Let $$f(x)=\sum\limits_{i=1}^{2020}\frac{i^2}{x-i}$$. Then, what is the number of solutions of $f(x)=0$?
I am struck at this problem. Specifically, would not $f(x)$ become undefined when $i=x$. Since the numerator is always positive, therefore the sum becomes zero only when there are points when $x>i $ and $x<i$. Then, would not a point when $x=i$ appear? Any hints? Thanks beforehand.
Multiplying both sides by $(x-1)\cdot\ldots\cdot(x-2020)$ and denoting $g(x)$, what you get is:
$$g(x) = \sum_{i=0}^{2020} g_i(x)$$
where $g_i(x)$ is a polynomial of degree $2019$ such that $g_i(n)=0$ for all $n\neq i$ and $g_i(i) = i!(2020-i)! \cdot (-1)^i i$. Hence, you have that $g(i)$ and $g(i+1)$ alternate in sign. This says by continuity of $g$ that there is a root in every interval $(i,i+1)$ for $i=1,\ldots,2019$. Thus, you have $2019$ roots, and $g(x)$ is a polynomial of degree $2019$. So you have all of them. Moreover, these are all the roots of your $f$, since we only introduced a numerator that is zero only in integers and our roots lie in open intervals $(i,i+1)$.