Super determinant

It can be proved that $d(x)$ is a constant function, such that its constant value is always different from 0. Therefore the number of real zeros is zero.

To prove that $d(x)$ is a constant function, we can prove a more general result. We can prove that for any set of polynomials $f_1, f_2, ...f_n$ , such that the degree of $f_k$ is equal to $k-1$ , for all $k$ in the set $\{1,2, ..., n\},$ the determinant $D(x)$ of the matrix $(f_i(x+j-1))_{1\leq i \leq n,\:1\leq j \leq n}$ is always constant. To prove it, we just have to prove that $D'(x)=0$ . This can be done in the following way. First, we need to notice that the derivative of $D(x)$ will be a sum of $n$ determinants, where the determinant number $k$ in the sum is the determinant of the matrix that is obtained replacing in the original matrix the $k$ th- row by $f'_k(x), f'_k(x+1), ...f'_k(x+n-1).$ Since the degree of $f'_k(x)$ is $k-2,$ and $\{f_1, f_2, f_3,..., f_{k-1}\}$ is a base of the vector space of all polynomials of degree $k-2,$ then there are certain real numbers $c_1, c_2, ..., c_{k-1},$ such that $f'_k(x)=\sum_{i=1}^{k-1}{c_i f_i(x)}.$ Then $f'_k(x+j-1)=\sum_{i=1}^{k-1}{c_i f_i(x+j-1)},$ for all $j$ such that $1\leq j \leq n$ . This implies that the $k$ -th row of this determinant is a linear combination of the previous $k-1$ rows and that is why that determinant is zero. Since this can be proved for any of the determinant of the sum, then $D'(x)=0$ for all $x.$ In a similar way, it can be proved that $d(x)$ is constant, and you can find its value evaluating at a particular value of $x$ . For example, evaluating at $x=0$ . When we do this, we obtain a triangular determinant whose entries below the main diagonal are all zeros and whose entries corresponding to the main diagonal are $0!, 1!, 2!, ..., (n-1)!$ , where $n=2018.$ So, the constant value of $d(x)$ is $0!* 1!* 2!* ...* (2017)!,$ which is different from zero.