二項式の因数分解(5)

$$x^{2n+1}-1=(x-1)\displaystyle \prod_{k=1}^{n}\left(x^2-2x \cos \frac{2kπ}{2n+1}+1\right)$$




<証明>

\(x^{2n+1}=1\) の解を求めるために \(x= \cos θ+i \sin θ\) とおくと$$x^{2n+1}= \cos (2n+1)θ+i \sin (2n+1)θ$$一方 \(1=1+0 \cdot i=1 \cdot ( \cos 0+i \sin 0)\)$$ \cos (2n+1)θ+i \sin (2n+1)θ=1 \cdot ( \cos 0+i \sin 0)$$となるので \(r=1, (2n+1)θ=0+2kπ\)$$ θ=\frac{2k}{2n+1}π (k=0,1,2, \cdots ,2n)$$よって \(x^{2n+1}=1\) の解は \(x= \cos θ+i \sin θ=e^{θi}=e^{\frac{2kπ}{2n+1}i}\) より$$x_0=e^{\frac{2 \cdot 0 \cdot π}{2n+1}i}=1     x_1=e^{\frac{2π}{2n+1}i}    x_2=e^{\frac{4π}{2n+1}i} \cdots  $$$$ \cdots x_{n+1}=e^{\frac{2nπ}{2n+1}i}     x_{n+2}=e^{\frac{(2n+2)π}{2n+1}i} \cdots  $$$$ \cdots x_{2n}=e^{\frac{(4n-2)π}{2n+1}i}    x_{2n+1}=e^{\frac{4nπ}{2n+1}i} $$以上より \(x^{2n+1}-1\) はこれら \((2n+1)\) 個の解を用いて因数分解できるので$$ x^{2n+1}-1=(x-1)(x-e^{\frac{2π}{2n+1}i})(x-e^{\frac{4π}{2n+1}i}) \cdots $$$$ \cdots (x-e^{\frac{2nπ}{2n+1}i})(x-e^{\frac{(2n+2)π}{2n+1}i}) \cdots (x-e^{\frac{(4n-2)π}{2n+1}i})(x- e^{\frac{4nπ}{2n+1}i})$$次に右辺の \((x-1)\) を除いた \(2n\) 個のうち\(2\) つを
「\(1\) 番目×\((2n)\) 番目」「\(2\) 番目×\((2n-1)\) 番目」…「\(n\) 番目×\((n+1)\) 番目」のように計算していくと
\begin{alignat}{2}
&(x-e^{\frac{2π}{2n+1}i})(x- e^{\frac{4nπ}{2n+1}i})=x^2-x(e^{\frac{2π}{2n+1}i}+e^{\frac{4nπ}{2n+1}i})+e^{\frac{2π}{2n+1}i+\frac{4nπ}{2n+1}i}\\
&                   =x^2-x(e^{\frac{2π}{2n+1}i}+e^{\frac{4nπ}{2n+1}i} \cdot e^{-2πi})+e^{2πi}\\
&                   =x^2-x(e^{\frac{2π}{2n+1}i}+e^{-\frac{2π}{2n+1}i})+1\\
&                   =x^2-2x\left(\frac{e^{\frac{2π}{2n+1}i}+e^{-\frac{2π}{2n+1}i}}{2}\right)+1\\
&                   =x^2-2x \cos \frac{2π}{2n+1}+1\\
\end{alignat}\begin{alignat}{2}
&(x-e^{\frac{4π}{2n+1}i})(x-e^{\frac{(4n-2)π}{2n+1}i})=x^2-x(e^{\frac{4π}{2n+1}i}+e^{\frac{(4n-2)π}{2n+1}i})+e^{\frac{4π}{2n+1}i+\frac{(4n-2)π}{2n+1}i}\\
&                    =x^2-x(e^{\frac{4π}{2n+1}i}+e^{\frac{(4n-2)π}{2n+1}i} \cdot e^{-2πi})+e^{2πi}\\
&                    =x^2-x(e^{\frac{4π}{2n+1}i}+e^{-\frac{4π}{2n+1}i})+1\\
&                    =x^2-2x\left(\frac{e^{\frac{4π}{2n+1}i}+e^{-\frac{4π}{2n+1}i}}{2}\right)+1\\
&                    =x^2-2x \cos \frac{4π}{2n+1}+1
\end{alignat}
$$\cdots$$
\begin{alignat}{2}
&(x-e^{\frac{2nπ}{2n+1}i})(x-e^{\frac{(2n+2)π}{2n+1}i})=x^2-x(e^{\frac{2nπ}{2n+1}i}+e^{\frac{(2n+2)π}{2n+1}i})+e^{\frac{2nπ}{2n+1}i+\frac{(2n+2)π}{2n+1}i}\\
&                    =x^2-x(e^{\frac{2nπ}{2n+1}i}+e^{\frac{(2n+2)π}{2n+1}i} \cdot e^{-2πi})+1\\
&                    =x^2-x(e^{\frac{2nπ}{2n+1}i} +e^{-\frac{2nπ}{2n+1}i})+1\\
&                    = x^2-2x\left(\frac{e^{\frac{2nπ}{2n+1}i} +e^{-\frac{2nπ}{2n+1}i}}{2}\right)+1\\
&                    =x^2-2x \cos \frac{2nπ}{2n+1}+1
\end{alignat}
以上より、右辺はこれら \(n\) 個と \((x-1)\) の積になるから$$x^{2n+1}=(x-1)\left(x^2-2x \cos\frac{2π}{2n+1}+1 \right)\left(x^2-2x \cos \frac{4π}{2n+1}+1 \right) \cdots $$$$ \cdots \left(x^2-2x \cos \frac{2nπ}{2n+1}+1 \right)$$ $$x^{2n+1}-1=(x-1)\displaystyle \prod_{k=1}^{n}\left(x^2-2x \cos \frac{2kπ}{2n+1}+1\right)$$

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