Unpacking the definition of negligible & non-negligible

Question

There are several threads on this topic including:

How is an epsilon of 1/1000 non-negligible?

How to calculate if probability is negligible or not

(and others)

but I do not fully understand the answers in those threads.

My question is:

I would like to see what the definition of negligible & non-negligible means in an intuitive sense and unpack the definitions and use an example to see how the definitions apply.

The definition I am using is:

$\epsilon$ is a function $\epsilon$: $Z^{+}$ $\rightarrow$ $R^{+}$ and

$\epsilon$ is non-neg: $\exists$d: $\epsilon$($\lambda$) $\geq$ 1/$\lambda^{d}$ inf often ($\epsilon$ $\geq$ 1/polynomials, for many $\lambda$)

$\epsilon$ is neg: $\forall$d, $\lambda \geq \lambda_{d}$: $\epsilon$($\lambda$) $\leq$ 1/$\lambda^{d}$ inf often ($\epsilon$ $\leq$ 1/polynomials, for large $\lambda$)

Several questions:

1. What do these definitions mean in an "intuitive" sense?
2. What does $\epsilon$, $\lambda$, d, $\lambda^d$ mean?
3. How can you use these definitions to show that $\epsilon$:($\lambda$)=1/$2^{\lambda}$ is negligible and $\epsilon$($\lambda$)=1/$\lambda^{1000}$ is non-negligible?

(I know one of these examples is shown in another thread but I do not understand their solution and I want to see what each piece in the definition means in 'english' and then see how the definition is applied to each of these examples to show that they are negligible and non-negligible. I am hoping someone can provide a fresh explanation to the definitions and show how the definitions are applied to these problems and problems in general.)

Thanks!

Answer 1

As @SEJPEM mentioned in the comment: intuitively, for every positive polynomial $p(n)$ ($n\in\mathbb{N}$) that you take, consider the function $q(n)=\frac{1}{p(n)}$, then we say that the function $\epsilon(n)$ is negligible if from some point $N\in\mathbb{N}$ it follows that $\epsilon(n)<q(n)$. This means that $\epsilon(n)$ reaches 0 very fast because it doesn't matter whether you take $p_1(n)=n$ or $p_2(n)=n^{100000}$, by the definition there would be $N_1$ from which $\epsilon(n)<\frac{1}{n}$ and $N_2$ from which $\epsilon(n)<\frac{1}{n^{100000}}$.

Now, to interpret the definition that you copied: $\lambda_d\in\mathbb{N}$ refers to $N$ in my explanation above, then the definition of negligible function requires that for every $\lambda\in\mathbb{N}$ such that $\lambda>\lambda_d$ (in my explanation above: for all $n$ that is bigger than this $N$) it follows that $\epsilon(\lambda)<q_d(\lambda)=\frac{1}{p_d(\lambda)}=\lambda^d$.

In order to show that $\epsilon(\lambda)=\frac{1}{2^\lambda}$ is negligible you need to show that for every polynomial $p(\lambda)$ it follows that $\epsilon(\lambda)=\frac{1}{2^\lambda}<q(\lambda)=\frac{1}{p(\lambda)}$ from some point $N\in\mathbb{N}$ (this point could be different for different $p(\lambda)$-s. Now take some arbitrary $p(\lambda)$, we can discover this $N$ from which $\epsilon(\lambda)<\frac{1}{p(\lambda)}$ as follows: let $p(\lambda)=\lambda^k$ then we want to find $\lambda$ such that $\frac{1}{2^\lambda}<\frac{1}{\lambda^k}$, thus, we want to find $\lambda$ such that $\lambda^k<2^\lambda$, or in other words, we want to prove that from some $\lambda$ the term $2^\lambda$ is bigger than $\lambda^k$, and this can be proven using the definition of a limit (see this proof).

Now take $\epsilon(\lambda)=\frac{1}{\lambda^{1000}}$, in order to show that it is not negligible you need to show that there exists some polynomial $p(\lambda)$ that violates the definition. Take the polynomial $p(\lambda)=\lambda^{2000}$, there exist no $\lambda_{2000}$ (or $N$) from which $\epsilon(\lambda)=\frac{1}{\lambda^{1000}} < q(\lambda)=\frac{1}{p(\lambda)}=\frac{1}{\lambda^{2000}}$ (i.e. for every $\lambda\in\mathbb{N}$ that you take, it follows that $\epsilon(\lambda)>\frac{1}{p(\lambda)}$.

I hope that it helps.