Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to use the following identities for poisson integral but i can't guz i don't know how to prove them.

$$\alpha^{2n}-1=\prod_{k=0}^{k=2n-1}(\alpha-e^{i\frac{2k\pi}{2n}})$$

$$\prod_{k=0}^{k=2n-1}(\alpha-e^{i\frac{2k\pi}{2n}})=(\alpha-1)(\alpha+1)\prod_{k=1}^{k=n-1}(\alpha-e^{i\frac{k\pi}{n}})(\alpha-e^{-i\frac{k\pi}{n}})$$

I would appreciate any help

what i did :

Let $\alpha >1\,\quad \alpha^{2n} =1$

\begin{align*} \alpha^{2n} &=1\\ (\rho.e^{i\phi})^{2n} &=1.e^{i.0} \text{ with $\rho \in \mathbb{R}^{+}_{*},\ \phi \in \mathbb{R}$}\\ \rho^{2n}.e^{i.2n.\phi}&=1.e^{i.0}\\ \rho^{2n}&=1, \quad e^{i.2n\phi}=e^{i.0}\\ \rho&=1, \quad 2n.\phi=0[2\pi]\\ \rho&=1, \quad \exists k\in \mathbb{Z} \ 2n.\phi=2.k.\pi\\ \rho&=1, \quad \phi=\frac{2.k.\pi}{2n} \text{ Then } \alpha_{k}=e^{i\frac{2.k.\pi}{2n}}\\ \text{Let } k &= t + 2nq \text{ or } t\in \{0,1,\ldots,2n-1\} \text{ and } q \in \mathbb{Z}.\\ &\text{ Then } \alpha_{k}=e^{i\dfrac{2.t.\pi}{2n}+2.\pi.q}=Z_{t} \\ S&=\{Z_{k}=e^{i\dfrac{2.k.\pi}{2n}}|k\in \{0,1,\ldots,2n-1\} \} \end{align*}

share|improve this question

2 Answers 2

up vote 2 down vote accepted

First Identity: Let $$p(x) = x^n + a_{n-1} x^{n-1}+ \ldots+a_0$$ a polynomial function of order $n$. Then, by the fundamental theorem of algebra, we have $$p(x) = \prod_{j=1}^n (x-x_j)$$ where $x_1,\ldots,x_n$ are the roots of $p$. Apply this to $p(x) := x^{2n}-1.$

Second Identity:

Write $$\prod_{k=0}^{2n-1} \left(\alpha - e^{\imath \frac{2k\pi}{2n}} \right) = \underbrace{\left(\prod_{k=0}^0 \dots \quad \cdot \prod_{k=n}^n \dots \right)}_{(\alpha-1)(\alpha+1)} \left( \prod_{k=1}^{n-1} \dots \quad \cdot \prod_{k=n+1}^{2n-1} \dots \right).$$

Now note that

$$\exp \left( 2\pi \imath \frac{2n-\ell}{2n} \right) = \exp \left( 2\pi \imath - 2\pi \imath \frac{\ell}{2n} \right) = \exp \left(- 2\pi \imath \frac{\ell}{2n} \right) \tag{1}$$

as $x \mapsto e^{\imath x}$ is periodic. For $k \in \{n+1,\ldots,2n-1\}$ we write $k = 2n-\ell$ where $\ell \in \{1,\ldots,n-1\}$. Applying $(1)$, we get

$$\prod_{k=n+1}^{2n-1} \left(\alpha - e^{\imath \frac{2k\pi}{2n}} \right) = \prod_{\ell=1}^{n-1} \left(\alpha - e^{- \imath \pi \frac{\ell}{n}} \right).$$

This finishes the proof.

share|improve this answer
    
Fundamental Theorem of Algebra--not Calculus. –  Jared Jun 7 at 6:55
    
@Jared Of course; thanks! –  saz Jun 7 at 6:55
1  
@Educ Let $k \in \{1,\ldots,n-1\}$. Then $$\exp \left(-\imath \frac{k\pi}{n} \right) = \exp \left(-\imath \frac{k\pi}{n} + \imath 2k\pi \right) = \exp \left(\imath \frac{(2k-1) \pi}{n}\right).$$ –  saz Jun 7 at 8:47
1  
@Educ I have added the proof of the 2nd identity. –  saz Jun 7 at 12:17
1  
@Educ Yes, but the multiplication is commutative, i.e. $a \cdot b = b \cdot a$. So $$\prod_{k=1}^n a_k = a_1 \cdot a_2 \cdots a_n = \prod_{k=n}^1 a_k.$$ –  saz Jun 7 at 14:01

The first line is a statement about roots of unity and factorization. The numbers $e^{2\pi i k / n}$ are precisely the $n$th roots of unity. (And the factor theorem, I suppose).

The second one is just a rearrangement of terms from the first. If you'd like an alternate view, you might think of $\alpha$ as a variable, and consider each side as polynomials in $\alpha$. Then since they have the same roots and the same value at $\alpha = 0$, they are the same polynomial.

In short: you should read up on roots of unity and possibly the factor theorem.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.