# prove : $\frac{b_{1}}{x-a_{1}}+\frac{b_{2}}{x-a_{2}}+...+\frac{b_{k}}{x-a_{k}}=0$

## prove that : $\frac{b_{1}}{x-a_{1}}+\frac{b_{2}}{x-a_{2}}+...+\frac{b_{k}}{x-a_{k}}=0$ with $b_{1},b_{2},...,b_{k}>0$ and $a_{1}<a_{2}<...<a_{k}$ has $k -1$ real solutions.

so i think i have managed to prove that though induction and the use of Bolzano's theorem my only problem is that in the end my solution got too big and at some points i think i might have some mistakes.Is there a faster and better way to solve this without induction because i think that bolzano's therorem is needed anyways?(i didnt post my solution since it is too big like 2 sheets of paper.)

• Notice $\frac{d}{dx}LHS < 0$ for all $x \not\in \{ a_1, \ldots, a_k \}$, LHS is strictly decreasing on the $k+1$ intervals $(-\infty,a_1), (a_1,a_2),\ldots,(a_k,\infty)$. Convince yourself there is a root in each of the $k-1$ bounded intervals. Dec 12, 2017 at 17:14
• The condition $a_1<a_2<\ldots<a_k$ seems almost irrelevant, given that there is no order for the $b_j$'s. The word "almost" is because the condition tells us that they are pairwise different. It can be useful to claim it WLOG, though. Dec 12, 2017 at 17:14

Consider $P(x) = (x-a_1)^{ b_1} \cdots (x-a_n)^{ b_n}$

By differentiating $\ln (P(x))$ or otherwise we can show

$$H(x) =\dfrac {P' ( x)}{P(x) } = \dfrac{ b_1}{(x-a_1)}+ \cdots \dfrac{b_n}{(x-a_n)}$$

Now $P(x)$ has $n$ distinct roots excluding multiplicities. By Rolle's theorem we have at $(n-1)$ roots of $P'(x)$ that are different from the roots of $P(x)$. Hence $H(x)$ has $(n-1)$ real roots and thus has all its roots real .

• Neat. I have a vested interest that my answer is accepted ;) but I really like this one!
– user491874
Dec 12, 2017 at 18:05
• how do we prove that each of these roots is unique so that the total of the roots is exactly k - 1? Aug 27, 2019 at 18:27
• @DimitrisPapadimitriou the roots of $P'(x)$ lie between the roots of $P(x)$. So in each interval formed by consecutive roots of $P(x)$ you have a root of $P'(x)$ thus netting us $n-1$ distinct roots of $P(x)$. Aug 29, 2019 at 5:11

Multiply with all denomiators. You get a polynomial \begin{align*} p(x) = \sum_{i=1}^k b_i\prod_{j\neq i} (x-a_j) \end{align*}

of degree $k-1$ on the left-hand side. None of the $a_i$ is a zero of that polynomial because for $x=a_i$ the left-hand side simplifies to $b_i \prod_{j\neq i} (x-a_j)$ and $a_i$ is not a zero of that simplified term.

Therefore, the zeros of the polynomial are the zeros of the original left-hand side.

Because of $a_i < a_{i+1}$ and $b_i>0$ you have \begin{align*} (-1)^{k-i} p(a_i) > 0. \end{align*} for $k=1,\ldots,k$ and therefore $k$ sign-changes for the continuous function $p(x)$. In between these sign changes there are $k-1$ zeros.

Set $f(x)=\frac{b_1}{x-a_1}+\frac{b_2}{x-a_2}+\cdots+\frac{b_k}{x-a_k}$.

One can easily see that the equation has no solutions in $(-\infty, a_1)$ or in $(a_k,+\infty)$ because $f(x)$ is strictly negative on the first interval and strictly positive on the second.

Also, $f(x)$ is strictly decreasing on all intervals where it is defined, so it can have at most one solution in each of the intervals $(a_i, a_{i+1})$ for $i=1,2,\ldots,k-1$. All that is left to be proven is that it actually has one solution in each of those $k-1$ intervals.

So, pick $i\in\{1,2,\ldots,k-1\}$. Note that: $$\lim_{x\to a_i+0}f(x)=+\infty$$ and $$\lim_{x\to a_{i+1}-0}f(x)=-\infty$$

Thus, we can choose $\varepsilon$, $0\lt\varepsilon\lt \frac{a_{i+1}-a_i}{2}$ such that $f(a_i+\varepsilon)\gt 0$ and $f(a_{i+1}-\varepsilon)\lt 0$. Thus, there is a point $\xi\in(a_i+\varepsilon, a_{i+1}-\varepsilon)\subset(a_i, a_{i+1})$ such that $f(\xi)=0$ (by Bolzano's theorem).

• why does the fact that f(x) is strictly decreasing on all intervals where it is defined mean that it can have at most one solution in each of the intervals (ai,ai+1) for i=1,2,…,k−1? Aug 27, 2019 at 18:14

Hint (following achille hui's comment):

Compare $\lim_{x\to a_j^+}$ and $\lim_{x\to a_{j+1}^-}$ of the expression.