finite polynomials satisfy $|f(x)|\le 2^x$

Solution 1:

Solution credits to Rui Yao:

According to a property of finite differences, if we set $c_n\in \mathbb{Z}$ to be the highest coefficient of $f(x)$, which has degree $n$, then $$n!c_n=\Delta ^n [f](x)=\sum_{i=0}^{n}\binom{n}{i} f(i)(-1)^{n-i}$$ Thus $$|n!c_n|=|\sum_{i=0}^{n}\binom{n}{i} f(i)(-1)^{n-i}|\le \sum_{i=0}^{n}|\binom{n}{i} f(i)(-1)^{n-i}|\le \sum_{i=0}^{n}2^i\binom{n}{i}=3^n$$ Combined with $c_n\in \mathbb{Z}/{0}$ this leads to $n\le 6$.

Since we've proved that for any given degree $n$ there's only finite number of polynomials satisfy the condition, we can conclude that the polynomial satisfy the condition is finite.