Difference between revisions of "Cyclotomic polynomial"

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==Roots==
 
==Roots==
 
The roots of <math>\Phi_n(x)</math> are <math>e^{\frac{2\pi i d}{n}}</math>, where <math>\gcd(d, n) = 1</math>. For this reason, due to the [[Fundamental Theorem of Algebra]], we have <math>\Phi_n(x) = \prod_{d: \gcd(d, n) = 1} (x - e^{\frac{2\pi i d}{n}})</math>.
 
The roots of <math>\Phi_n(x)</math> are <math>e^{\frac{2\pi i d}{n}}</math>, where <math>\gcd(d, n) = 1</math>. For this reason, due to the [[Fundamental Theorem of Algebra]], we have <math>\Phi_n(x) = \prod_{d: \gcd(d, n) = 1} (x - e^{\frac{2\pi i d}{n}})</math>.
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Therefore, <math>x^{n}-1</math> can be factored as <math>\Phi_{d_{1}}x*\Phi_{d_{2}}x*\Phi_{d_{3}}x*\cdots*\Phi_{d_{k}}</math> where <math>d_1, d_2, d_3\cdots, d_k</math> are the positive divisors of <math>n</math>.
  
 
==Examples==
 
==Examples==

Revision as of 16:00, 31 May 2020

Definition

The cyclotomic polynomials are recursively defined as $x^n-1=\prod_{d \vert n} \Phi_n (x)$, for $n \geq 1$. All cyclotomic polynomials are irreducible.

Roots

The roots of $\Phi_n(x)$ are $e^{\frac{2\pi i d}{n}}$, where $\gcd(d, n) = 1$. For this reason, due to the Fundamental Theorem of Algebra, we have $\Phi_n(x) = \prod_{d: \gcd(d, n) = 1} (x - e^{\frac{2\pi i d}{n}})$.

Therefore, $x^{n}-1$ can be factored as $\Phi_{d_{1}}x*\Phi_{d_{2}}x*\Phi_{d_{3}}x*\cdots*\Phi_{d_{k}}$ where $d_1, d_2, d_3\cdots, d_k$ are the positive divisors of $n$.

Examples

For a prime $p$, $\Phi_p (x)=x^{p-1}+x^{p-2}+ \cdots + 1$, because for a prime $p$, $\Phi_p (x) \cdot \Phi_1 (x)=x^p - 1$ and so we can factorise $x^p - 1$ to obtain the required result.

The first few cyclotomic polynomials are as shown: \begin{align*} \Phi_1(x)&=x-1 \\ \Phi_2(x)&=x+1 \\ \Phi_3(x)&=x^2+x+1 \\ \Phi_4(x)&=x^2+1 \\ \Phi_5(x)&=x^4+x^3+x^2+x+1 \\ \Phi_6(x)&=x^2-x+1 \\ \Phi_7(x)&=x^6+x^5+\cdots + 1 \\ \Phi_8(x)&=x^4+1 \\ \Phi_9(x)&=x^6+x^3+1 \\ \Phi_{10}(x)&=x^4-x^3+x^2-x+1\\ \end{align*}

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