TITLE: Spivak Calculus chapter 7 theorem 9
QUESTION [7 upvotes]: I am working through Spivak Calculus chapter 7 theorem 9. There is one statement that I can't quite understand.
The theorem states:
If $n$ is odd, then any equation
$$
\ x^n+a_{n-1}x^{n-1} +\cdots+a^0
$$
has a root.
proof: we would like to prove that $f$ is sometimes positive and sometimes negative. The
intuitive idea is that for large $|x|$, the function is very much like $g(x) = x^n$ and,
since $n$ is odd, this function is positive for large positive $x$ and negative for large
negative $x$. A little algebra is all we need to make this intuitive idea work.
$$
f(x) = x^n+a_{n-1}x^{n-1} +\cdots+a^0 = x^n \left(1+\frac{a_{n-1}}{x}+\cdots+\frac{a_0}{x^n}\right)
$$
Note that 
$$
\left|\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_0}{x^n} \right|\le \frac{|a_{n-1}|}{|x|}+\frac{|a_{n-2}|}{|x^2|}+\cdots+\frac{|a_{0}|}{|x^n|}
$$
Consequently if we choose $x$ satisfying
$$
|x|>1,2n|a_{n-1}|,\ldots,2n|a_0| \tag{*}
$$
I am not sure how he comes to $(*)$
Thanks in advance

REPLY [3 votes]: You need to keep reading in the same page. He wants to make 
$$
\left|\frac{a_{n-1}}{x}+\frac{a_{n-2}}{x^2}+\cdots+\frac{a_{0}}{x^n} \right|\le \frac12,
$$
in order to guarantee that the term multiplying $x^n$ is positive (one plus a number of absolute value less than a half will be positive). 
So he chooses $x$ such that $|a_{n-k}/|x^k|\leq 1/2n$ (for all $k$), and then the sum will be less than $1/2$. He needs $|x|\geq1$ so that $|x^k|\geq|x|$.