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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?