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Student Support Forum: 'Polynomial roots' topicStudent Support Forum > General > "Polynomial roots"

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Paolo Rossi
10/15/99 11:11pm

Hi:
I am trying to obtain the roots for the secular equation listed below as a fuction of the eta parameter ''e'' using the student version of Mathematica 3.0. If I give ''e'' a certain numerical value, between 0 and 1 (actually, I tried e=2,10 and it still works) I get the correct eigenvalues (at least for e=0). However, if I don't specify the e value I get the nonsense listed below. It seems to me that if Mathematica can solve this ''special'' kind of quintic equation, it should be able to provide me with the expressions as a function of ''e'' that I need.
For secular equations of order 4 and below such expression are given.
I have tried some other random quintic polynomial which, even without the adjustable parameter, should normally not have analytical solutions (right??) and, as expected, the only thing I get is the pure function expression in the same form as listed below.
Could you tell me what is happening?

thanks
Paolo Rossi


In[88]:=

Solve[(x^5)-11*(3+e^2)*(x^3)-44*(1-e^2)*(x^2)+(44/3)*((3+e^2)^2)*x+48*(3+e^2)*(1-e^2)==0,x]''
Out[88]=

{{x Root[432 - 288 e^2 - 144 e^4 + 396 #1 + 264 e^2 #1 +
44 e^4 #1 - 132 #1^2 + 132 e^2 #1^2 - 99 #1^3 -
33 e^2 #1^3 + 3 #1^5&, 1]},
{ x Root[432 - 288 e^2 - 144 e^4 + 396 #1 + 264 e^2 #1 +
44 e^4 #1 - 132 #1^2 + 132 e^2 #1^2 - 99 #1^3 -
33 e^2 #1^3 + 3 #1^5&, 2]},
{ x Root[432 - 288 e^2 - 144 e^4 + 396 #1 + 264 e^2 #1 + 44 e^4 #1 - 132 #1^2 + 132 e^2 #1^2 - 99 #1^3 -
33 e^2 #1^3 + 3 #1^5&, 3]},
{ x Root[432 - 288 e^2 - 144 e^4 + 396 #1 + 264 e^2 #1 +
44 e^4 #1 - 132 #1^2 + 132 e^2 #1^2 - 99 #1^3 -
    33 e^2 #1^3 + 3 #1^5&, 4]},
{ x Root[432 - 288 e^2 - 144 e^4 + 396 #1 + 264 e^2 #1 +
44 e^4 #1 - 132 #1^2 + 132 e^2 #1^2 - 99 #1^3 -
33 e^2 #1^3 + 3 #1^5&, 5]}}

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