## Why Do Mathematicians Do Something Different?

In case my salty ranting on the previous page couldn’t clue you in, I will now take the time to explain what the notation of $\mathcal{O}, \Omega, \Theta$ truly mean in mathematics. Remember that for the purposes of not confusing y’all, I will be denoting the true versions as $\mathcal{O}^T, \Omega^T, \Theta^T$.

The original case was for purposes of analytic number theory, however I think its useful to show it in terms of an example from approximation analysis. Remember back in physics class when your professor said

for small enough values of $\theta$, you can use $\sin\theta\approx\theta$?

Well lets see how good of an approximation this is we can provide the Taylor Expansion of $\sin\theta$ (don’t worry if you don’t know where this comes from, just take it on faith here) to show that the error of the approximation for any value of $\theta$ is $$ \begin{align} \text{Error}(\theta)&=\theta - \sin\theta \\ &= \theta - \sum_{n=0}^\infty\frac{(-1)^{n+1}(\theta)^{2n+1}}{(2n+1)!}\\ &= \theta - \left[\theta - \frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}+\ldots\right] \\ &= \frac{\theta^3}{3!}-\frac{\theta^5}{5!}+\frac{\theta^7}{7!}-\ldots \end{align} $$

Great, now we have an expression for our error! Plugging this expression into a calculator though would take an infinite amount of time though, so we really just want to get a good sense for how this error changes for various values of $\theta$. If $\theta<1$, then we can see that raising $\theta$ to higher powers actually makes it smaller, so the biggest term in our error is actually the first one! In fact, we can show that as we make $\theta$ get closer and closer to $0$, our first term of $\frac{x^3}{3!}$ will dominate our expression more and more. We would notate this as $$ \sin\theta = \theta + \mathcal{O}^T(\theta^3). $$ In this we are saying that for our limit that we are looking at, that our function effectively looks like $\theta$ with some extra term that is kinda like $\theta^3$.

That was probably really confusing so lets actually get to the definition.

### The Biggest of True Os

The idea of Big $O$ is that we want to be able to answer the question *“I have some function $f(n)$, will it eventually be smaller than this other function $M\cdot g(n)$ where $M$ is a constant?”*. The reason why we are asking is because we want to be able to provide ourselves with some worst case rate for how we are growing over time so that we can say “ok yeah this is good enough”.

This is actually going to be best explained using examples and the definition, so I will do that so we can move into it.

Definition: We say a function $f(n)\in\mathcal{O}^T(g(n))$, also notated $f(n)=\mathcal{O}^T(g(n))$ iff there exists an $N>0$ and $M>0$ such that $\forall n > N$, $f(n)\leq M\cdot g(n)$.

What this definition is saying is that if we multiply our function $g(n)$ by some constant of **our choice** $M$, we can find a value of $N$ such that for any value of $n$ that we choose past that, our function times $M$ will be larger than $f(n)$.

Let’s first show an example where we don’t need $M$ so we can get a good idea of whats going on here. In later examples, choices of $M$ will actually make proofs easier though.

Example: Prove $\frac{1}{2}n^2 + 5n + 10\in\mathcal{O}^T(n^2)$.

## Proof

We can directly find our value of $N$ by solving the inequality $$\begin{align} \frac{1}{2}n^2 + 5n + 10 &\leq n^2 \\ 0 &\leq \frac{1}{2}n^2 -5n -10 \end{align}$$ We can solve this using the quadratic formula to find our positive root is $N=5+3\sqrt{5}\approx 11.71$. As such, so long as $n\geq 12$, we can see that $n^2$ will be bigger than $\frac{1}{2}n^2 + 5n + 10$, which means that our claim is true!QED

Quite obviously, this doesn’t really give a super good approximation of any meaningful rate of error, but it shows us that if $n^2$ is considered “good enough” for us, well then the full expression will too, and we don’t need to bother our efforts working on it.

But then why do we need to have the factor of $M$ be considered at all? Can’t we just always do this? Well not always, consider the example below.

Example: Prove $2n\in\mathcal{O}^T(n)$

Well these two functions grow at the same rate, so it would be nice if we could do the same thing as before and handwave to say “yeah its just $\mathcal{O}^T(n)$ no worries”, but in this case $2n\not\leq n$ for any value of $n>0$, so we need to use our value of $M$. If we set $M=3$ for example, then we can say that $2n\leq 3n$ for every $n>0$, which would prove us true. Note though, that our choice of $M$ was not unique, and we could have made infinite selections. In fact, sometimes a choice of $M$ can depend on your choice of $N$!

Example: Prove $2n^2+2n-10\in\mathcal{O}^T(n^2)$

## Proof

In order to prove this we cannot do the same method before, as $2n^2+2n-10\geq n^2$ for large values of $n$. Instead we can apply a super clever trick. First lets factor out $n^2$ from our $2n^2+2n-10$ to get $$n^2\left(2+\frac{2}{n}-\frac{10}{n^2}\right).$$ Now something interesting to notice is that $\frac{2}{n}\leq 2$ if $n\geq 1$ and $-\frac{10}{n^2} < 0$ since its negative. As such we can say that if we choose $N=1$, for any value of $n\geq 1$ we can see $$2+\frac{2}{n}-\frac{10}{n^2} < 2+2=4.$$ Because of this we can say that if $n\geq 1$ then $$2n^2+2n-10 \leq 4n^2$$ which proves our claim.QED

Now notice though, if we didn’t really care about the smaller values of $n$, we could see that if $n\geq 1000$ then we would have that $\frac{2}{n}\leq .002$, and by extension
$$
2n^2+2n-10\leq 2.002n^2
$$
for values of $n\geq 1000$. These values of $M$ are the *constant factors*, and we can see that if we get larger and larger values of $N$ selected, then our values get closer and closer to $2$.

### The Biggest of True $\Omega$s and $\Theta$s

The true definition of Big $\Omega^T$ is actually exactly the same as $\mathcal{O}^T$, except now we use $\geq$ instead of $\leq$.

Definition: We say a function $f(n)\in\Omega^T(g(n))$, also notated $f(n)=\Omega^T(g(n))$ iff there exists an $N>0$ and $M>0$ such that $\forall n > N$, $f(n)\geq M\cdot g(n)$.

This is saying that, if we find a large enough value of $N$, eventually $g(n)$ will be smaller than $f(n)$. Since we did previous examples with $\mathcal{O}^T$ already, and its super late and I need to shower and I have work tomorrow, I think Imma just leave it here for y’all to try your own examples for now.

Definition: A function $f(n)\in\Theta^T(g(n))$ iff $f(n)\in\mathcal{O}^T(g(n))$ and $f(n)\in\Omega^T(g(n))$.

So, same definition as before, just using the true versions instead.

## How the True Version Relates to our Version

It turns out that we can relate our true and practical versions of notation through some theorems, so you can tailor how you are talking to different people, and actually understand how they are related to each other.

In this section we will be comparing the $mathcal{O}$ and $\Omega$ of the operations function $T(x)$, however, from the perspective of theoretical sense, we do not need to do this. You can define the $\mathcal{O}$ and $\Omega$ of whatever you want. For example, you can consider the worst case function $W(n)$ and then say that $W(n)\in\Omega^T(f(n))$ which means the lower bound of the worst case is $f(n)$. Similarly, you will often see the best case denoted as $\mathcal{O}^T(f(n))$

I personally dislike this, as I feel it muddies the waters, however you can think of it like this.

### Relating the Practical to the Mathematical

Theorem: Let $T(x)$ be an operations function with worst case $W(n)$ and best case $B(n)$. Then

- If $T(x)\in\mathcal{O}(g(n))$ then $T(x)\in\mathcal{O}^T(g(n))$
- If $T(x)\in\Omega(g(n))$ then $T(n)\in\Omega^T(g(n))$
- If $T(x)\in\Theta(g(n))$ then $T(n)\in\Theta^T(g(n))$

## Proof

Let us first prove claim $1$. For this we will be using the technical definition of a limit from analysis (so you may not know this if you haven't taken that course). Since we know $T(x)\in\mathcal{O}(g(n))$, we know that $$ \lim_{n\rightarrow\infty}\frac{W(n)}{g(n)}=L $$ where $0< L < \infty$. In the technical definition we can say that for every value of $\varepsilon > 0$, we can find an $N$ such that if $n > N$ then $$ \lvert\frac{W(n)}{g(n)}-L\rvert < \varepsilon. $$ We only need for our purposes that $$ \frac{W(n)}{g(n)} < \varepsilon + L \implies W(n) < (\varepsilon+L)g(n). $$ Now if we choose some particular value of $\varepsilon$ and get the assosiated $N$, since $T(n)\leq W(n)$ we know that if $M=(\varepsilon+L)$ and $n>N$ that $$ T(n) < Mg(n) \implies T(n)\in\mathcal{O}^T(g(n)). $$For claim $2$ assume that $T(n)\in\Omega(g(n))$, then we can say that for every $\varepsilon$ there is a value of $N$ such that if $n>N$ then $$ \lvert\frac{B(n)}{g(n)}-L\rvert < \varepsilon $$ with $L$ being the limit. We can then say that $$ L-\varepsilon < \frac{B(n)}{g(n)} $$ and then perform the same steps as in case $1$ to prove our claim. This shows that $T(n)\in\Omega^T(g(n))$.

Claim $3$ then follows from claims $1,2$.

QED

### Relating the Mathematical to the Practical.

The forward direction that we just showed is actually quite easy because when we already know that the limit is defined, it makes things very straightforward to work with. The technical definition is much less restrictive, so we will need to be very mindful of different scenarios that we can encounter. I will prove each claim independently

Theorem: If $T(n)\in\mathcal{O}^T(g(n))$ then $T(n)\in\mathcal{O}(g(n))$ or $T(n)\in o(g(n))$.

## Practice Problems

Problem: Prove that $\mathcal{O}^T$ isnotan equivalence relation. This is the biggest difference in the two versions!