Polynomial roots and beyond

Question

Ok, so we have $$f=1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)\cdots (x-k).$$ If $S$ is equal to sum of all real roots of $f$ and $T$ is equal to sum of all real roots of $f'$, than $S-T$ is equal to? I really can't find anything.

Answer 1

Let's define $$ f_n(x)=1-\sum_{k=0}^n \frac{x(1-x)\cdots(k-x)}{(k+1)!} $$ so that the $f(x)$ of interest is $f_{100}(x)$.

If we compute the first few $f_n(x)$ for $n=0,1,\ldots$, we may observe that $$ f_n(x) = \frac{(1-x)\cdots(n+1-x)}{(n+1)!}. $$ This can be proved through induction as, assuming it holds true for $f_{n-1}(x)$, we get $$ f_n(x) = f_{n-1}(x) - \frac{x(1-x)\cdots(n-x)}{(n+1)!} = \frac{(1-x)\cdots(n-x)}{(n+1)!}\cdot\bigl[(n+1) - x\bigr]. $$

Since the roots of $f_n(x)$ are $1,\ldots, n+1$, which are all real, the sum of the real roots is $S_n=1+\cdots+(n+1)=(n+1)(n+2)/2$.

As the roots of $f_n(x)$ are all real, so will the roots of $f'_n(x)$. Basically, $f'_n(x)$ will have one root in each of the intervals $(1,2)$, $\ldots$, $(n,n+1)$.

We could now compute the top degree coefficients of $f'_n(x)$ and use Vieta's formula to get the sum of the roots. However, there's another approach which is simpler.

If $u_1<\cdots<u_n$ are the roots of $f'_n(x)$, we have $u_i+u_{n+1-i}=n+2$. The reason for this is that $f_n(n+2-x)=\pm f_n(x)$. Therefore, the sum of the roots must be $T_n=n(n+2)/2$.

So, for any $n$, the roots of $f_n(x)$ and $f_{n+1}(x)$ are all real and sum to $S_n=(n+1)(n+2)/2$ and $T_n=n(n+2)/2$, which makes the difference $S_n-T_n=(n+2)/2$.

For $n=100$, this makes $S-T=51$.