Algebra equation with symbolic
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HYEOKJUNE LEE
on 3 May 2018
Answered: Walter Roberson
on 3 May 2018
Hello,
I try to solve the 4th order equation with symbolic, but the answer is not numerical numbers, see below:
My code is
m1 = 100; %kg
m2 = 10; %kg
c1 = 1;
c2 = 1;
c3 = 1;
k1 = 100;
k2 = 100;
k3 = 100;
%
M(1,1) = m1;
M(2,2) = m2
%
C(1,1) = (c1+c2);
C(1,2) = -c2;
C(2,1) = -c2;
C(2,2) = (c2+c3)
%
K(1,1) = (k1+k2);
K(1,2) = -k2;
K(2,1) = -k2;
K(2,2) = (k2+k3)
%
a0 = M(1,1)*M(2,2)
a1 = M(1,1)*C(2,2) + M(2,2)*C(1,1)
a2 = M(1,1)*K(2,2) + C(1,1)*C(2,2) + M(2,2)*K(1,1) - C(1,2)*C(2,1)
a3 = C(1,1)*K(2,2) + K(1,1)*C(2,2) - C(1,2)*K(2,1) - C(2,1)*K(1,2)
a4 = K(1,1)*K(2,2) - K(1,2)*K(2,1)
%
syms w
%
func = a0*w^4 + a1*w^3 + a2*w^2 + a3*w^1 + a4
%
wsol = solve(func,w)
then, the matlab give me a solution which format is root(σ1, z, 1). The sigma is the above function.
How can I get the solution?
Thank you.
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Accepted Answer
John D'Errico
on 3 May 2018
Edited: John D'Errico
on 3 May 2018
4 roots, all of which are complex.
vpa(wsol)
ans =
- 0.10269696007084728245763107930116 - 4.53087688516932263934459193381i
- 0.10269696007084728245763107930116 + 4.53087688516932263934459193381i
- 0.0073030399291527175423689206988387 + 1.208534091963622145606298796162i
- 0.0073030399291527175423689206988387 - 1.208534091963622145606298796162i
If you plot func, you will see that it never crosses zero.
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More Answers (1)
Walter Roberson
on 3 May 2018
wsol = simplify(solve(func, w, 'MaxDegree', 4));
This will give you the numeric solutions, such as
((-1)^(1/4)*10^(1/4)*230339930457^(3/4)*(13099491187973 + 159912003^(1/2)*1638000000i)^(1/4)*(- 159912003^(1/2)*1638000000i - 13099491187973)^(1/6)*(2*33315^(1/2)*(- 5456997*30^(1/2)*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/2) - 4218680045*10^(1/2)*(8437360090 + 10*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(2/3) - 439697*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/3))^(1/2) - 439697*10^(1/2)*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/3)*(8437360090 + 10*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(2/3) - 439697*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/3))^(1/2) - 5*10^(1/2)*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(2/3)*(8437360090 + 10*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(2/3) - 439697*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/3))^(1/2))^(1/2) + 10^(3/4)*2221^(1/2)*(33*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/6)*(8437360090 + 10*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(2/3) - 439697*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/3))^(1/4) + 3^(1/2)*(8437360090 + 10*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(2/3) - 439697*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/3))^(3/4)))*(159912003^(1/2)*8793940000000i + 10*(293032087997 + 159912003^(1/2)*20000000i)*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(1/3) + 133040111*(- 13099491187973 + 159912003^(1/2)*1638000000i)^(2/3) + 16594597703959910)^(1/4)*1i)/7769568131425052256545514000
You should consider whether you actually want the numeric solutions, or if you want approximate results instead, such as the ones John showed.
If what you want is the approximate results then:
wsol_approx = vpasolve(func);
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