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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?

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I think you misprint index on R.H.S, should be $b_j$ instead of $b_i$. I do not see how to proceed directly, need to think of it. – Neeraj Jun 5 '12 at 11:45
Thanks, corrected now. – Felix Goldberg Jun 5 '12 at 11:53
I don't know what an "approximate explicit function" is. – Gerry Myerson Jun 5 '12 at 12:31
Arnold Ross used to ask, "What is an approximation to 5?" and he would answer, "Any number except 5." So I guess you can take $f$ to be any function whatsoever, so long as it doesn't always give the exact answer. If you want better than that, you have to pin down what you mean by those wavy lines. – Gerry Myerson Jun 6 '12 at 3:50
I'd consider $5$ to be an excellent approximation to $5$. – Andreas Blass Aug 4 '13 at 15:51

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.)

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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.

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Suppose the $\alpha_i$ and $\beta_j$ are real, and sorted in increasing order. In any interval $(\alpha_i, \alpha_{i+1})$ that contains no $\beta_j$ there will be at least one solution. Similarly in any interval $(\beta_j, \beta_{j+1})$ that contains no $\alpha_i$. – Robert Israel Jun 5 '12 at 22:06
So the only case where you can have no real solutions is where the distinct $\alpha_i$ and $\beta_j$ are interlaced. – Robert Israel Jun 5 '12 at 22:09
For example, with $m=1$ and $n=2$ (with $\alpha_1 < \alpha_2$) there are no real solutions iff $\alpha_1 \le \beta_1 \le \alpha_2$. – Robert Israel Jun 5 '12 at 22:12

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