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How would I solve this for $\lambda$?

$\tan{\sqrt{\lambda}}=\frac{\sqrt{\lambda}(10e+1)}{\lambda-10e}$

It came from an eigenvalue problem I have been working on.

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  • $\begingroup$ @Eoin it has $\tan{(\sqrt{\lambda})}$ how is it quadratic? $\endgroup$
    – sci-guy
    Commented Mar 8, 2015 at 7:03

1 Answer 1

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Beside the trivial root $\lambda=0$, there is no explicit solution to this equation and only numerical methods could solve the problem.

Plotting the function, you probably noticed that the first root is close to $\lambda=4$. So, let us use Newton method which is one of the simplest root-finder. Starting from a "reasonable" guess $\lambda_0$, it will ypdate it according to $$\lambda_{n+1}=\lambda_n-\frac{f(\lambda_n)}{f'(\lambda_n)}$$ So, let us try with $$f(\lambda)=\tan \left(\sqrt{\lambda }\right)-\frac{(1+10 e) \sqrt{\lambda }}{\lambda -10 e}$$ $$f'(\lambda)=\frac{1}{2\sqrt \lambda}\Big(\frac{(1+10 e) (\lambda +10 e)}{(\lambda -10 e)^2}+\sec ^2\left(\sqrt{\lambda }\right)\Big)$$ and let us start with $\lambda_0=4$. The method will then generates the following iterates : $3.86703$, $3.87506$, $3.87510$ which is the solution for six significant digits.

Surprizingly, according to $RIES$, this solution seems to be very close to $$\lambda_*=\frac{8} {\pi }\tan ^{-1}\left(\frac{64}{\pi }\right)$$ for which $f(\lambda_*)=4.6 \times 10^{-6}$ while $f(3.87510)=4.1 \times 10^{-6}$. So, using $\lambda_0=\lambda_*$, a single iteration of Newton method would lead to $\lambda \approx 3.8750980471$ to which would correspond $f(\lambda)=-5.5 \times 10^{-12}$.

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  • $\begingroup$ How can I get the next iteration? Should be around 22.5351... and the next, and the next... $\endgroup$
    – sci-guy
    Commented Mar 8, 2015 at 7:59
  • $\begingroup$ Just do the same but you need to find a "good" starting point. Zoom the graph. They would be $22.5$, $59.5$, $115$ and so on. $\endgroup$ Commented Mar 8, 2015 at 8:07

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