Simplify $\sum_{i=1}^n\frac{a}{\theta-x_i}$ where $\{x_i\}\in(0,\theta)$ and $a\in(-1,0]$ I have this problem where I need to solve $-\frac{n(1+a)}{\theta}+\sum_{i=1}^n\frac{a}{\theta-x_i}=0 $ for $\theta$ (the solution is allowed to be in terms of the mean of $x_1,..., x_n$), provided  that $\{x_i\}\in(0,\theta)$ and $a\in(-1,0]$.
I really have no idea how to simplify the summation. Maybe I'm just missing something simple?
 A: This equation is a variant of the so-called Underwood equation (have a look at my answer here).
Taking into account the specificity of this one, there are $n$ solutions each of them between a pair of vertical asymptotes corresponding to the $x_i$.
Now, if your condition is that you look for $0 <\theta < x_i^{min}$, then, instead of considering the function
$$f(\theta)=-\frac{n(1+a)}{\theta}+\sum_{i=1}^n\frac{a}{\theta-x_i}$$ consider
$$g(\theta)=\theta(\theta - x_i^{min})f(\theta)$$ which removes the first asymptotes.
Now, for an approximation, use Newton method with $\theta_0=0$ and you will see all $x_i$'s appearing in $x_1$.
This is a problem I worked for decades and I have published may papers on this topic. If you need more, just tell.
Edit
Since it is the first time I face a problem with an asymptote at exactly $\theta=0$, I reworked the equations.
$x_1$ is the samllest of the list and no other $x_i$ is equal to it.
What we have is
$$g(0)=(a+1) n x_1 > 0 \qquad \text{and} \qquad g(x_1)=a x_1 <0$$
$$g'(0)=a-(1+a)n+a x_1 \sum_{i=2}^n \frac 1 {x_i}$$
$$g''(0)=2a \Big[x_1 \sum_{i=2}^n \frac 1 {x_i^2} -\sum_{i=2}^n \frac 1 {x_i}\Big]$$
The first iterate of Newton method is then
$$\theta_1^{(2)}=\frac{(a+1) n x_1}{(1+a)n-a-a x_1 \left(\sum _{i=2}^n \frac{1}{x_i}\right)}$$
Let us try for $n=5$, $a=-\frac 1 2$, $x_i=\left\{\frac{1}{6},\frac{1}{5},\frac{1}{4},\frac{1}{3},\frac{1}{2}\right\}$. This gives $\theta_1^{(2)}=\frac 1 {10}=0.1000$ while the exact solution is $0.1127$.
Using Halley's method, the first iterate would be $\theta_1^{(3)}=\frac 5 {47}=0.1064$ while the exact solution is $0.1127$.
Using Householder's method, the first iterate would be $\theta_1^{(4)}=\frac {47} {430}=0.1093$ while the exact solution is $0.1127$.
Notice that using the secant method, we should have
$$\theta_1^{(1)}=\frac{ (1+a)n x_1}{(1+a)n-a}$$ which totally hides the contribution of the other $x_i$'s. For the worked case, this would give $\theta_1^{(1)}=\frac {5} {36}=0.1389$ while the exact solution is $0.1127$.
We could continue with higher-order method which, as a function of their order, would generate the sequence
$$\left\{\frac{1}{10},\frac{5}{47},\frac{47}{430},\frac{2150}{19407},\frac{19407}{173
   918},\frac{173918}{1552301},\frac{1552301}{13822834},\frac{69114170}{614604843}\right\}$$ which are closer and closer to the solution (the last in the table is $0.1125$).
A: I think there is no solution. Let's take the simple case where $n=1$. Then we have the following :
$$\frac{1+a}{\theta} = \frac{a}{\theta - x}$$
$$\frac{\theta}{\theta - x } = \frac{1+a}{a}$$
Subtracting one from both sides,
$$\frac{x}{\theta -x} = \frac{1}{a}$$
$$x = \frac{\theta}{1+a}$$
Now, since $a \in (-1,0]$, the denominator is less than one, therefore $x$ is larger then $\theta$ for any allowed value of $a$.
