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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

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    $\begingroup$ He wants to pin down the interval eventually by making this sum in $-1/2$ and $1/2$. $\endgroup$ – Shuhao Cao May 30 '13 at 23:05
  • $\begingroup$ So the choice of 2n is arbitrary, could he have chosen 3n etc? $\endgroup$ – ano May 30 '13 at 23:23
  • $\begingroup$ Yeah, he could have chosen $3n|a_{n-1}|,\ldots$ in (*), and the proof still holds. $\endgroup$ – Shuhao Cao May 30 '13 at 23:27
  • $\begingroup$ It makes sense, thanks $\endgroup$ – ano May 30 '13 at 23:31
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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 $|x^k|\geq|x|$.

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