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I want to prove that for every positive polynomial $p(n)$ and any negligible function $negl(n)$ the product $negl(n)p(n)$ is also negligible.

This is what I tried so far:

$\forall n>n_0:negl(n)<1/k(n),k=$any positive polynomial. $\Rightarrow p(n)negl(n)<p(n)1/z(n) $ With $z(n)=p(n)k(n)$ $\Rightarrow p(n)negl(n)<p(n)1/z(n)=p(n)\frac{1}{p(n)k(n)}=1/k(n)$

Question : I am not sure if I am allowed to substitute $z(n)$ with the product of $p(n)$ and $k(n)$.

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  • $\begingroup$ What is "negligible" in your context? $\endgroup$
    – nir shahar
    Nov 14 '21 at 16:35
  • $\begingroup$ A function $f(n)$ is negligible if for any $n>n_0 \in N$ $f(n)$ is smaller then $1/p(n)$ for any positive polynomial $p(n)$. $\endgroup$
    – UniX
    Nov 14 '21 at 16:39
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You are allowed, but you have to make sure to keep it formal.

The property of $negl$ guarantees you that $\exists n_0 : \forall n>n_0: negl(n)<\frac{1}{p(n)k(n)}$ for some polynomial $k(n)$.

Now, for any $n>n_0$, you know that this holds and thus $p(n)negl(n)<p(n)\frac{1}{p(n)k(n)}=\frac{1}{k(n)}$ if $p(n)$ is positive (it will be for all $n>n_1$ for some $n_1$).

Thus, for all $n>\max\{n_0,n_1\}$, we know that $p(n)negl(n)<\frac{1}{k(n)}$ and by definition we get that $p(n)negl(n)$ is also negligible.

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We have $negl(n)$ as a negligible function, then by definition, it is smaller than the inverse of any polynomial, for all sufficiently large $n$. In particular, given any polynomial $q(n)$, $negl(n)$ is smaller than $1/(p(n) \cdot q(n))$

$$negl(n) <1/(p(n) \cdot q(n)) $$ Now, we have

$$negl(n)p(n) < q(n).$$


Limit definition also easy to use;

$f_1(n)$ is negligible than for every polynomial $q(n)$ we have;

$$\lim_{n \rightarrow \infty} q(n) f_1(n) =0$$

Now, we need to show that for every $q(n)$, $negl(n)p(n)$ is negligible;

$$\lim_{n \rightarrow \infty} q(n) negl(n)p(n)= 0$$

This is true; since $negl(n)$ is negligible implies for every polynomial;

$$\lim_{n \rightarrow \infty} [q(n) p(n)] negl(n) =0,$$ where $q(n) p(n)$ is a polynomial.

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