Is there any way to solve the following system of non linear equations

Question

$$a\,(k_1-1)+b\,(k_2-1)+c\,(k_3-1)=-1/2$$ $$a\,(k_1-k_1^2)+b\,(k_2-k_2^2)+c\,(k_3-k_3^2)=1/6$$ $$a\,(k_1-k_1^3)+b\,(k_2-k_2^3)+c\,(k_3-k_3^3)=2/8$$ $$a\,(k_1-k_1^4)+b\,(k_2-k_2^4)+c\,(k_3-k_3^4)=3/10$$ $$a\,(k_1-k_1^5)+b\,(k_2-k_2^5)+c\,(k_3-k_3^5)=4/12$$ $$a\,(k_1-k_1^6)+b\,(k_2-k_2^6)+c\,(k_3-k_3^6)=5/14$$ I tried to solve them by Wolframalpha but I couldn't. I wish to get the solution even though without details because the next step in my work depends on these values Thanks.

Answer 1

According to Maple, a Groebner basis of your system is $$ \eqalign{ 35\,{{\it k_3}}^{3}&-45\,{{\it k_3}}^{2}+15\,{\it k_3}-1,\cr 7\,{{\it k_2}}^{2}&+7\,{\it k_2}\,{\it k_3}+7\,{{\it k_3}}^{2}-9\,{\it k_2}- 9\,{\it k_3}+3,\cr 7\,{\it k_3}&+7\,{\it k_1}+7\,{\it k_2}-9,\cr 140\,{{\it k_3}}^{2}&+144\,c-145\,{\it k_3}-20,\cr -140\,{\it k_2}\,{\it k_3}&-140\,{{\it k_3}}^{2}+144\,b+35\,{\it k_2}+180\, {\it k_3}-80,\cr 140\,{\it k_2}\,{\it k_3}&+144\,a-35\,{\it k_2}-35\,{\it k_3}-35\cr}$$ Thus $k_3$ can be any of the three roots of the first polynomial, which are approximately $0.08858795951$, $0.4094668644$, and $0.7876594618$ (the cubic can be solved in terms of radicals, but it's not pretty). Then solve the next polynomials successively for $$ k_2 = -\frac{\it k_3}2+{\frac{9}{14}}\pm\frac{\sqrt {-147\,{{\it k_3}}^{2}+126\,{ \it k_3}-3}}{14}$$ $$ k_1 = -k_2 - k_3 + \frac{9}{7}$$ $$ c = -\frac{35}{36} k_3^2 + \frac{145}{144} k_3 + \frac{5}{36} $$ $$ b = \frac{35}{36} k_2 k_3 + \frac{35}{36} k_3^2 - \frac{35}{144} k_2 - \frac{5}{4} k_3 + \frac{5}{9}$$ $$ a = -\frac{35}{36} k_2 k_3 + \frac{35}{144} k_2 + \frac{35}{144} k_3 + \frac{35}{144}$$

Answer 2

Considering the symmetric nature of the system, it turns out we only need 2 equations. If we define the 3 roots of, $$35x^3 - 45x^2 + 15x - 1=0\tag1$$

as, $$k_1,k_2,k_3\, \approx\, 0.0885,\; 0.4094,\; 0.7876$$

in that order, then the three roots of,

$$n^6y^3-135n^4y^2+5925n^2y-84035=0\tag2$$

with $n=12$ are, $$a,b,c\, \approx\, 0.2204,\; 0.3881,\; 0.3288$$

in that order. These give your 6 unknowns $a,b,c,k_1,k_2,k_3$.