How to find the real factors of $$ (1+x^{2n}) $$ Solution given is : $$\prod_{k=0}^{n-1} (x^{2} -2x\cos((2k+1)π/2n) +1 ) $$

How to prove the solution? Please note that I tried equating the equation to 0. So I have $$x^{2n} = -1.$$ I don't know whether my approach is correct

  • 1
    $\begingroup$ So, find the (complex) solutions of $x^{2n}=-1$ and pair off each with its complex conjugate. $\endgroup$ – Lord Shark the Unknown Jul 7 '18 at 10:29
  • $\begingroup$ Can you show it in your answer please $\endgroup$ – David Jul 7 '18 at 10:30
  • $\begingroup$ You can continue like this: $x^{2n}=-1=e^{\pi i}$. Then $x=e^{(\pi i + 2k\pi)/(2n)}$, for $k=0,1,...,2n-1$. Therefore, $x^{2n}+1=\prod_{k=0}^{2n-1}(x-e^{(\pi i + 2k\pi)/(2n)})$. Now, when you multiply the pairs of factors $(x-e^{(\pi i + 2k\pi)/(2n)})(x-e^{(\pi i + 2l\pi)/(2n)})$, where $k+l+1=2n$, you get those factors with real coefficients, because the roots $e^{(\pi i + 2k\pi)/(2n)}$ and $e^{(\pi i + 2l\pi)/(2n)}$ are conjugate complex numbers (their product is $1$ and their sum is $2$ times their real part). To get their real part use Euler's formula $e^{ir}=\cos(r)+i\sin(r)$. $\endgroup$ – user566930 Jul 7 '18 at 10:46
  • $\begingroup$ @minghan Thank you $\endgroup$ – David Jul 7 '18 at 11:16

The roots of $x^{2n}+1$ are the roots of $-1=e^{\pi i}$, which come in pairs: $$\exp\left(\pm\frac{\pi i}{2n}\right),\exp\left(\pm\frac{3\pi i}{2n}\right),\ldots,\exp\left(\pm\frac{(2n-1)\pi i}{2n}\right).$$In each pair you have a complex number and its conjugate. So, $x^{2n}+1$ is the product of the quadratic polynomials$$\left(x-\exp\left(\frac{(2k+1)\pi i}{2n}\right)\right)\left(x-\exp\left(-\frac{(2k+1)\pi i}{2n}\right)\right),\tag1$$with $k\in\{0,1,\ldots,n-1\}$. And $(1)$ is equal to $x^2-2\cos\left(\frac{(2k+1)\pi}{2n}\right)x+1$

  • $\begingroup$ thanks sir. very helpful $\endgroup$ – David Jul 7 '18 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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