Hi Max,

OK, I will try again: As I proposed above one can use the equation

$$x_i = \frac{a_i \sum_k x_k}{b_i+\sum_k x_k}$$ for an iterative process. It might be not unreasonable to assume that

$$\big(\sum_k x_k\big) >(\gg) \;b_i \quad \text{(for any i)} $$

witch results as a first approximation: $$x_i \approx \frac{a_i \sum_k x_k}{\sum_k x_k} = a_i$$ This gives a first estimate for the sum which can be used for a better approximation for the $x_i$, and so on.

You did not give us the coefficient vectors a and b, so I am using random numbers.

dim = 15;
a = RandomReal[10, dim];
b = RandomReal[10, dim];

The iteration itself is pretty much straightforward:

newX[x_List] := With[{sum = Total[x]}, sum a/(sum + b)];
xList = FixedPointList[newX, a];

and one can watch the system converging:

ListPlot[Transpose[xList], Joined -> True, ImageSize -> Large, GridLines -> Automatic]

and finally one can check the result using the formula from your original post:

x = Last[xList];  (* final result! *)
Chop[(x - a)*Total[x] + b x] == ConstantArray[0, dim]
(*  Out:   True  *)

Does/will this work with your coefficients?

Hello Richard,

typically n is not very high, in the range of 3-10. But since the ai and bi are going to be fit parameters and are different for every calculation of the xi, calculation speed plays an important role. Therefore, I am searching for a simple formula with a low number of operations.

Also, I am generally wondering if there is a formalism or tool to convert expressions like the c[i] to simpler formulas. FullSimplify is doing well if the result is just a product of sums and differences of the ai and bi (as for my example c[n+1]), but not if there are terms like ai/bi involved, as for c[1] and c[2] as above.

Max

I don't know why that would necessarily make more sense-- it depends on where the system arises. Also such a change would have at least two notable effects: the origin would (generically) not be a solution, and the solution count would be two rather than n+1.

Hello Henrik, thanks you for this idea, it is really good and simple. Although there is no possibility to find sum, and the above equation still requires solving a polynomial equation of the same order as before, it has the significant advantage that it is already posed in the simplified form that allows a quick calculation of the coefficients. Since sum is x[1]+...+x[n], the solution

x[i] == sum*a[i]/(sum+b[i])

can be used to re-write the problem as

sum - Sum[sum*a[i]/(sum+b[i]),{i,1,n}] == 0

In contrast to my above attempts, the solution is already in a simple form. Now I can hopefully directly calculate the coefficients for solving

c[0] + c[1]*sum + c[2]*sum^2 + ... + c[n+1]*sum^(n+1) == 0

Best, Max

Max, For the phenomena you are modeling, how large might n be?

Ok, you would not use tools for linear systems on a vector equation containing the quadratic form x.M.x

Is your original formula correct?

Table[(x[i] - a[i])Sum[x[k], {k, 1, n}] + b[i]x[i] == 0, {i, 1, n}]

It produces n equations of the form:

xi•(x1 + … + xn) - ai•(x1 + … + xn) + bi•xi == 0,  i = 1 to n.

"how can this system be solved algebraically?"

I am thinking of it as a vector equation solved by vector algebra. The way you're looking at it, you have rows of:

xi•(x1 + … + xn) - ai•(x1 + … + xn) + bi•xi == 0.

As a vector equation, the 3rd term is a matrix of all zeroes except b_i's down the diagonal, multiplying a column vector of x.

Hi Max, I believe your equation can be put in quadratic vector form

x^t.M.x

(x^t == x transpose) and then solved algebraically?