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From: TRan Thanh on 28 Jan 2010 05:35
I writed this code in Matlab.I don't know why results have '_Z'.I have never seen them before.
syms m1 m2 c a l ;
g=10;
M=[m1+m2 m2*l 0 0 0 0;0 0 m1+m2 m2*l 0 0;0 0 0 0 m1+m2 m2*l; 1 l 0 0 0 0;0 0 1 l 0 0;0 0 0 0 1 l ]
K=[2*c 0 -c 0 0 0;-c 0 2*c 0 -c 0;0 0 -c 0 c 0; 0 g 0 0 0 0;0 0 0 g 0 0;0 0 0 0 0 g]
A=inv(M)*K;
[V,D]=eig(A);
omega1=sqrt(D(1,1));
omega3=sqrt(D(3,3))
omega5=sqrt(D(5,5));
And results return:
omega3 =

RootOf(_Z^6*l^3*m1^3+(-30*l^2*m1^2*m2-30*l^2*m1^3-5*l^3
*m1^2*c)*_Z^5+(100*l^2*m1*c*m2+6*l^3*m1*c^2+150*l^2
*m1^2*c+600*l*m1^2*m2+300*l*m1*m2^2+300*m1^3*l)*_Z^4
+(-60*c^2*l^2*m2-2000*c*l*m1*m2-500*c*l*m2^2-180*m1*c^2*l^2-1500
*m1^2*c*l-1000*m1^3-3000*m1^2*m2-3000*m1*m2^2-1000
*m2^3-c^3*l^3)*_Z^3+(5000*m1^2*c+5000*m2^2*c+10000*m2*m1*c
+30*c^3*l^2+1200*c^2*m2*l+1800*c^2*l*m1)*_Z^2+(-6000*c^2*m1-6000
*c^2*m2-300*c^3*l)*_Z+1000*c^3,index = 3)^(1/2)

Everyone help me,please!
Th(a)nh cô nương.

From: Steven Lord on 28 Jan 2010 09:35

"TRan Thanh" <xt2005_1208dx(a)yahaoo.com> wrote in message
news:hjrp8s$pu9$1(a)fred.mathworks.com...
>I writed this code in Matlab.I don't know why results have '_Z'.I have
>never seen them before.
>
> syms m1 m2 c a l ;
> g=10;
> M=[m1+m2 m2*l 0 0 0 0;0 0 m1+m2 m2*l 0 0;0 0 0 0 m1+m2 m2*l; 1 l 0 0 0 0;0
> 0 1 l 0 0;0 0 0 0 1 l ]
> K=[2*c 0 -c 0 0 0;-c 0 2*c 0 -c 0;0 0 -c 0 c 0; 0 g 0 0 0 0;0 0 0 g 0 0;0
> 0 0 0 0 g]
> A=inv(M)*K;
> [V,D]=eig(A);
> omega1=sqrt(D(1,1));
> omega3=sqrt(D(3,3))
> omega5=sqrt(D(5,5));
> And results return:
> omega3 =
>
> RootOf(_Z^6*l^3*m1^3+(-30*l^2*m1^2*m2-30*l^2*m1^3-5*l^3
> *m1^2*c)*_Z^5+(100*l^2*m1*c*m2+6*l^3*m1*c^2+150*l^2
> *m1^2*c+600*l*m1^2*m2+300*l*m1*m2^2+300*m1^3*l)*_Z^4
> +(-60*c^2*l^2*m2-2000*c*l*m1*m2-500*c*l*m2^2-180*m1*c^2*l^2-1500
> *m1^2*c*l-1000*m1^3-3000*m1^2*m2-3000*m1*m2^2-1000
> *m2^3-c^3*l^3)*_Z^3+(5000*m1^2*c+5000*m2^2*c+10000*m2*m1*c
> +30*c^3*l^2+1200*c^2*m2*l+1800*c^2*l*m1)*_Z^2+(-6000*c^2*m1-6000
> *c^2*m2-300*c^3*l)*_Z+1000*c^3,index = 3)^(1/2)

You're trying to symbolically compute the eigenvalues of a 6-by-6 symbolic
matrix. That would require solving a 6th order polynomial. Galois proved
that there's no general formula for the solutions to a general 6th order
polynomial in radicals, so the Symbolic Math Toolbox returns a result that
indicates that the result you're looking for is the root of a 6th order
polynomial in a "placeholder" variable _Z. If you were to substitute values
into that expression for your parameters (m1, m2, c, a, and l) you could
then try to evaluate this using DOUBLE or VPA to obtain a numeric result.

--
Steve Lord
slord(a)mathworks.com
comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ


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