solution to $\sum_{i=1}^{n}\frac{1}{a_{i}x+b_{i}} = 0$

Question

Is there any general procedure to solve the equation $$ \sum_{i=1}^{n}\frac{1}{a_{i}x+b_{i}}=0 $$ with respect to $x$ for given $a_{i}$ and $b_{i}$, with $i=1,\dots,n$?

Answer 1

As said in comments, for $n>4$, a numerical method will be required.

In fact, as you wrote it, your equation is just a variant of the so-called Underwood equation which write $$F(x)=\sum_{i=1}^n \frac {a_i}{b_i -x}+c =0$$ which is widely used in chemical engineering.

In my former research group, we developed rapid and robust methods for obtaining the solutions. Have a look at this paper.

Still closer to your specific problem, look at this paper which addresses the problem of the solution of the so-called Rachford-Rice equation $$G(x)=\sum_{i=1}^n \frac {z_i}{x-b_i } =0$$ which is also widely used in chemical engineering.

Illustration

I shall not use any of the above methods to show how you could treat the problem with a spreadsheet. Let $$H(x)=\sum_{i=1}^n \frac {1}{a_ix+b_i } =0$$ and I shall admit that terms have been sorted such that $$\frac{b_1}{a_1} >\frac{b_2}{a_2}> \cdots >\frac{b_n}{a_n}$$ and that we search for the solution between two vertical asymptotes, say $$x_1=-\frac{b_1}{a_1} < x <-\frac{b_2}{a_2}=x_2$$ In a first step, remove this asymptotes and consider instead the function $$\tilde H(x)=(a_1x+b_1)(a_2x+b_2) \sum_{i=1}^n \frac {1}{a_ix+b_i } $$ So, we have $$\tilde H(x_1)=b_2-\frac{a_2}{a_1}b_1 \qquad \qquad \tilde H(x_2)=b_1-\frac{a_1}{a_2}b_2\qquad \qquad \tilde H(x_1)\,\tilde H(x_2)<0$$ Draw the straight line joining the two points and compute the $x$ intercept. This defines a new point $$x_3=-\frac{b_1+b_2}{a_1+a_2}$$ Computing $\tilde H(x_3)$ defines now either $x_1 <x<x_3$ or $x_3<x<x_2$ and you could continue just with bisection.

To make an example : $n=6$, $a_i=p_i$ and $b_i=\sqrt{i!}$

$$\left( \begin{array}{ccccc} i &x_1 & x_2 & x_3 & x_{\text{solution}} \\ 1 & -0.500000 & -0.471405 & -0.482843 & -0.479100 \\ 2 & -0.471405 & -0.489898 & -0.482963 & -0.493248 \\ 3 & -0.489898 & -0.699854 & -0.612372 & -0.672910 \\ 4 & -0.699854 & -0.995859 & -0.880746 & -0.961714 \\ 5 & -0.995859 & -2.064063 & -1.574469 & -1.978404 \end{array} \right)$$ and this is the first iteration of the most simplistic numerical method.