I have a question about this Big $\mathcal{O}$ problem. I have the question down $90\%$, but the other $10\%$ isn't getting to me. I will write out the entire question and I'll point out the step, which I am confused about. This is an exercise in the book too so it might seem easy but nevertheless, I am slow at math...
Q: Let $f$, $g$ be mappings from positive integers to real numbers, given by $$f(n) = 5n\,,\quad g(n) = n^2$$ for all positive integers, $n$. If we compute $f(n)$ and $g(n)$ for numbers $1$, $2$, $3$, $4$ we find that $$f(n)>g(n)$$
However, with any number greater than or equal to $5$ we find that $$g(n)≥f(n)$$
So, with $m = 1$ and $k= 5$, we find that:
$$ n≥k\,,\quad|f(n)|≤ m|g(n)|$$
Can someone explain to my why $m = 1$ because, according to my textbook, $m$ is supposed to bound $f$ but I don't understand how $1$ could possibly be the upper bound of $f(n)$.
$m$ is not an upper bound on $f(n)$, but $m \cdot g(n)$ is. By the definition of Big-Oh notation, if we want to say $f(n) = O(g(n))$, then we need to show there exists some $m$ such that $m \cdot |g(n)|$ is an upper bound on $|f(n)|$ for all $n \geq k$.
In the given example, $m=1$ and $k=5$ lead to a perfectly valid proof, but they're not the only possible combination of values. For example, if we choose $m=6$ and $k=1$, we see that for all $n \geq k$, $|f(n)| \leq m|g(n)|$.
That is, for all $n\geq 1$, $5n \leq 6n^2$.