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I have a set of 29 coupled quadratic equations, with 29 unknown variables.

Can anyone offer any advice on how I could go about solving this?

3 days of staring at a wall has so far given me no thoughts on how to do this at all.

EDIT: $ T_1 = X_1^{2} X_2 X_3 X_4 X_5 X_6 \\$

$ T_2 = X_2^{2} X_1 X_3 X_4 X_5 X_6 \\$

$ T_3 = X_3^{2} X_1 X_2 X_4 X_5 X_6 \\$

$ T_4 = X_4^{2} X_1 X_2 X_3 X_5 X_6 \\$

$ T_5 = X_5^{2} X_1 X_2 X_3 X_4 X_6 \\$

$ T_6 = X_6^{2} X_1 X_2 X_3 X_4 X_5 \\$

$T_7 = X_1 X_2 X_3 X_4 X_5 X_6 X_7^2 X_8 X_9 X_{10} (1-X_5) \\$

$T_8 = X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8^2 X_9 X_{10} (1-X_5) \\$

$T_9 = X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9^2 X_{10} (1-X_5) \\$

$T_{10} = X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10}^2 (1-X_5) \\$

$T_{11} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11}^2 X_{12} X_{13} X_{14} X_{15} \\$

$T_{12} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12}^2 X_{13} X_{14} X_{15}\\$

$T_{13} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13}^2 X_{14} X_{15} \\$

$T_{14} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14}^2 X_{15} \\$

$T_{15} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15}^2\\$

$T_{16} = X_1 X_2 X_3 X_4 X_5 X_6 (1-X_6)(1-X_9)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16}^2 X_{17}\\$

$T_{17} = X_1 X_2 X_3 X_4 X_5 X_6 (1-X_6)(1-X_9)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17}^2\\$

$T_{18} = X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18}^2 X_{19} X_{20} X_{21}\\$

$T_{19} = X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19}^2 X_{20} X_{21}\\$

$T_{20} = X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19} X_{20}^2 X_{21}\\$

$T_{21} = X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19} X_{20} X_{21}^2\\$

$T_{22} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_5) X_7 X_8 X_9 X_{10} [(1-X_{17} + (1-X_9)(1-X_7)X_{16}X_{17}] + (1-X_2)\} X_{22}^2 X_{23}$

$T_{23} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_5) X_7 X_8 X_9 X_{10} [(1-X_{17} + (1-X_9)(1-X_7)X_{16}X_{17}] + (1-X_2)\} X_{22} X_{23}^2$

$T_{24} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_3) + (1-X_5)(1-X_8)X_7 X_8 X_9 X_{10} \} X_{24}^2 X_{25}$

$T_{25} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_3) + (1-X_5)(1-X_8)X_7 X_8 X_9 X_{10} \} X_{24} X_{25}^2$

$T_{26} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_4) + \{ (1-X_6) + (1-X_{10})(1-X_5))X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15} \}(1-X_{12}) + X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}(1-X_{20})(1-X_5)(1-X_9)(1-x_16) +(1-X_25)\{(1-X_3) + (1-X_5)X_7 X_8 X_9 X_{10} (1-X_8) \}X_{24}X_{25} \} X_{26} $

$T_{27} = X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_4) + \{ (1-X_6) + (1-X_{10})(1-X_5))X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15} \}(1-X_{12}) + X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}(1-X_{20})(1-X_5)(1-X_9)(1-x_16) +(1-X_25)\{(1-X_3) + (1-X_5)X_7 X_8 X_9 X_{10} (1-X_8) \}X_{24}X_{25} \} X_{27} $

$T_{28} = \{ X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_4) + \{ (1-X_6) + (1-X_{10})(1-X_5))X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15} \}(1-X_{12}) + X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}(1-X_{20})(1-X_5)(1-X_9)(1-x_16)+(1-X_25)\{(1-X_3) + (1-X_5)X_7 X_8 X_9 X_{10} (1-X_8) \}X_{24}X_{25} \} \}(1-X_{27}+ X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19} X_{20} X_{21}(1-X_20) +(1-X_{13})X_{28}X_{29} X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} \} X_{28} $

$T_{29} = \{ X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_4) + \{ (1-X_6) + (1-X_{10})(1-X_5))X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15} \}(1-X_{12}) + X_7 X_8 X_9 X_{10} X_{11}X_{12} X_{13} X_{14} X_{15}X_{16}X_{17}X_{18}X_{19}X_{20}X_{21}(1-X_{20})(1-X_5)(1-X_9)(1-x_16)+(1-X_25)\{(1-X_3) + (1-X_5)X_7 X_8 X_9 X_{10} (1-X_8) \}X_{24}X_{25} \} \}(1-X_{27}+ X_1 X_2 X_3 X_4 X_5 X_6 X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} X_{16} X_{17} (1-X_9)(1-X_5)(1-X_{16}) X_{18} X_{19} X_{20} X_{21}(1-X_20) +(1-X_{13})X_{28}X_{29} X_1 X_2 X_3 X_4 X_5 X_6 \{ (1-X_6) + (1-X_{10})(1-X_5)X_7 X_8 X_9 X_{10} \} X_{11} X_{12} X_{13} X_{14} X_{15} \} X_{29} $

Here are the equations, my unknowns are the X terms. T terms are known.

T and X terms are both real and positive.

Thanks for the comments, still attempting to understand what a "Groebner basis" is so far... EDIT END

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    $\begingroup$ @Taladris Coupled means that in each equation one may have several of the variables present. $\endgroup$ – Pp.. Jan 22 '15 at 18:25
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    $\begingroup$ The key words to read about the topic are: Groebner basis, regular chain, elimination of variables. Alternatively, or also, approximation methods. You can star from here $\endgroup$ – Pp.. Jan 22 '15 at 18:31
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    $\begingroup$ That example is not what is normally called quadratic. A property that does stand out is the small number of monomials. $\endgroup$ – Pp.. Jan 22 '15 at 18:33
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    $\begingroup$ Dare I ask what physical system? $\endgroup$ – John Jan 22 '15 at 19:27
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    $\begingroup$ The X terms are actually exponentials of the form $e^{-t/\tau}$, which are supposed to represent lifetime decays of excited electron states in a crystal structure. There are 29 state transitions, where the T values are transition rates which have been measured. Apparently measuring things is much easier than modelling them. $\endgroup$ – user209848 Jan 22 '15 at 20:31
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In the following, I assume that the $(X_i)$ are real numbers.

Following the Pp.. comment about Grobner basis, we can solve the system constituted with the first $6$ equations when the $(T_i)_{i\leq 6}$ are generically chosen.

We obtain a result in the form $a{X_1}^7+b=0$ and, for every $i\leq 6$, $X_i=c_iX_1$. If the $(T_i)$ and $(X_i)$ are real, then we obtain a sole (and explicit) solution for $(X_i)_{i\leq 6}$.

EDIT 1. Since the $(X_i)_{i\leq 6}$ are known, solve the following blocks of equations (we obtain exactly the same type of equations as above)

i) $\{7,\cdots,10\}$, $a{X_7}^5+b=0$, one solution in $X_7,\cdots,X_{10}$.

ii) $\{11\cdots 15\}$, $aX_{11}^6+b=0$, $0$ or $2$ solutions in $X_{11},\cdots,X_{15}$..

iii) $\{16,17\}$, $aX_{16}^3+b=0$, one solution.

iv) $\{18,\cdots,21\}$, after $\{22,23\}$, after $\{24,25\}$. Equations $26,27$ have degree $1$ in $X_{26},X_{27}$. Each of these equations admits generically a unique solution.

v) Equations $28,29$ are in the form $aX_{28}X_{29}+bX_{28}=c,dX_{28}X_{29}+eX_{29}=f$, $0$ or $2$ solutions.

EDIT 2. Grobner basis theory is useless. Indeed the solution(s) of a system of $n$ equations in the form of our first $6$ equations is: ${X_1}^{n+1}=\dfrac{{T_1}^n}{T_2\cdots T_n}$ and, for every $i\leq n$, $X_i=\dfrac{T_i}{T_1}X_1$.

EDIT 3. (answer to Respawned Fluff). The set of solutions of our system is zero-dimensional over $\mathbb{C}$. Then it is easy to keep only the real solutions. Here the system is block-triangular ; moreover each block (using the solutions of the previous blocks) admits an "effective solution" in the following sense: there is $i$ and a polynomial $P$ of degree $d$ s.t. $P(x_i)=0$ and, for every $j\not= i$, there are polynomials $P_j$ of degre $<d$ s.t. $x_j=P_j(x_i)$. Finally, our system has generically $0$ or $4$ real solutions. This is a simple system and clearly it can be easily solved by the standard softwares under the condition that the chosen order for the unknown is essentially $X_1,\cdots,X_{29}$ ; otherwise, the calculation time is likely to be very long.

At least, 2 teams of researchers are working about this subject: the LIP 6 laboratory (J.C. Faugère) and a group around M. Moreno Maza. The first one studies the so-called semi-regular systems over $\mathbb{C}$ and the second one studies the so-called regular semi-algebraic systems over $\mathbb{R}$.

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    $\begingroup$ Sorry, I simply forgot to write $T_{29}$. It's tagged matrices because I was initially trying to write them out as a matrix and then see if that would make it clearer to me on how to solve them. Also I have to have at least one tag to post the question, no currently existing tags that I could see are applicable to the question ,and I do not have enough whatever points to create a new tag.... $\endgroup$ – user209848 Jan 22 '15 at 20:26
  • $\begingroup$ Does this triangularization eventually result in a regular semi-algebraic system? If so the authors of that (fairly recent) paper would probably be happy to hear their method applies to some concrete physics problem that's hard to solve otherwise... $\endgroup$ – Fizz Jan 23 '15 at 10:27

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