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if I have a discrete system x(n+1)= 0+y , y(n+1)= -0.5x +0
The fixed point (0,0) is center by Jacobian., I do not know how Plot this by using Mathematica code to get center. I am thankful if any one write the code
$\begingroup$Maybe this will give you some idea: {x[n],y[n]}/.RSolve[{x[n+1]==0+y[n], y[n+1]== -1/2x[n] +0},{x[n],y[n]},n][[1]] which returns {2^(-1-n/2)*((-I)^n+I^n)*C[1]+I*2^(-1/2-n/2)*((-I)^n-I^n)*C[2],(-I)*2^(-3/2-n/2)*((-I)^n-I^n)*C[1]+2^(-1-n/2)*((-I)^n+I^n)*C[2]} Make a table of values Table[%,{n,1,12}] which returns {{C[2],-C[1]/2},{-C[1]/2,-C[2]/2},{-C[2]/2,C[1]/4},{C[1]/4,C[2]/4},{C[2]/4,-C[1]/8},{-C[1]/8,-C[2]/8},{-C[2]/8,C[1]/16},{C[1]/16,C[2]/16},{C[2]/16,-C[1]/32},{-C[1]/32,-C[2]/32},{-C[2]/32,C[1]/64},{C[1]/64,C[2]/64}} where C[1] and C[2] are unknown coeffs due to no initial conds.$\endgroup$
{x[n],y[n]}/.RSolve[{x[n+1]==0+y[n], y[n+1]== -1/2x[n] +0},{x[n],y[n]},n][[1]]
which returns{2^(-1-n/2)*((-I)^n+I^n)*C[1]+I*2^(-1/2-n/2)*((-I)^n-I^n)*C[2],(-I)*2^(-3/2-n/2)*((-I)^n-I^n)*C[1]+2^(-1-n/2)*((-I)^n+I^n)*C[2]}
Make a table of valuesTable[%,{n,1,12}]
which returns{{C[2],-C[1]/2},{-C[1]/2,-C[2]/2},{-C[2]/2,C[1]/4},{C[1]/4,C[2]/4},{C[2]/4,-C[1]/8},{-C[1]/8,-C[2]/8},{-C[2]/8,C[1]/16},{C[1]/16,C[2]/16},{C[2]/16,-C[1]/32},{-C[1]/32,-C[2]/32},{-C[2]/32,C[1]/64},{C[1]/64,C[2]/64}}
where C[1] and C[2] are unknown coeffs due to no initial conds. $\endgroup$