Prove that a polynomial of odd degree has a root.

Question

I don't understand the yellow highlighted parts in this proof.

First, is $g(n) = |\frac {b_{d-1}n^{d-1}+...+b_1n+b_0}{n^d}|$ ?? why is it divided by $n^d$, and why does the term, $x^d \rightarrow n^d$, on the numerator disappear??

On the second part highlighted with yellow, why does it have negative sign ??

Thank you in advance!!

Answer 1

First, is $g(n) = |\frac {b_{d-1}n^{d-1}+...+b_1n+b_0}{n^d}|$ ??

No, $\,g(n)=n^d+b_{d-1}n^{d-1}+\ldots+b_1n+b_0\,$ per the line just above.

why is it divided by $n^d$

Because the sign (which is what we/they are interested in at that point) does not change if dividing by a positive quantity, such as $\,n^d\,$ for $\,n \in \mathbb{N}\,$.

and why does the term, $x^d \rightarrow n^d$, on the numerator disappear??

It doesn't "disappear" but note that $\,n^k \le n^{d-1}\,$ for $\,k \le d-1\,$, so:

$$ |b_{d-1}|n^{d-1}+|b_{d-2}|n^{d-2}+\ldots+|b_1|n+|b_0| \\ \le |b_{d-1}|n^{d-1}+|b_{d-2}|n^{d-1}+\ldots+|b_1|n^{d-1}+|b_0|n^{d-1} \\ = n^{d-1}\big(|b_{d-1}|+|b_{d-2}|+\ldots+|b_1|+|b_0|\big) $$

On the second part highlighted with yellow, why does it have negative sign ??

Because $\displaystyle \left|\frac{u}{v}\right| \lt 1 \iff -1 \lt \frac{u}{v} \lt 1$ implies $\,-u \lt v\,$ when $\,v \gt 0\,$, and in the context that's what's needed to conclude the proof..

Answer 2

$$g(n) \neq \left|\frac {b_{d-1}n^{d-1}+\ldots+b_1n+b_0}{n^d}\right|$$

The purpose of $\left|\frac {b_{d-1}n^{d-1}+\ldots +b_1n+b_0}{n^d}\right|$ is to compare the magnitude of the lower order term with the dominant term and show that as $n \to \infty$, the lower order term is insignificant.

From $\left|\frac {b_{d-1}M^{d-1}+\ldots+b_1M+b_0}{M^d}\right|< 1$

we have $|b_{d-1}M^{d-1}+...+b_1M+b_0|< M^d$ which implies that

$$-(b_{d-1}M^{d-1}+...+b_1M+b_0)< M^d$$

The motivation to write it in that way (that is put the negative in front ) is so that we have

$$0 < M^d + b_{d-1}M^{d-1}+...+b_1M+b_0=g(M)$$

Answer 3

Assume the polynomial $p(x)$ is of odd degree, and WOLOG that the leading coefficient is positive. As $x$ goes to $\infty$ $p(x)$ goes to $\infty$, and as $x$ goes to $-\infty$ $p(x)$ goes to $-\infty$. Because $p(x)$ is continuous there must be some $x_0$ such that $p(x_0)=0$.