# [SOLVED]closed form solution for large eigenvalues

#### dwsmith

##### Well-known member
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
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.