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For what real values does $\phi(x):=1+x+ \dots + x^{2m-1}$ take the value $0$? What can you say about the sign as $x$ varies?


I need help adding rigor to my observation to create a formal proof.

$\phi(x):=1+x+x^2+x^3+ \dots +x^{2m-2}+ x^{2m-1}$

$=(1+x)+x^2(1+x)+ \dots + x^{2m-2}(1+x)$

I think that this is fairly straightforward and obvious. How can I more rigorously prove that this re-forming is true?

From this re-forming, obviously $\phi(x)=0$ if $x=-1$. Additionally, $\phi(x)<0$ if $x<-1$ and $\phi(x)>0$ if $x>-1$.

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  • $\begingroup$ I think that it is perfect just like this. $\endgroup$
    – ajotatxe
    Commented Apr 22, 2015 at 19:02

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What you've done is quite fine if you're only interested in real zeros. For what it's worth though, the geometric series approach gives a concise answer:

$$\phi(x)=\frac{1-x^{2m}}{1-x},$$

which has zeros when $x^{2m}=1$, except $x=1$, so $x=e^{2\pi k i/ 2m}$ for $k=1,2,\cdots,2m-1$.

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Another way of tackling this: Observe (or recall from geometric series) that $(x-1)\phi(x)=x^{2m}-1$. The latter is zero (over the reals) iff $x=\pm1$. Clearly $\phi(1)=2m-1>0$ so $x=-1$ is the only possible real root of $\phi(x)$ (and since $(x-1)$ is nozero there, it is indeed a root).

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