Number of real solutions for $f(x)=0$

Question

How many real solutions does the equation $f (x) = 0$ have, where $f (x)$ is defined as follows?

$$f(x)=\sum_{i=1}^{2020}\dfrac{i^2}{x-i}$$

My try:

I observed $f(x)$ is undefined for $x=1,2,\cdots,2020$ and that the deg$(f)=2019$ so that it can have atmost $2019$ and atleast one real root.

I tried rewriting $f(x)=0 $ as $\displaystyle\sum_{i=1}^{2020}\dfrac{i^2}{x-i}=\sum_{i=1}^{2020}\left(i^2\prod_{j=1\\j\neq i}^{2020}(x-j)\right)=0$.

But I wasn't able to take it anywhere from this.

I tried to show continuity of the function in the intervals excluding the above endpoints. I showed the following

for $x,y\in(1,2)$ and $ y>x$ $$|f(x)-f(y)|\leq\dfrac{|y-x|}{|x-1|^2}\sum_{i=1}^{2020}i^2<\varepsilon$$ If $|y-x|<M|x-1|^2$ where $M=\displaystyle\sum_{i=1}^{2020}i^2$, then $f(x)$ is continuous in $(1,2)$. Similar proof can be shown for intervals $(j,j+1)$ for $j=2,3,\cdots,2019$.

By intermediate value theorem, we see that $f$ has a root in each of the above intervals and hence $f$ has $2019$ real roots.

I also found this link Prove that all roots of $\sum_{r=1}^{70} \frac{1}{x-r} =\frac{5}{4} $ are real

Are there any other simpler ways of estimating the number of real roots? Any hints would be appreciated.

Answer 1

For each $k=1,2,\dots,2019$, we have $$\lim_{x\to k^+}f(x)=+\infty\\ \lim_{x\to(k+1)^{-}}f(x)=-\infty$$ Since $f$ is continuous in $(k,k+1)$, we have a real root in each of those intervals. On the other hand, as you noted, there can be no more than $2019$ roots.

Answer 2

$\sum_{i=2}^{2020}{\frac{i^{2}}{x-i}}=\frac{\sum_{i=1}^{2020}{\ \ i^{2}\prod_{j\neq i}{(x-j)}}}{\prod_{i=1}^{2020}{(x-i)}}$

Since the numerator is a $2019$ - degree polynom, $f(x)$ has AT MOST $2019$ real roots.

We know $f(x)$ is continuous everywhere except at $x=i$ for integer $i$, $1\leq i \leq 2019$. Furthermore, for integer $i$, $1\leq i \leq2019$, $\lim_{x\to i^{+}}f(x)=\infty$ and $\lim_{x\to i^{-}}f(x)=-\infty$ and therefore, $f(x)$ has AT LEAST one real roots in each of $(i , i+1)$, totaling AT LEAST $2019$ real roots.

The conclusion: $f(x)$ has exactly $2019$ real roots.