an equation in fractions I have an equation of the form $\sum_{i=1}^{m}{\frac{1}{a_{i}-x}}=\sum_{j=1}^{n}{\frac{1}{b_{j}-x}}$ and would like to express $x$ as a an approximate explicit function of the $a_{i},b_{j},m,n$. Have you encountered such a problem?
 A: Solving this equation is equivalent to finding the zeros of the derivative of a rational function, based only on knowing only the rational function's factorization.  
Let  $$Q(x)=\prod_{i=1}^{n}\left(x-\alpha_{i}\right)\prod_{i=1}^{m}\left(x-\beta_{i}\right)^{-1}.$$  Then taking the logarithmic derivative we find that $$\frac{Q'(x)}{Q(x)}=\sum_{i=1}^{n}\frac{1}{x-\alpha_{i}}-\sum_{i=1}^{m}\frac{1}{x-\beta_{i}},$$ and so the equation $$\sum_{i=1}^{n}\frac{1}{x-\alpha_{i}}=\sum_{i=1}^{m}\frac{1}{x-\beta_{i}}$$ is solved if and only if $\frac{Q'(x)}{Q(x)}=0$.  At any point where the denominator has a pole of degree $k$, the numerator will have a pole of degree $k+1$, and so the numerator has no contribution to the number of zeros of $\frac{Q'(x)}{Q(x)}$.
Thus, we find that $\sum_{i=1}^{n}\frac{1}{x-\alpha_{i}}=\sum_{i=1}^{m}\frac{1}{x-\beta_{i}}$ if and only if $Q'(x)=0$, and there are many existing resources for this kind of problem. (For example to do this numerically one can use Newtons method.)
A: In the proof of Eric Naslund:   I would like to add few more things.
I like the answer of Eric.
As Eric pointed out  $\sum_{i=1}^{n}\frac{1}{x-\alpha_i}=\sum_{i=1}^{m}\frac{1}{x-\beta_i}$ if and only if ${Q}^{\prime}(x)=0$. What I want to say is ${Q}^{\prime}(x)$ may or may not have solution. And so, one may or may not have solution of $\sum_{i=1}^{n}\frac{1}{x-\alpha_i}=\sum_{i=1}^{m}\frac{1}{x-\beta_i}$.
For example: Take $m=1$ and $n=2$. one has ${Q}(x)=\frac{(x-\alpha_1)(x-\alpha_2)}{(x-\beta_1)}$. Then ${Q}^{\prime}(x)=\frac{x^2-2\beta_1 x+\beta_1(\alpha_1+\alpha_2)-\alpha_1\alpha_2}{(x-\beta_1)^2}$. So ${Q}^{\prime}(x)=0$ if and only if $x^2-2\beta_1 x+\beta_1(\alpha_1+\alpha_2)-\alpha_1\alpha_2=0$. The real roots of this depends on the choice  of $\alpha_1,\alpha_2,\beta_1$, which makes discriminants greater than or equal to zero. 
I have considered an elementary case, and still hope for finding roots is so dificult. In general, It seems  difficult to find solution of $\sum_{i=1}^{n}\frac{1}{x-\alpha_i}=\sum_{i=1}^{m}\frac{1}{x-\beta_i}$. 
However, theoretically answer lies on the roots of ${Q}^{\prime}(x)=0$, but practically, It seems hard. 
