【分圆多项式(Cyclotomic Polynomial)】的计算,逐步讲解,一目了然

1 分圆多项式的理解

互素或互质(Coprime Integers): 若整数 a a a与整数 b b b的唯一公约数为 1 1 1,则称 a a a b b b互素或互质(coprime integers)。因此,任何能整除 a a a的质数都不能整除 b b b,反之亦然。也即两者的最大公约数(Greatest Common Divisor, GCD) 1 1 1


简单描述:分圆多项式可以简单理解为多项式 x n − 1 x^n - 1 xn1分解后,所得的不可约多项式。

正式定义为:在数论中,对于任意正整数 n n n n n n次分圆多项式(nth cyclotomic polynomial)是一个具有整数系数的唯一不可约多项式 ϕ n ( x ) \phi_n(x) ϕn(x)(the unique irreducible polynomial with integer coefficients)。其中,对于任何 k < n k < n k<n, ϕ n ( x ) \phi_n(x) ϕn(x) x n − 1 x^n-1 xn1的一个因子,但不是 x k − 1 x^k-1 xk1的一个因子。该分园多项式的根为n次原语单元根(nth primitive roots of unity) e 2 i π k n e^{2i\pi \frac{k}{n}} e2nk,其中, k k k为不大于 n n n的正整数,且 k k k n n n为互质数。换句话说, n n n次分圆多项式(nth cyclotomic polynomial)为:

ϕ n ( x ) = ∏ 1 ≤ k ≤ n g c d ( k , n ) = 1 ( x − e 2 i π k n ) \phi_n(x) = \displaystyle\prod_{1≤k≤n \atop gcd(k,n)=1}(x - e^{2i\pi \frac{k}{n}}) ϕn(x)=gcd(k,n)=11kn(xe2nk)

2 举例说明计算方式

因计算过程中,常使用到 c o s / s i n cos/sin cos/sin的计算,结合下图可快速得出它们的结果。

【分圆多项式(Cyclotomic Polynomial)】的计算,逐步讲解,一目了然_第1张图片

复数计算1:由于 e i θ = c o s ( θ ) + i ⋅ s i n ( θ ) e^{i\theta} = cos(\theta) + i \cdot sin(\theta) eiθ=cos(θ)+isin(θ),可得 e 2 i π k n = c o s ( 2 π k n ) + i ⋅ s i n ( 2 π k n ) e^{2i\pi \frac{k}{n}}=cos(2\pi \frac{k}{n}) + i \cdot sin(2\pi \frac{k}{n}) e2nk=cos(2πnk)+isin(2πnk)
复数计算2: i 2 = − 1 i^2=-1 i2=1

2.1 演算 ϕ 1 ( x ) = x − 1 \phi_1(x) = x - 1 ϕ1(x)=x1

  • 仅存在一种情况: n = 1 , k = 1 n=1, k=1 n=1,k=1
    c o s ( 2 π ) + i ⋅ s i n ( 2 π ) = 1 cos(2\pi) + i \cdot sin(2\pi) = 1 cos(2π)+isin(2π)=1,所以 ϕ 1 ( x ) = x − 1 \phi_1(x) = x - 1 ϕ1(x)=x1

2.2 演算 ϕ 2 ( x ) = x + 1 \phi_2(x) = x + 1 ϕ2(x)=x+1

  • 仅存在一种情况: n = 2 , k = 1 n=2, k=1 n=2,k=1
    c o s ( π ) + i ⋅ s i n ( π ) = − 1 cos(\pi) + i \cdot sin(\pi) = -1 cos(π)+isin(π)=1,所以 ϕ 2 ( x ) = x + 1 \phi_2(x) = x + 1 ϕ2(x)=x+1

2.3 演算 ϕ 3 ( x ) = x 2 + x + 1 \phi_3(x) = x^2 + x + 1 ϕ3(x)=x2+x+1

  • 情况 1 1 1 n = 3 , k = 1 n=3, k=1 n=3,k=1
    c o s ( 2 π 3 ) + i ⋅ s i n ( 2 π 3 ) = − 1 2 + 3 2 ⋅ i cos(\frac{2\pi}{3}) + i \cdot sin(\frac{2\pi}{3}) = -\frac{1}{2} + \frac{\sqrt{3}}{2} \cdot i cos(32π)+isin(32π)=21+23 i
  • 情况 2 2 2 n = 3 , k = 2 n=3, k=2 n=3,k=2
    c o s ( 4 π 3 ) + i ⋅ s i n ( 4 π 3 ) = − 1 2 − 3 2 ⋅ i cos(\frac{4\pi}{3}) + i \cdot sin(\frac{4\pi}{3}) = -\frac{1}{2} - \frac{\sqrt{3}}{2} \cdot i cos(34π)+isin(34π)=2123 i

所以 ϕ 3 ( x ) = ( x + 1 2 − 3 2 ⋅ i ) ⋅ ( x + 1 2 + 3 2 ⋅ i ) = ( x + 1 2 ) 2 − ( 3 2 ⋅ i ) 2 = x 2 + x + 1 4 − ( − 3 4 ) = x 2 + x + 1 \phi_3(x) = (x + \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot i) \cdot (x + \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot i) ={(x + \frac{1}{2})^2} -{(\frac{\sqrt{3}}{2} \cdot i)^2} =x^2 + x +\frac{1}{4} -{(-\frac{3}{4})}=x^2 + x + 1 ϕ3(x)=(x+2123 i)(x+21+23 i)=(x+21)2(23 i)2=x2+x+41(43)=x2+x+1

2.4 演算 ϕ 4 ( x ) = x 2 + 1 \phi_4(x) = x^2 + 1 ϕ4(x)=x2+1

  • 情况 1 1 1 n = 4 , k = 1 n=4, k=1 n=4,k=1
    c o s ( π 2 ) + i ⋅ s i n ( π 2 ) = i cos(\frac{\pi}{2}) + i \cdot sin(\frac{\pi}{2}) = i cos(2π)+isin(2π)=i
  • 情况 2 2 2 n = 4 , k = 3 n=4, k=3 n=4,k=3
    c o s ( 3 π 2 ) + i ⋅ s i n ( 3 π 2 ) = − i cos(\frac{3\pi}{2}) + i \cdot sin(\frac{3\pi}{2}) = -i cos(23π)+isin(23π)=i

所以 ϕ 4 ( x ) = ( x − i ) ⋅ ( x + i ) = x 2 − ( − 1 ) = x 2 + 1 \phi_4(x) = (x - i) \cdot (x + i) =x^2 -{(-1)}=x^2 + 1 ϕ4(x)=(xi)(x+i)=x2(1)=x2+1

同理,可得 ϕ 5 ( x ) , ϕ 6 ( x ) . . . \phi_5(x),\phi_6(x)... ϕ5(x),ϕ6(x)...

综述,从 1 1 1 30 30 30的分圆多项式为:

【分圆多项式(Cyclotomic Polynomial)】的计算,逐步讲解,一目了然_第2张图片


参考材料

  • Wikipedia: Cyclotomic Polynomial

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