1
$\begingroup$

Hello guys I have a simple question to ask. For example I have the equation :

$$x^n + x^{n-1} + x^{n-2} + ... + 1 = 0$$

I read somewhere that the number of solutions to an equation is given by the biggest power in the equation. So in the equation above, there should be $n$ solutions. Is this correct ? If it is, can anyone provide a proof?

$\endgroup$
  • $\begingroup$ are you interested in purely real solutions? $\endgroup$ – oldrinb Jan 23 '15 at 20:19
  • 1
    $\begingroup$ Sometimes, there are so-called "multiple" roots. For example, $1$ is a "double root" of $x^2 - 2x + 1 = 0$, and there are no other roots. So the theorem, called the "fundamental theorem of algebra" is only valid if roots are taken into account with their multiplicities. Also, the theorem doesn't work if you only want real solutions; you need to allow complex roots. The fundamental theorem of algebra is a difficult theorem to prove. In the specific example you gave, however, the $n$ roots are easy to find. They're the complex numbers $e^{2\pi k i/(n+1)}$ for $k = 1, 2, \dots, n$. That's... $\endgroup$ – user208259 Jan 23 '15 at 20:41
  • 1
    $\begingroup$ ...because if you multiply both sides of the equation by $x-1$, you get $x^{n+1} - 1 = 0$. So the roots are all $(n+1)$st complex roots of unity besides $1$. If you're only interested in real roots, the answer is that there are none of $n$ is even, and just $x=-1$ if $n$ is odd. $\endgroup$ – user208259 Jan 23 '15 at 20:43
2
$\begingroup$

The solutions are counting multiplicity and include complex solutions.

It's called the fundamental theorem of algebra. Low-level proofs are not easy to come by, however.

$\endgroup$
  • $\begingroup$ Working over a field different from $\mathbb{C}$ the situation could be totally different. $\endgroup$ – Nicky Hekster Jan 23 '15 at 20:19
  • 1
    $\begingroup$ Presumably OP is interested in the reals, or the complex numbers. For this particular equation, we do not need the Fundamental Theorem of Algebra. The complex solutions can be identified explicitly. $\endgroup$ – André Nicolas Jan 23 '15 at 20:27

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.