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This is a classical problem in numerical methods research: evaluating the zeros of special functions. Many years of research have gone into devising efficient methods. The canonical starting point for anything related to this is the classic book by Abramowicz and Stegun. Scanned versions of the book can easily be found on the internet (I'm not sure if these ...


3

As with most questions about the computation of special functions, the Digital Librarary of Mathematical Functions is a good place to start. In particular, see chapter 6, which deals with the exponential integral and $\mbox{li}(x)$. You'll find that different methods (e.g. the power series for small $x$ versus asymptotic expansion for large $x$) work best ...


3

Yes, it is very well arguable that $\frac{e^x-1}{x}$ is ``as right as'' expm1. It appears frequently in applications; see for instance Sections 2.1 and 10.7.4 of Higham's Function of Matrices, which mention exponential integrators and matrix functions; this function is called $\psi_1(x)$ there. Your example, integrating exponentials, is another one, and it ...


2

expm1 is a primitive because for small $x$, the Taylor series for $e^x=1+x+x^2+\cdots$ adds small terms to one -- which implies that if $x$ is small enough, you lose almost all digits of accuracy other than the leading one and so the expression $e^x-1$ can not be accurately computed by first evaluating the exponential. The fact that you happen to have an ...


2

It looks like the eigenvalues depend smoothly on your $r$ parameter. Since the problem seems to be easy to solve for small values of $r$, you can compute the eigenvalues for a set of radii $r_1,r_2,r_3$, and then extrapolate each of these eigenvalues using a quadratic function to some $r^\ast>r_3$. This extrapolation should then be an excellent starting ...


2

Using the recurrence step and the normalized incomplete gamma function definition I simplified the formula to: $$\theta e^{k \log \left(\frac{o}{\theta }\right)-\frac{o}{\theta }- \ln{\Gamma (\kappa)}} +\theta \kappa Q \left(\kappa,\frac{o}{\theta }\right)-o Q \left(\kappa,\frac{o}{\theta }\right)+o+s$$ In this equation $Q$ stands for the normalized ...


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