How to solve this nonlinear equation?

I tried to solve my equation using Solve[-(c1/g1) - c2/g2 - c3/g3 - c4/g4 - c5/g5 ==0 && -(c1/g1^2) - c2/g2^2 - c3/g3^2 - c4/g4^2 - c5/g5^2 == 1/2 && -(c1/g1^3) - c2/g2^3 - c3/g3^3 - c4/g4^3 - c5/g5^3 ==0 && -(c1/g1^4) - c2/g2^4 - c3/g3^4 - c4/g4^4 - c5/g5^4 == -3/8 && -(c1/g1^5) - c2/g2^5 - c3/g3^5 - c4/g4^5 - c5/g5^5 == 0 && -(c1/g1^6) - c2/g2^6 - c3/g3^6 - c4/g4^6 - c5/g5^6 == 1/8 &&c1 + c2 + c3 + c4 + c5 == 1 &&c1*g1 + c2*g2 + c3*g3 + c4*g4 + c5*g5 == 0 &&c1*g1^2 + c2*g2^2 + c3*g3^2 + c4*g4^2 + c5*g5^2 == -1/2 &&c1* g1^3 + c2*g2^3 + c3*g3^3 + c4*g4^3 + c5*g5^3 == 0,{c1,c2,c3,c4,c5,g1,g2,g3,g4,g5}], but it doesn't work. How can I solve it?

• You didn't specify what variable(s) you were solving for. – J. M. will be back soon Oct 23 '17 at 14:16
• You might want to have a look at Reduce. – Boson Oct 23 '17 at 14:20
• @J.M. variables are c1,c2,c3,c4,c5,g1,g2,g3,g4,g5 – BINGNAN Oct 23 '17 at 14:39
• Your question is not well formulated. Please consider defining it more clearly. – PureLine Oct 23 '17 at 15:06

Numeric solver NSolve says: NO solution.

a = 100;(*A dummy number.Bigger better:P *)

NSolve[-(c1/g1) - c2/g2 - c3/g3 - c4/g4 - c5/g5 ==
0 && -(c1/g1^2) - c2/g2^2 - c3/g3^2 - c4/g4^2 - c5/g5^2 ==
1/2 && -(c1/g1^3) - c2/g2^3 - c3/g3^3 - c4/g4^3 - c5/g5^3 ==
0 && -(c1/g1^4) - c2/g2^4 - c3/g3^4 - c4/g4^4 - c5/g5^4 == -3/
8 && -(c1/g1^5) - c2/g2^5 - c3/g3^5 - c4/g4^5 - c5/g5^5 ==
0 && -(c1/g1^6) - c2/g2^6 - c3/g3^6 - c4/g4^6 - c5/g5^6 == 1/8 &&
c1 + c2 + c3 + c4 + c5 == 1 &&
c1*g1 + c2*g2 + c3*g3 + c4*g4 + c5*g5 == 0 &&
c1*g1^2 + c2*g2^2 + c3*g3^2 + c4*g4^2 + c5*g5^2 == -1/2 &&
c1*g1^3 + c2*g2^3 + c3*g3^3 + c4*g4^3 + c5*g5^3 == 0 && -a < c1 <
a && -a < c2 < a && -a < c3 < a && -a < c4 < a && -a < c5 <
a && -a < g1 < a && -a < g2 < a && -a < g3 < a && -a < g4 <
a && -a < g5 < a, {c1, c2, c3, c4, c5, g1, g2, g3, g4, g5}, Reals]

(* {} *)

From Mathematica Documentation Center: • I forgot to mention, c1~c5 are reals, g1~g5 might have imaginary parts. – BINGNAN Oct 23 '17 at 16:00
• @BINGNAN. It does not change anything. – Mariusz Iwaniuk Oct 23 '17 at 16:07

You can even show analytically very fast, that this equations have no solution.

What you have is a linear matrix equation

matrix.cvector == bvector

where

matrix = {-{1/g1, 1/g2, 1/g3, 1/g4, 1/g5}, -{1/g1, 1/g2, 1/g3, 1/g4,
1/g5}^2, -{1/g1, 1/g2, 1/g3, 1/g4, 1/g5}^3, -{1/g1, 1/g2, 1/g3,
1/g4, 1/g5}^4, -{1/g1, 1/g2, 1/g3, 1/g4, 1/g5}^5, -{1/g1, 1/g2,
1/g3, 1/g4, 1/g5}^6, {1, 1, 1, 1, 1}, {g1, g2, g3, g4, g5}, {g1,
g2, g3, g4, g5}^2, {g1, g2, g3, g4, g5}^3};

cvector = {c1, c2, c3, c4, c5};

bvector = {0, 1/2, 0, -3/8, 0, 1/8, 1, 0, -1/2, 0};

Compare wiht your equations

eqs == And @@ Thread[(matrix.cvector == bvector)] // Simplify

(*     True     *)

LinearSolve says, that it has no solution:

LinearSolve[matrix, bvector]

(*  LinearSolve::nosol: Linear equation encountered that has no solution. >>  *)