proof that odd polynomial has root I'm an Engineer (so blindingly obvious things to Mathematicians are not to me) trying to understand mathematics after applying it for many years. So I'm slowly reading Spivak's Calculus on a journey to read some basic Mathematics degree material.
I have finally gotten stuck on theorem 9 proof see extract below from the chapter Three Hard Theorems.
1) I do not understand how $\overset{n terms}{\frac{1}{2n} + \ldots + \frac{1}{2n}} = 1/2$ in the proof.
2) I'm also not sure why $|x| > 2n|a_n-1|,\ldots,2n|a_0|$,
Any help appreciated.


 A: The condition 
$$|x|>1,2n|a_{n-1}|,\ldots,2n|a_0|$$
is just a choice aims to guarentee that
$$f(x)=x^n\left(1+\frac{a_{n-1}}{x}+\cdots+\frac{a_0}{x^n}\right)$$ be positive. Notice that this is possible and just we can take for example $|x|$ is the maximum of these reals plus $1$.
For your first question:
$$\underbrace{\frac1{2n}+\cdots+\frac1{2n}}_{n\;\text{times}}=n\times \frac1{2n}=\frac12$$
A: As an Engineer, you should appreciate when I tell you that they're reinventing too much of the wheel. You should build on what you know. You have to assume the Intermediate Value Theorem, which is a big deal. So, why not short-cut some of the confusion and assume properties of limits that you know from Calculus, too. For example,
$$
                  \lim_{x\rightarrow\pm\infty}\frac{p(x)}{x^{n}}=1.
$$
So, there exists $R > 0$ such that $|p(x)/x^{n}-1| < 1/2$ for $x > R$, or
$$
               1/2 < p(x)/x^{n} < 3/2,\;\;\; x > R\\
                       \implies 1/2 < p(2R)/(2R)^{n} < 3/2\\
                       \implies p(2R) > 0.
$$
Similarly, there exists $R' > 0$ such that $|p(x)/x^{n}-1| < 1/2$ for $x < -R'$, and you get $p(-2R') < 0$. Now apply the Intermediate Value Theorem because nobody wants to re-invent that wheel unless absolutely necessary.
