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I don't know the solution, but the thing that most bothers me about your preprint is the non-explicit nature of Theorem 2 and your inferences from it. Each $O(\,)$ has some "constant" implied in it, …

Close to $n/2$ is possible. I'll do odd $n$ and leave even $n$ for someone else's pleasure. Let $m=(n-1)/2$. For $i=0,\ldots,n-1$ and $j=1,\ldots,m$, let $M(i,j)$ be the pair $\{i-j,i+j\}$ (all val …

It is a terrible idea to divide out roots as they are found. There will be examples where the later roots are lost almost completely. See this wikipedia article for a famous and remarkably simple exa …

Modern mathematics libraries should be able to find $\log x$ precisely for all floating-point numbers, as the algorithms for doing that have long been known and adopted. My experiments on a fairly rec …

It seems fundamentally ill-conditioned. Since $\int_{-1}^{+1} x^rP_n(x),dx=0$ for $r=0,1,\dots,n-1$, your integral is unchanged if you subtract a polynomial of degree $n-1$ from $f(x)$. I'm guessing …

As Pietro says, $u=\exp(W(x))$ satisfies $u=x/\ln u$. This is a contraction mapping for large enough $u$, so just start with any old approximation, like $u=x/\ln x$, and do $u:=x/\ln u$ until it conv …

I suggest you try Gauss-Hermite integration. You can guess the precision by increasing the number of abscissae. Tables of abscissae and weights are here.

There are some issues that are not emphasised enough in the previous comments and answers. Having the source code used by an author does not let you check that the author's theorems are correct. It on …

Maple (2018 edition) can compute this sum numerically quite fast. I don't know the method it uses. > dP := unapply(diff(binomial(n+k,k),n),n,k); > Lsum := (k,s) -> evalf(Sum(dP(n,k)/binomial(n+k,k)^ …

Here is a little Maple. Note that using "sum" on the inside causes it to find an algebraic expression for the sum over $\ell$ and using "Sum" on the outside tells it to not try to sum that algebraica …

Federico's answer gives you the theory. Since you only ask for two abscissae, it is easy to solve this case by first principles. Just put in the abscissae as variables and do the integral. $f(-3^{-1/ …

The numbers 36,29,62 were obtained as the best match between the received signals and the output of computer simulations. The 90% confidence intervals on these numbers are about $\pm 4$. The details …

Version 2 Here is some experimental evidence. Like most people here I'm using $-f(x)$ and assuming $0<b<a<1$. Consider $b=1/100$ and $a=i/100$. For $2\le i\le 3$, there is a quadratic polynomial. F …