I'm learning a MIT Session about computation complexity
At about 6:26 into the video linked above, the speaker says "basically, we look over here at the polynomial function." The link starts at 6:21.
A polynomial function is usually in this form
$f(x)= a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{2}x^{2}+a_{1}x+a_{0}$
where n is a non-negative integer and a0, a1, a2, ..., an are constant coefficients.
When considering $$f(n) = n \log n$$ a polynomial function, there is only one term which has a coefficient of 1. Is my understanding correct?
I know how to count the degree of a regular polynomial function. For example, the degree of the polynomial function $f(x) = 5x^{2} + x$ is 2.
What's the degree of the polynomial function $f(n) = n \log n$?
A few seconds later, the speaker says "… the log factor is smaller than a polynomial factor." I guess "the log factor" refers to $\log n$ though I can't even make a guess about what "polynomial factor" refers to.
I've watched few minutes of that video, and the explanation here is indeed a bit messy.
Lets make things clear first: $n\log n$ is not a polynomial function.
But it does have polynomial factor and logarithmic factor. The lecturer wants to explain why it grows slower than $n^2\log n^{2019}$ and $n^3$. He starts with "if we factor $n$ out of that then we're comparing $\log$ with a, you know..." and he seems to be lost for a second (which is a pity, I think that was a correct start). Then he restarts with the "basically, we look over here at the polynomial function". But I think he was refering to the polynomial factor of $f_5$ here, not entire $f_5$ (his hand is even directly over $n$). Because then he says: "this one" (I assume he refers to $n$) "is the smallest out of them" (I think he refers to polynomial pieces of other functions) and finally he says: "and the $\log$ factor grows slower than linear" making $n\log n$ smaller than $n^2\log n^{2019}$ and $n^3$.
I don't see any other logical explanation of his words.