1

$a_i$, $i=1,2018$ are distinct real numbers. Find number of zeroes of a function: $$f(x) = \frac{1}{x-a_1} + \frac{1}{x-a_2} + ... + \frac{1}{x-a_{2018}}$$

Problem is in the limits and continuity section ...

2 Answers2

4

Hint From

$$P(x)=\prod\limits_{i=1}^{2018}(x-a_i) \text{ and } P'(x)=\sum\limits_{i=1}^{2018}\prod\limits_{k=1,k\ne i}^{2018}\left(x-a_k\right)$$ then $$f(x)=\frac{P'(x)}{P(x)}=\sum\limits_{i=1}^{n}\frac{1}{x-a_i}$$ and the answer is in $P'(x)$ and Rolle's theorem.

Another, not as easy as the previous one, hint is to use intermediate value theorem (aka IVT). E.g. let's assume $a_1<a_2<...<a_{2018}$ and let's consider 2 adjacent $a_{i}<a_{i+1}$. It's not too difficult to see that $$\lim\limits_{x\rightarrow a_i^{+}}f(x)\rightarrow +\infty$$ $$\lim\limits_{x\rightarrow a_{i+1}^{-}}f(x)\rightarrow -\infty$$ or, there $\exists \varepsilon>0$ s.t. $f(a_i +\varepsilon)>0$ and $f(a_{i+1}-\varepsilon)<0$. Also, $f(x)$ is continuous in $[a_i +\varepsilon, a_{i+1}-\varepsilon]$. By IVT, $f(x)$ has a zero in $[a_i +\varepsilon, a_{i+1}-\varepsilon]$. It remains to show this zero is unique, between $a_i$ and $a_{i+1}$.

rtybase
  • 16,907
  • I think it's supposed to be done without derivatives or using more advanced tools –  Jan 16 '19 at 20:11
  • Are you familiar with the Rolle's theorem? – rtybase Jan 16 '19 at 20:13
  • No, I'm not. What is the point of that question ? –  Jan 16 '19 at 20:17
  • 1
    It's a fairly simple theorem. I learned it as part of my high school curriculum, many years ago though ... so I assumed you probably learned it, but (as it happens) forgot it. Never mind ... – rtybase Jan 16 '19 at 20:24
  • Yes but even if I know it, I'm interested in how to do this just using limits (doesn't make much sense to use limits here) and continuity, because problem is in that section of textbook .. –  Jan 16 '19 at 20:25
  • 1
    Since $P(x)$ has 2018 distinct zeroes, $P'(x)$ must have at least 2017 distinct zeroes (by Rolle's theorem) at points $\ne a_i$. On the other hand, as $P'(x)$ is a polynomial of degree 2017, by the fundamental theorem of algebra is HAS exactly 2017 roots which can be complex and counted with multiplicity. Taking both statements together, $P'(x)$ has exactly 2017 distinct zeroes at points $\ne a_i$. Since they are not cancelled by the zeroes of the denominator (which are at $a_i$), the zeroes of $P'(x)$ are the zeroes of $f(x)$ and this is what you want. – Andreas Jan 16 '19 at 20:29
  • 1
    @someone the way to figure this out with minimal theory is to draw a careful graph for, say, $\frac{1}{x-1} + \frac{1}{x-2},$ on graph paper by hand if you can, but in any case figuring out how to prove that what the picture indicates really does happen for any two real $a_1, a_2.$ Then the same thing for $\frac{1}{x-1} + \frac{1}{x-2}+ \frac{1}{x-3}.$ – Will Jagy Jan 16 '19 at 20:34
  • 1
    @someone I may have another approach, but I need to know if you are familiar with intermediate value theorem? – rtybase Jan 16 '19 at 20:37
  • Yes, of course. Apparently, I was not being clear enough about this problem being in continuity section of my textbook. –  Jan 16 '19 at 20:39
  • I drew a picture, posted. Scale is not the best for seeing the vertical asymptotes. – Will Jagy Jan 16 '19 at 21:08
  • That's it, thanks! –  Jan 16 '19 at 21:14
1

enter image description here

============================================

Will Jagy
  • 139,541