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Question : Let $\{a_i\}$ be a sequence of real numbers such that $0<a_1<a_2\cdots <a_n$. Show that the equation : $$\frac{a_1}{a_1−x}+\cdots+\frac{a_n}{a_n−x}=2015$$ has exactly $n$ real solutions.

My try:

I know that this is an nth degree polynomial. But I really have no idea how to show the required.

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    $\begingroup$ There are vertical asymptotes at $x=a_i$. Consider the derivative of the function : it is always positive. Then,$???$ $\endgroup$ – Claude Leibovici Apr 4 '17 at 11:33
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    $\begingroup$ It does not look like a polynomial to me $\endgroup$ – uniquesolution Apr 4 '17 at 11:56
  • $\begingroup$ It might help if you sketch a graph for $1/(a_1-x)$, then for $1/(a_1-x) + 1/(a_2-x)$. $\endgroup$ – CiaPan Apr 4 '17 at 14:09
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Hint: the LHS is continuous in each $(a_i,a_{i+1})$ interval. Note that for a small enough $\varepsilon$, $\dfrac{a_i}{a_i-x} \ll 0$ for an $x\in (a_i,a_i+\varepsilon)$ and $\dfrac{a_{i+1}}{a_{i+1}-x} \gg 0$ for an $x\in (a_{i+1}-\varepsilon,a_{i+1})$.

Hint2: The fact that the RHS is nonzero and the fact that the LHS gets close to $0$ for $x \ll a_1$ or $x\gg a_n$ should yield the last real solution.

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    $\begingroup$ Thanks! So to conclude there is an $x \in [a_i + \epsilon, a_{i+1}-\epsilon]$ which "crosses the x-axis" i.e., a root. Is this enough for a "complete" proof? And I guess there's no significance to the number $2015$, right? $\endgroup$ – Random-generator Apr 4 '17 at 11:58
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    $\begingroup$ Indeed. The fact that is "crosses the $x$-axis" is just Bolzano's Theorem. You might want to fill in the details, but in essence it is a complete proof. $\endgroup$ – Darth Geek Apr 4 '17 at 12:08
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    $\begingroup$ Sorry, it was a typo. It was ment to be LHS (left hand side). $\endgroup$ – Darth Geek Apr 4 '17 at 12:23
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    $\begingroup$ @Lohith-kumar Check my edit. The original solution only yields $n-1$ solutions. There is some significance to $2015$ after all. $\endgroup$ – Darth Geek Apr 4 '17 at 13:50
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    $\begingroup$ @DarthGeek there's no significance to the $2015$ whatsoever. Check my answer to see why. Any positive works, and it can even be tweaked for negatives $\endgroup$ – vrugtehagel Apr 4 '17 at 13:54
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We see that the original equation is equal to

$$2015\prod_{i=1}^n(a_i-x)$$

Now define

$$P(x)=\left(\sum_{k=1}^n\frac{a_k}{a_k-x}\prod_{i=1}^n(a_i-x)\right)-2015\prod_{i=1}^n(a_i-x)$$

We need to show that $P$ has exactly $n$ real roots. We can compute $P(a_l)$ to see that

$$P(a_l)=\sum_{k=1}^n\frac{a_k}{a_k-a_l}\prod_{i=1}^n(a_i-a_l)$$

which is

$$P(a_l)=a_l\prod_{1\leq i\leq n, i\neq l}(a_i-a_l)$$

(note that $P(a_l)\neq 0$. This is important, because if $a_l$ would've been a root, it wouldn't have been a solution to the original equation). Since $a_i<a_l$ iff $i<l$, we know all the terms $(a_i-a_l)$ with $i<l$ are negative, and all other factors are positive (also the $a_l$), so the sign of this is $(-1)^{l-1}$. This means $P$ is wobbling up and down, crossing the x-axis between every $a_i$ and $a_{i+1}$. This gives us $n-1$ roots (one between $a_1$ and $a_2$, one between $a_2$ and $a_3$, ..., one between $a_{n-1}$ and $a_n$). But where is the last root? No worries! If we have a complex root $\alpha$, then $\bar{\alpha}$ is also a (different and complex) root (this only holds if $P$ has real coefficients, but it does, so we're good). This means, if we have $n-1$ real roots of a polynomial degree $n$, then we can't have the last root complex, because complex roots come in pairs.

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