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In an algorithms lecture in school theres a proof for asymptotic tight bound like:

Take $C=a_k/2$ and show that $f(n) \ge \frac{a_k}{2} n^k$ when $n > N$ for some $N$.

$$\begin{align} f(n) &= a_k n^k + a_{k-1} n^{k-1} + ... + a_1 n + a_0 \\ &= \frac{a_k}{2} n^k + (\frac{a_k}{2k} n^k + a_{k-1} n^{k-1})+ ... + (\frac{a_k}{2k} n^k + a_{1} n) + (\frac{a_k}{2k} n^k + a_0) \\ &= \frac{a_k}{2} n^k + (\frac{a_k}{2k} n + a_{k-1})n^{k-1} + ... + (\frac{a_k}{2k} n^{k-1} + a_{1}) n + (\frac{a_k}{2k} n^k + a_{0}) \end{align}$$

We want all terms $(\frac{a_k}{2k} n + a_{k-1}), ..., (\frac{a_k}{2k} > n^k + a_0)$ to be positive. Thus we take $N=max(−2ka_{k−1} /a_k , ..., > −2ka_1 /a_k , −2ka_0 /a_k , 1)$. For any $n>N$, all previous terms are positive. Thus $f(n) > \frac{a_k}{2} n^k$ for all $n>N$ and $f(n) \in > \Omega(n^k)$

But can anyone explain simply the purpose of this? Just in a previous slide theres a simpler proof just taking the absolute value of each coefficient and setting all powers to be the max power in the expression. Why is that insufficient?

Also I dont understand whats it trying to do actually ...

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You can't just set all powers to the max power, because one of the coefficients of a smaller power might be equal to and of opposite sign of the coefficient of the largest power.

The purpose of the result is to show that the function grows, for large enough argument, to the same order as the term with largest exponent.

This allow you to show that the function is non-zero for large arguments and, eventually, prove the fundamental theorem of algebra.

You readily show that, by similar means, if $f(x) = \sum_{k=1}^m a_k x^{c_k}$ where $a_m > 0$, and $c_k > c_{k-1}$ for all $k$, then, for any $\epsilon > 0$, $f(x) > (a_m-\epsilon)x^{c_m}$ for all large enough $x$.

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