TITLE: Spivak's proof that every polynomial of odd degree has a root
QUESTION [15 upvotes]: 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 $|x| > 0$?

REPLY [4 votes]: 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 $\frac12$?