k} = z^{3 (2^n-1)} + \sum_{k=0}^{2^n-2} z^{3k} = 1 + \sum_{k=0}^{2^n-2} z^k = 1, which gives \displaystyle \prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right)=1. Now, let n be even. Then z^{2^n-1}-1=(z^3-1)(z^{2^n-4}+z^{2^n-7}+\ldots+z^3+1)=0. If z is a third root of unity, then z^3=1 and
\prod_{k=0}^{n-1} (z^{3\cdot2^k}+1) = \sum_{k=0}^{2^n-1} z^{3k} = 2^n, which gives \displaystyle \prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right)=2^n. If z is not a third root of unity, since z^{2^n-(3k+1)}=z^{3(2^n-1-k)}, then
\prod_{k=0}^{n-1} (z^{3\cdot2^k}+1) = \sum_{k=0}^{2^n-1} z^{3k} = z^{3k(2^n-1)} + \sum_{k=1}^{2^n-1} z^{3(2^n-1-k)} = 1 + 3 \sum_{k=1}^{\frac{2^n-1}{3}} z^{2^n-(3k+1)} = 1.
In summary,
\prod_{k=0}^{n-1}\left(z^{2^k}+\dfrac{1}{z^{2^k}}-1\right)=\left\{ \begin{array}{lll} 2^n & \textrm{if } n \textrm{ is even and } z = e^{\frac{2i\pi}{3}},e^{\frac{4i\pi}{3}} \\ 1 & \textrm{otherwise.} \end{array} \right.
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