# Spivak's proof that every polynomial of odd degree has a root

I have the second edition of Spivak. Consider

Can someone tell me why he considers $2n|a_{n-1}| \dots$? Later he shows everything is squeezed between -1/2 and 1/2 and he gets the desired result. I am perplexed as to why $2n$? He could have just done it wit $n$ and it would work out.

Also he never considered the case for $|x| < 1$, what happens then?

EDIT: On another matter, why does he write $|x| > 1, 2n|a_{n-1}| \dots$ instead of just writing $|x| > \max \{1, 2n|a_{n-1}|, \dots \}$?

EDIT2:What is more interesting is that he assumes that $\dfrac{|a_{n-k}|}{|x|} < \dfrac{|a_{n-k}|}{2n|a_{n-k}|}$. Is there a reason why he it will not be just $\dfrac{|a_{n-k}|}{|x|} < |a_{n-k}|$ and why does he even consider $|x| > 1$, why not $|x| > 0$?

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The factor $2$ doesn't seem necessary, given the sharp inequalities involved (I mean $<$ instead of $\le$). As for $|x|<1$, he does not need to look at that because all he's interested in is showing that the polynomial takes both positive and negative values somewhere. And he finds those for large $|x|$. – Harald Hanche-Olsen May 4 '13 at 19:22
Oh so he is considering a root, not $n$ number (different $n$ in my context) of root(s) that happens to lie $x > 1$ – Hawk May 4 '13 at 19:24

The point is not just to ensure that $|1+\sum_{k=1}^{n}a_{n-k}/x^k|$ is positive but that it is actually bounded away from zero, independently of $x$. Actually, the factor $2$ is overkill, but it makes the analysis quick and easy. Harald's comment answers your second question.
Edit: (In response to the OP's query) Yes, it's sufficient to take $x\ge n|a_k|+\varepsilon\;(k=1,\dots,n)$, along with $x>1$, where $\varepsilon>0$ is independent of $x$. Then the multiplying factor $|1+\sum_{k=1}^{n}a_{n-k}/x^k|$ exceeds a constant, which can be written in terms of $\varepsilon$ and the $a_k$, which can be seen to be positive. But why bother when you can just write $\frac12$?