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

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

Dimensionally it would make more sense as

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

Even if you obtain such formulas the result will not likely be very useful for solving the system. The formulas are large and substitution of values for the a and b parameters will likely give rise to cancellation error.

Hello Richard,

Yes the formula is correct, it is

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

, so bi*xi and not just bi. One solution is xi = 0 for all i = 1...n.

I also found a formula for c[2]:

c[2] == a[n]^(n - 2) Product[b[i], {i, 1, n}] ((a[n] n)/b[n] + (b[n] - a[n]) Sum[1/b[i], {i, 1, n - 1}] - (n - 1) (1 - Sum[a[i]/b[i], {i, 1, n - 1}]) - b[n] (Sum[a[i]/b[i], {i, 1, n - 1}] Sum[1/b[i], {i, 1, n - 1}] - Sum[a[i]/b[i]^2, {i, 1, n - 1}]))

which is complicated, but the number of operations to calculate it is of order n and not n^2. I assume that there are also simple formulas for the other c[i], because when I extract all powers of an, bn, ai, bi in the coefficients c[i], I see a pattern that is similar for all c[i], see attached notebook. It gives a table for all terms that appear in the coefficients.

The columns are the powers of ai, bi (all i including n), an, bn, and the coefficient in front of each term. The last column gives the multiplicity of how many terms of this type appear, which is in this case the number of permutations that the indices i different from n allow.

E.g. for n=5, c[5] is

+b[4]^3

-b[1] b[4]^2
-b[2] b[4]^2
-b[3] b[4]^2

+b[1] b[2] b[4]
+b[1] b[3] b[4]
+b[2] b[3] b[4]

-b[1] b[2] b[3]

which gives the table entries

0   3 0   3 1   1
0   3 0   2 -1  3
0   3 0   1 1   3
0   3 0   0 -1  1

It is easy to find a rule for how all these terms are constructed for any n; the binomial coefficients in column 5 and their sign are determined by the powers of the a's and b's in columns 1-4. But as discussed above, my goal is to find a simpler rule for constructing them, with a lower number of operations.

Best, Max

Attachments:

eqn1.nb

Hello Richard,

sure, I see that the system can be written in a vector form. But since there are quadratic terms xi*(x1+...+xn), I thought it cannot be solved by the tools for linear equations. But it has been a while since I studied these things in detail. If you are aware of a practical way to solve it algebraically, please post it.

Thanks,

Max

Hello Richard,

this is true, but how can this system be solved algebraically? I think it still requires solving a polynomial equation of order n+1.

I was able to find a formula for determining the coefficients for this polynomial equation, but it is quite complex. E.g. when you set n=5 and look at c[3], it contains 50 terms, each being a product of 7 a[i] or b[i]. For n=6 it is already 576 terms in total for all coefficients. Although I could figure out the pattern how to construct these coefficients for any n, I was wondering if there is a simpler way to express them, e.g. as products of a few terms, as the above examples for c[2] and c[n+1].

To make the example more explicit, for n=4, c[n+1] is:

-b[1] b[2] b[3] + b[1] b[2] b[4] + b[1] b[3] b[4] + b[2] b[3] b[4] - b[1] b[4]^2 - b[2] b[4]^2 - b[3] b[4]^2 + b[4]^3

which is 16 multiplications and 7 additions, but it can be simplified to

(b[4] - b[1]) (b[4] - b[2]) (b[4] - b[3])

which is 3 multiplications and 3 subtractions.

If it is possible to find a similar general rule for all coefficients, the calculation for higher n would be much accelerated.

Max