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For $c \in \mathbb{R}$, the equation $\dfrac{1}{x-a_1} + \dfrac{1}{x-a_2} + \ldots + \dfrac{1}{x-a_n} = c$ has $n-1$ real solutions

The follow up to this is that prove that the equation has $n$ solutions if $c \neq 0$. Here $a_1 < a_2 < \ldots < a_n$ and $c \in \mathbb{R}$. How does this equation even have solutions, I'm really confused with this!

Winther
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onknc
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  • Try to plot it for say $n=3$ to see what is going on: https://www.wolframalpha.com/input/?i=plot+1%2Fx+%2B+1%2F(x-1)+%2B+1%2F(x-2)+for+-2+%3C+x+%3C+4 . When $x\to a_i$ from "the left" the term $\frac{1}{x-a_i}$ goes to $-\infty$ where when we approach if from "the right" it goes to $+\infty$. Since it's continuous on $a_i < x < a_{i+1}$ the intermediate value theorem gives you exisitence of a zero. – Winther Oct 19 '16 at 17:24
  • This is a similar question you could check out. – StubbornAtom Oct 19 '16 at 17:36
  • @JackD'Aurizio, sorry, I need more coffee.... – Paul Oct 19 '16 at 19:00

2 Answers2

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The function $$ f(x)=\sum_{k=1}^{n}\frac{1}{x-a_k} $$ is a continuous and differentiable function over its domain, $\mathbb{R}\setminus\{a_1,\ldots,a_n\}$. Since $$ f'(x)=-\sum_{k=1}^{n}\frac{1}{(x-a_k)^2} \leq 0 $$ $f(x)$ is a decreasing function on the connected components of the domain $I_0=(-\infty,a_1), I_1=(a_1,a_2),\ldots,I_n=(a_n,+\infty)$. Since for any $k\in[1,n]$ we have $$ \lim_{x\to a_k^+}f(x)=+\infty,\qquad \lim_{x\to a_k^-}f(x)=-\infty $$ and $$\lim_{x\to \pm\infty}f(x)=0,$$ by the mean value property of continuous functions we have that there is exactly a solution of $f(x)=c$ in $I_1,I_2,\ldots,I_{n-1}$ and, assuming $c\neq 0$, a solution of $f(x)=c$ either in $I_0$ or in $I_n$ according to the sign of $c$.

Jack D'Aurizio
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Hint let $f 1(x)=\frac {1}{(x-a_1) }$ differentiating we have $\frac {-1}{(x-a_1)^2} $ thus function is decreasing hence $f1 (x) $ cuts x axis at least one time then the given function is summation of all such functions thus given function has atleast n solutions .

Archis Welankar
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