Closed form solution for large eigenvalues

[Dustinsfl]

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]

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.