Can someone please help me understand and break down this estimation lemma for polynomials. I don't understand what the conclusion is saying and what it means. I'm really confused with the logical steps used from (11) onwards. I don't understand how the writer has concluded each step from the previous. I've used $p(x)=5x^3+2x^2+7x+1$ as a random example but it's not illuminating the statement and proof for me.
What does it mean to estimate polynomials for large values of $x$?
Where does $|x|^r ≤ x^{n−1}$ for $r = 0,1,...,n−1$ and the logical equivalence (13) come from?
Allow me try to explain the proof with you example.
What we want to do is to make an estimate of $p(x)$ when $|x|$ is large, where $p(x)$ is a polynomial. We may notice that the first term $a_n x^n$ grows faster that all of the rest of terms, so it is reasonable to conjecture that when $x$ is really large, the $p(x)$ is relatively close to $a_n x^n$. For example, at this time $\frac 1 2 a_n x^n \le p(x) \le \frac 3 2 a_n x^n$, or we can write equivalently $\left|\frac{p(x) - a_n x^n}{a_n x^n} \right|\le \frac 12$. So how large is enough? Like all of the other proofs in mathematical analysis, we cannot point out an actual number such as $100$, $100000$, or something else like that. For example, $100$ may not be good enough for $p(x) = x+101$. Whatever number you provide, there is always a polynomial such that your number is not good enough. Instead, for all polynomial $p(x)$, the inequality holds when $x$ is sufficiently large means that there exist a boundary (which varies with $p(x)$), such that when $x$ is even larger than that, we may conclude that the inequality holds.
To prove this, we would like to find a positive number $N$ such that $x>N$ implies the previous inequality holds.
Firstly, let’s observe the things that we would like to compare:
$$\begin{align} | p(x) - a_n x^n | &= |a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 | \\ |a_n x^n | &= |a_{n} x^{n} + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 | \\ \end{align} $$
When the first one is smaller than a half of the second one? Remember that what we want to do is just try to find a such $N$, and it doesn’t matter how large it is. So we can estimate them a little bit roughly. Use the triangular inequality, and we have:
$$\begin{align} | p(x) - a_n x^n | &\le |a_{n-1}| |x|^{n-1} + \ldots + |a_1| x + |a_0 | \\ &\le |a_{n-1}| |x|^{n-1} + \ldots + |a_1| x^{n-1} + |a_0 |x^{n-1}\\ &= (|a_n|+\ldots+|a_1|+|a_0|)|x|^{n-1} \\ \end{align} $$
The second inequality holds when we choose $N>0$, which implies $x\le1$. It is reasonable to do that because we want to make the whole thing simpler, so that we may reach the conclusion at some point. Following your book, donate $K=(|a_n|+\ldots+|a_1|+|a_0|) $. Note that $K$ does not rely on $x$, so $N$ could have some to do with the number $N$. By now, we have
$$ \begin{align} |p(x)|&=|p(x)-a_n x^n + a_n x^n| \\ &\ge |a_n x^n|-|p(x)-a_n x^n| \\ &= |a_n| |x|^n-|p(x)-a_n x^n| \\ &\ge |a_n| |x|^n-Kx^{n-1} \end{align} $$
Here again the triangular inequality is applied. Review the inequality again:
$$ \left|\frac{p(x) - a_n x^n}{a_n x^n} \right| \le \frac {K|x|^{n-1}} {|a_n| |x|^n-K|x|^{n-1}} = \frac K {a_n x - K} $$
To make LHS < 1/2, $x>\frac {2K} {a_n}$ is sufficient. Thus the $N$ we have been looking for is $\max\{1, \frac {2K} {a_n}\}$.
In your example, let $K=2+7+1$ and $N=2\cdot(2+7+1)/5=4$. $p(4) = 381$, and $5\cdot 4^3=320$. It is greater than $\frac {381} 2$.