## 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?

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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 at 11:45
Thanks, corrected now. – Felix Goldberg Jun 5 at 11:53
I don't know what an "approximate explicit function" is. – Gerry Myerson Jun 5 at 12:31
I mean, write $x \approx f(a_{i},b_{j},m,n)$ where $f$ is some explicit function. – Felix Goldberg Jun 5 at 13:10
Multiply by the product of the denominators, and you have a polynomial of degree $m+n-1$ if $m \ne n$, somewhat less if $m = n$. I doubt that the result will be any easier to solve than the general polynomial of that degree. In fact, I suspect that you can choose the $a_i$ and $b_j$ to get a nonzero constant multiple of any polynomial of that degree. – Robert Israel Jun 5 at 19:34
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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 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 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 at 22:12