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I am having this equation: $$ \frac{1}{x-a_1} + \frac{1}{x-a_2} + \cdots + \frac{1}{x-a_n}=0 $$ where $a_1 < a_2 < \cdots < a_n$ are real numbers.

Now I want to prove with the intermediate value theorem that this equation has $n-1$ solutions in the real numbers.

My thoughts:

With $a_1 < a_2 < \cdots< a_n $, you can see that every summand gets smaller than the summand before.

My other thought was that about the $n$-summands, with the intermediate value theorem you know that every zero (point) is in the interval and is located between the $n$-summands. So there are $n-1$ solutions for this equation!

Questions:

How can I prove my thoughts in a formal correct way? (Are my thoughts generally correct?)

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  • $\begingroup$ The first idea seems irrelevant (and false). The second is not clear at all. If it is indeed the case that every zero is "between the $n$-summands", that is something you need to prove carefully, it is not enough to simply say it. By the way, "between the $n$-summand" is too imprecise to be useful. Use inequalities. Write explicit inequalities indicating exactly what you mean. Once you do that, go further and prove that the inequalities must be satisfied by appropriate numbers. $\endgroup$
    – Andrés E. Caicedo
    Commented Dec 30, 2014 at 16:01
  • $\begingroup$ @AndresCaicedo Why is my first idea wrong? Second idea: You mean that I should pick up two summands or what did you mean? $\endgroup$
    – basti12354
    Commented Dec 30, 2014 at 16:11
  • $\begingroup$ @user3433232 The first idea is not correct because the denominator doesn't get bigger just because one of the terms in it gets bigger. In particular, in this problem each term blows up somewhere, which is actually the key to properly solving the problem. I don't even know what you meant by the second idea. $\endgroup$
    – Ian
    Commented Dec 30, 2014 at 16:23

1 Answer 1

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Let's call your function $f$. We'll look at it moving left to right along the line. When $x<a_1$, $f(x)<0$, because all the summands are negative. So there are no roots there.

Now let's look at $(a_1,a_2)$. In particular let's look at $x \to a_1^+$. All but the first term is negative, but the first term is going to $+\infty$ while the others are remaining finite. So $\lim_{x \to a_1^+} f(x) = +\infty$. In particular, $f$ is positive somewhere on $(a_1,a_2)$. Similarly we get that $f$ is negative somewhere on $(a_1,a_2)$ by considering $x \to a_2^-$.

Can you take it from here?

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  • $\begingroup$ Great explanation, but I don't understand how I can use the intermediate value theorem with your beginning. $\endgroup$
    – basti12354
    Commented Dec 30, 2014 at 16:37
  • $\begingroup$ @user3433232 On $(a_1,a_2)$, approaching $a_1$ gives us $x_1$ with $f(x_1)>0$. Approaching $a_2$ gives us $x_2$ with $f(x_2)<0$. Then the intermediate value theorem gives us $x_3 \in (x_1,x_2)$ with $f(x_3)=0$. $\endgroup$
    – Ian
    Commented Dec 30, 2014 at 16:39
  • $\begingroup$ Some visualization may help. A familiar function which is very much like your function is $\cot(x)$. $\cot$ has vertical asymptotes at each multiple of $\pi$ and blows up to $+\infty$ on one side of each asymptote and $-\infty$ on the other. Your function does the same, although it is not periodic and only does this finitely many times. Do you see why visually this means that there must be a root? $\endgroup$
    – Ian
    Commented Dec 30, 2014 at 16:43
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    $\begingroup$ To show there cannot be more than one solution between consecutive $a_i$, we can note the derivative is negative where defined. $\endgroup$
    – André Nicolas
    Commented Dec 30, 2014 at 17:04

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