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Periodic Functions

  • Thread starter Nusc
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  • #1
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Homework Statement


What kind of conditions do eigenvalues impose to ensure periodicity?
Is it plausible to say that irrational multiples of eigenvalues imply no harmonic oscillations, if so why?

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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No, that doesn't make a whole lot of sense. Eigenvalues per se have nothing to do with periodicity.
 
  • #3
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[tex]

\left[\begin{array}{cccc}0 & -a & 0 & 0 \\ -a & 0 & -b & 0 \\ 0 & -b & 0 & -c \\ 0 & 0 & -c & 0\end{array}\right]

[/tex]

For any fixed a, I want to find b and c in terms of a such that lambda_i / lambda_ j is a rational number for every i,j=1,2,3,4 .

See, before I was working with

[tex]


\left[\begin{array}{cccc}0 & -a & 0 & 0 \\ -a & 0 & -a & 0 \\ 0 & -a & 0 & -a \\ 0 & 0 & -a & 0\end{array}\right]


[/tex]

But that gave me irrational e-values if I put it into Maple and I don't want that.
 
Last edited:
  • #4
Dick
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[tex]

\left[\begin{array}{cccc}0 & -a & 0 & 0 \\ -a & 0 & -b & 0 \\ 0 & -b & 0 & -c \\ 0 & 0 & -c & 0\end{array}\right]

[/tex]

For any fixed a, I want to find b and c in terms of a such that lambda_i / lambda_ j is a rational number for every i,j=1,2,3,4 .

See, before I was working with

[tex]


\left[\begin{array}{cccc}0 & -a & 0 & 0 \\ -a & 0 & -a & 0 \\ 0 & -a & 0 & -a \\ 0 & 0 & -a & 0\end{array}\right]


[/tex]

But that gave me irrational e-values if I put it into Maple and I don't want that.
Now that makes more sense. It also looks like a hard question. How about a=1, b=0, c=1? Is that good enough? What's this for anyway?
 
  • #5
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c and b must be in terms of a.

But forget that for now, how would I show that the eigenvalues are irrational for n>3 ? That's why I asked this guy in this thread https://www.physicsforums.com/showthread.php?t=224954&page=3

how to find an equation for the eigenvalues.
 
  • #6
Dick
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You can certainly get an expression for the eigenvalues in terms of a,b and c. You get a quartic equation to solve, but it only has l^4, l^2 and a constant term. So you can solve it with the quadratic equation. But how you enforce the condition that the ratio of roots is rational, I have no idea.
 

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