split the set of Eigenvalues

55 Views Asked by At

I have a question on how to split a set of eigenvalues.

I have a set of $n$ eigenvalues denoted by $\lambda_{m}$ for $=0...n-1$ given as

$$ \lambda_{m}=1 -\dfrac{1}{K} \left( \dfrac{\sin(\frac{m \pi}{n}(K+1))}{\sin(\frac{m \pi}{n})} - 1 \right), $$ where $K$ and $n$ are finite numbers and I assume here that $K <n$ .

I am interested only on examining the eigenvalues $0<\lambda_{m} \ll 1$, how can I represent them? Should I just select $m$ for which the eigenvalues will be small? But I don't know at which $m$ I should stop.

Is it possible to find a general expression for the small eigenvalues without the $m$ inside?

1

There are 1 best solutions below

0
On

There is no general definition of $\lambda_m \ll 1,$ therefore, the $m$ at which to stop is very application specific. However, one way to get a good intuition for the behaviour of small values of $\lambda_m$ is by

  1. Relaxing $\frac{m\pi}{n} \to x.$
  2. Noting that the small values lie around zero and $\pi.$
  3. Obtaining the Taylor expansion of the relaxed $\lambda_x$ around these points and approximating the Taylor series to the first few terms.