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 Jul 15 comment Second derivative numerical estimate - stability and approach …nice only if the underlying function is smooth. If OP's discrete samples have some error in them, Richardson does no good. Jul 15 revised Why is the Gamma function off by 1 from the factorial? edited tags Jul 15 comment Derivation of approximation of Error function I'm now telling you to try the economization procedure. Again: Maclaurin, convert to Chebyshev, truncate, and go back to the monomials. You should then see something like what Mr. Hastings had, or better. Abramowitz and Stegun should have the table for converting monomials to Chebyshev. Jul 15 comment Derivation of approximation of Error function Well, binomial won't get you anywhere near Mr. Hastings. If you'd tried looking up the things I told you to look up, you'd have already read about the "equi-ripple" and "minimax" properties of a Chebyshev approximation… Jul 15 comment Derivation of approximation of Error function Well, have you tried expanding the transformed function as a Maclaurin series, and re-expressing the monomials in Chebyshev terms? (As I said earlier, look up Chebyshev economization.) Jul 15 comment Derivation of approximation of Error function With respect to Winitzki: the reasoning is that you want to build a rational function (or any approximant, really) whose qualitative behavior at $0$ and $\infty$ is similar to $\mathrm{erf}(x)$; polynomials certainly cannot do that, since they cannot exhibit asymptotes as in this case. Jul 15 comment Derivation of approximation of Error function Ah, that parameter I am not too sure of; Cecil Hastings was (in)famous for coming up with clever approximations that seemed to have just come from thin air. Jul 15 comment Conditions on Matrix invertibility I was nudging you to search for them yourself, as I have now given you the words you should be looking for. Jul 15 comment Derivation of approximation of Error function Let's call the polynomial in your expression $p(t)$; you can rearrange your formula to obtain $\exp(x^2)\mathrm{erfc}(x)\approx p(t)$. Replace the $x$ on the left with an expression in terms of $t$. That is the function which you will be approximating as a series in Chebyshev polynomials. You might want to look up the literature on this, including "economization" of series. Jul 15 comment Derivation of approximation of Error function That was one of his papers; there was another one that specially dealt with $\mathrm{erf}$. Jul 15 comment Conditions on Matrix invertibility There are methods for updating the QR decomposition and singular value decomposition when a new column or row is added to the original matrix; you might want to look them up. Jul 15 comment Derivation of approximation of Error function These are, if memory serves, Chebyshev fits of a transformed version of the error function. There are nicer and more analytically tractable approximations now; you might want to search for the work of Serge Winitzki on this. Jul 15 revised Derivation of approximation of Error function edited tags Jul 15 comment Logarithmic derivative of Polygamma functions Well, your logarithmic derivative is merely $\frac{\psi^{(k+1)}(x)}{\psi^{(k)}(x)}$; as to why one would be interested in ratios of (shifted, generalized) harmonic numbers, I've no idea. Jul 15 revised Logarithmic derivative of Polygamma functions edited tags Jul 14 comment How to learn cryptography Wouldn't crypto.SE be a better place for this? Jul 14 revised What's the area of the shape defined by all points whose distances from two focal points multiply to give the same product? edited tags Jul 14 comment What's the area of the shape defined by all points whose distances from two focal points multiply to give the same product? What you have called a "multiplicoid" is classically called a Cassinian oval, after the astronomer. The link I have given has formulae for the area of Cassinian ovals depending on shape. In your particular case, the formula involves the complete elliptic integral of the second kind. Jul 14 comment Find the roots of this 6th degree polynomial Well, you know what the three cube roots of $1$ look like, no? You can multiply those with the cube root you already have. Jul 14 comment Divergent series whose terms converge to zero In any event: have you seen this?