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I am given that:

$$\sum_{i=1}^{n}{w_{i}}=1\\$$

and that a set of numbers $e_i$, where $i$ can range from $1$ to $n$.

Now I need to find a number $u$, such that

$$\sum_{i=1}^{n}{\left(\frac{w_{i}}{u-e_{i}}\right)^{2}}=1$$

My questions are: are there systematic way of finding all possible solutions $u?$

And is the number of solution related to $n?$

I am thinking of maybe for $n=2$, the number of solutions $u$ is $1?$

$u$ is unconstrained... all the rest are given...

and yes, $w_i >$ or = $0$ for all $i$...

And for general $n$, the number of solutions $u$ is $n-1$?

Thanks a lot!

[Edit]

Now I need to find a number $u$, such that

$$\sum_{i=1}^{n}{\left(\frac{w_{i}}{u-e_{i}}\right)^{2}}=1$$

And I am looking for real numbers $u$...

And after finding all these roots $u$'s,

I would like to compare all of the following:

$$\sum_{i=1}^{n}{\left(\frac{w_{i}}{u-e_{i}}\right)^{2}/e_{i}^{2}}$$

and find one of the roots u* which maximizes the above expression?

Any possible shortcuts?

Thanks

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1  
It looks like you forgot something. What should $\sum_{i = 1}^n \frac{w_i}{(u-e_i)^2}$ be? –  Calle Apr 18 '12 at 17:27
    
If you are given the result of the sum, $\sum_{i = 1}^n \frac{w_i}{(u-e_i)^2}=k$, for $n=1$ you have a quadratic, so will generally have two solutions. Depending upon the parameters, they may be complex. –  Ross Millikan Apr 18 '12 at 17:35
    
I've added some new info... any more thoughts? thx –  Luna Apr 18 '12 at 19:30
    
No constraints on the $e_i$? –  Calle Apr 18 '12 at 21:17

3 Answers 3

If any of the $w_i$ are zero, they don't affect either sum, so we may assume $w_i\gt0$ for all $i$.

As $u\to\infty$, the sum approaches zero. If $u$ is just a hair bigger than the biggest $e_i$, then the sum is enormous. Between the biggest $e_i$ and $\infty$, the sum is decreasing. It follows that there is a unique solution $u$ strictly between the biggest $e_i$ and $\infty$.

Two questions remain: how to find that $u$, and whether there are any solutions less than the biggest $e_i$.

Both of these seem difficult. Clearing denominators yields an equation of degree $2n$ in $u$, so I think only numerical methods (e.g., Newton's Method) apply. And what happens between the $e_i$ is not obvious to me.

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Also the sum goes to $0$ as $u \to -\infty$, so you'll have at least two real solutions. –  Robert Israel Apr 19 '12 at 0:16
1  
If there is a big enough gap between some of the $e_j$, the sum will also be less than $1$ in the middle of such a gap, and then you'll get another pair of real solutions, one near each end of the gap. –  Robert Israel Apr 19 '12 at 0:52
    
Thank you! I've added some info. Any more thoughts? Thanks again! When n is large (say thousounds), what's the systematic way to guide a numerical solver to iteratre through all roots? –  Luna Apr 19 '12 at 2:53

I will rewrite your equations to be more readable:

$$\sum_{i=1}^{n}{w_{i}}=1\\ \sum_{i=1}^{n}{\left(\frac{w_{i}}{u-e_{i}}\right)^{2}}=1$$

I thought about least squares solution, but it seems the problem is quite different...

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I've added some new info... any more thoughts? thx –  Luna Apr 18 '12 at 19:30
    
If you would know the $w_{i}$s, I would say this is weighted least squares problem and as such it has one unique solution. But if you don't know the $w_{i}$s, I am afraid there is infinitely many solutions (one for every combination of $w_{i}$s). –  Libor Apr 18 '12 at 20:04

This question reminds me of the 1-D problem of computing the electrostatic force between a test charge $q$ and a collection of charges $q_i$, where $i = 1,...,n$. The electrostatic force is $$ F(r) = \frac{q}{4 \pi \epsilon_0} \sum_{i=1}^n \frac{q_i}{(r-r_i)^2} \; . $$

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