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[SOLVED] closed form solution for large eigenvalues

dwsmith

Well-known member
Feb 1, 2012
1,673
The eigenvalues are found by
$$
\tan\lambda_n = \frac{1}{\lambda_n}
$$
For large eigenvalues, the intersection get closer and closer to $\lambda_n = \pi k$ where $k\in\mathbb{Z}^+$ and $k > 15$.
Is this correct? Without arbitrary picking a $k$, is there a better way to determine a $k$ for when $\lambda\to\pi k$?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
The eigenvalues are found by
$$
\tan\lambda_n = \frac{1}{\lambda_n}
$$
For large eigenvalues, the intersection get closer and closer to $\lambda_n = \pi k$ where $k\in\mathbb{Z}^+$ and $k > 15$.
Is this correct? Without arbitrary picking a $k$, is there a better way to determine a $k$ for when $\lambda\to\pi k$?
\[\tan\lambda_n = \frac{1}{\lambda_n}\]

\[\Rightarrow \lambda_n=k\pi+\tan^{-1}\frac{1}{\lambda_n}\mbox{ where }k\in\mathbb{Z}\]

For large positive or negative values of \(\lambda_n\), \(\tan^{-1}\frac{1}{\lambda_n}\approx 0\)

\[\therefore \lambda_n\approx k\pi\mbox{ where }k\in\mathbb{Z}\]

The larger \(k\) value you use the approximation will be better. Choosing a bound for \(k\) depends on the accuracy that you need for your approximation.