Talk:Carlson symmetric form

Bille C. Carlson
shouldn't there be bio of the one who invented this form of Elliptic Integrals? Bille Carlson. He did very important work. According to http://dlmf.nist.gov/about/bio/BCCarlson, he is Professor Emeritus in the Department of Mathematics and Associate of the Ames Laboratory (U.S. Department of Energy) at Iowa State University, Ames, Iowa.

Bille Carlson recently passed away (June 27, 1924 - August 16, 2013). He has a doctorate in Physics from Oxford. — Preceding unsigned comment added by 70.35.36.66 (talk) 19:23, 11 February 2014 (UTC)

Duplication formula
Copied from User talk:47.53.231.196:


 * I suspect that the following formula is uncorrect (see the article for the notations).


 * $$\begin{align}R_{J}(x,y,z,p) & = 2 R_{J}(x + \lambda,y + \lambda,z + \lambda,p + \lambda) + 6 R_{C}(d^{2},d^{2} + (p - x) (p - y) (p - z)) \\

& = \frac{1}{4} R_{J}\left( \frac{x + \lambda}{4},\frac{y + \lambda}{4},\frac{z + \lambda}{4},\frac{p + \lambda}{4}\right) + 6 R_{C}(d^{2},d^{2} + (p - x) (p - y) (p - z)) \end{align}$$


 * where $$d = (\sqrt{p} + \sqrt{x}) (\sqrt{p} + \sqrt{y}) (\sqrt{p} + \sqrt{z})$$ and $$\lambda = \sqrt{x y} + \sqrt{y z} + \sqrt{z x}$$


 * I found a different statement of the same "duplication theorem" in Carlson's paper "Computing Elliptic Integrals by Duplication" (Numerische Mathematik 33, 1979). The article is avalaible at http://www.digizeitschriften.de/index.php?id=resolveppn&PID=PPN362160546_0033 and the alternative formula is cited also in https://dlmf.nist.gov/19.26#iii. Conversely, I was not able to find any (open access) reference for the formula above.
 * If anyone is interested in expanding this page, maybe it would be better to use this version of the formula, being it openly retievable in literature. Anyway, at the moment the page seems fairly stubbish.


 * B. C. Carlson NUMERICAL COMPUTATION OF REAL OR COMPLEX ELLIPTIC INTEGRALS has (eq. 22)


 * $$R_{J}(x,y,z,p) = \frac{1}{4} R_{J}\left( \frac{x + \lambda}{4},\frac{y + \lambda}{4},\frac{z + \lambda}{4},\frac{p + \lambda}{4}\right) + \frac{6}{d} R_{C}\left( 1,1 + \frac{(p - x) (p - y) (p - z)}{d^{2}}\right)$$


 * $$\lambda = \sqrt{x y} + \sqrt{y z} + \sqrt{z x}$$


 * $$d = (\sqrt{p} + \sqrt{x}) (\sqrt{p} + \sqrt{y}) (\sqrt{p} + \sqrt{z})$$


 * which is the same (given $$R_{C}(x,y) = k R_{C}(k^{2} x,k^{2} y)$$)


 * Also, the NIST Digital Library of Mathematical Functions has (19.26.22)


 * $$R_{J}(x,y,z,p) = 2 R_{J}(x + \lambda,y + \lambda,z + \lambda,p + \lambda) + 3 R_{C}(\alpha^{2},\beta^{2})$$


 * $$\lambda = \sqrt{x} \sqrt{y} + \sqrt{y} \sqrt{z} + \sqrt{z} \sqrt{x}$$


 * $$\alpha = p (\sqrt{x} + \sqrt{y} + \sqrt{z}) + \sqrt{x} \sqrt{y} \sqrt{z}$$


 * $$\beta = \sqrt{p} (p + \lambda)$$


 * but also has (19.26.26)


 * $$R_{C}(x,y) = 2 R_{C}(x + y + 2 \sqrt{x} \sqrt{y},2 y + 2 \sqrt{x} \sqrt{y})$$


 * so


 * $$\begin{align}R_{J}(x,y,z,p) & = 2 R_{J}(x + \lambda,y + \lambda,z + \lambda,p + \lambda) + 3 R_{C}(\alpha^{2},\beta^{2}) \\

& = 2 R_{J}(x + \lambda,y + \lambda,z + \lambda,p + \lambda) + 6 R_{C}(\alpha^{2} + \beta^{2} + 2 \alpha \beta,2 \beta^{2} + 2 \alpha \beta) \\ & = 2 R_{J}(x + \lambda,y + \lambda,z + \lambda,p + \lambda) + 6 R_{C}((\alpha + \beta)^{2},(\alpha + \beta)^{2} + \beta^{2} - \alpha^{2}) \\ & = 2 R_{J}(x + \lambda,y + \lambda,z + \lambda,p + \lambda) + 6 R_{C}(d^{2},d^{2} + (p - z) (p - y) (p - x)) \end{align}$$


 * (using


 * $$(\alpha + \beta)^{2} = ((\sqrt{p} + \sqrt{x}) (\sqrt{p} + \sqrt{y}) (\sqrt{p} + \sqrt{z}))^{2} = d^{2}$$


 * $$\begin{align}\beta^{2} - \alpha^{2} & = p^{3} + p x y + p y z + p z x - p^{2} x - p^{2} y - p^{2} z - x y z \\

& = (p - z) (p - y) (p - x) \end{align}$$


 * either from first principles or 19.26.22) --catslash (talk) 01:00, 25 July 2018 (UTC)

Unfortunately, there is no clear definition for the Legendre and Carlson Elliptic integrals for all inputs. As a result, both duplication formulas can technically be used interchangeably, provided you know where all jump discontinuities are so you don't run into any issues.

I'm currently trying to implement the elliptic integrals into a calculator I'm programming with JavaScript, so this is actually very frustrating for me. As of writing this, I have found a well behaved version of the elliptic F and E functions. I haven't, however, had the same luck with the elliptic Π function. :( Math Machine 4 (talk) 01:14, 29 July 2020 (UTC)

UPDATE: I now know why they're called duplication formulas, and know how to prove that they work. There exists an identity:
 * $$\begin{align} F (\theta | k ) = 2 csgn(\sin(\theta)) F \left(\sin ^{(-1)} \left( \frac{\sqrt{1-\sqrt{1-k \sin^{2}(\theta)}}}{\sqrt{k(1+\cos(\theta))}}\right)|k\right) +4 K(k) \left[\frac{Re(\theta)}{2 \pi} \right] \end{align}$$

With csgn being a variant of the sgn function, returning 1 if the real part is positive or if the real part is 0 and the imaginary part is positive, returning -1 otherwise. In other words, it returns 1 if the input is the square root of something, -1 if it's the negative square root of something.

This can be proven without too much difficulty to be true. If you take the derivative of both sides, you'll find them to be equal, and they both yield 0 if \theta is 0. In addition, if we accept the premise that
 * $$\begin{align} F (\theta | k) = \sin(\theta) R_{F} (\cos ^{2}(\theta), 1-k\sin ^{2}(\theta),1) \end{align}$$

then it is not too easy to prove the duplication formula for R_{F}.

Please note, I had to figure this out myself. As you pointed out, most places where they tell you about these duplication formulas are only available for paid access. With that said, I can assume these are called duplication formulas because they are derived from identities whereby you take the elliptic integral of one angle, double it, and it yields the elliptic integral of another angle (with the k term remaining the same).

Likewise, it's also not too difficult to prove the same is true for the duplication formulas for the R_{D} and R_{J} functions. There exists an identity similar to the elliptic F function's identity for both the E and \PI functions, whereby the only difference is that you add on an extra term at the end. Likewise, given the premise that the R_{F}, R_{D}, and R_{J} functions can be used to find the E and \PI functions, we find that that extra term translates into that extra R_{C} term at the end of the duplication formulas. Keep in mind, I say these aren't too hard to prove. When I say that, I mean you could prove them given the information I've given you, but even I didn't have the time on my hands to do so myself.

Now, in terms of there being two different variants of the duplication formula for the R_J function, here's what I have to say. Basically, the R_C term at the end of the R_J duplication formula supplied by Wikipedia, when reduced using arctangents, looks like this:


 * $$\begin{align} \frac{6 \tan ^{(-1)} \left( \frac{\sqrt{(p-x)(p-y)(p-z)}}{(\sqrt{p}+\sqrt{x})(\sqrt{p}+\sqrt{y})(\sqrt{p}+\sqrt{z})} \right)}{\sqrt{(p-x)(p-y)(p-z)}} \end{align}$$

whereas the equivalent supplied by DLMF yields:


 * $$\begin{align} \frac{3 \tan ^{(-1)} \left( \frac{\sqrt{(p-x)(p-y)(p-z)}}{p(\sqrt{x}+\sqrt{y}+\sqrt{z})+\sqrt{x}\sqrt{y}\sqrt{z}} \right)}{\sqrt{(p-x)(p-y)(p-z)}} \end{align}$$

If we apply the tangent angle addition formulas, we see that these two expressions are equivalent. Or rather, they're sometimes equivalent, other times, they will be off by a factor of ±3π/(the ugly denominator in both above formulas).

I'd say the DLMF formula is better overall, as it's less computationally expensive and yields much, much, MUCH fewer branch cuts. However, they both yield branch cuts no matter what. Technically, the elliptic PI function (as well as the R_J function alone) must have branch cuts, due to their derivative being discontinuous over the complex plane. However, even with that said, both variations of the duplication formula yield unnecessary branch cuts which are very difficult to remove.

Overall, while they are both equivalent (mod 2π), I actually would recommend switching out the duplication formula currently in the Wikipedia page (July 2020) with the one given in the DLMF page.

I would do it myself, but I'm not comfortable making large edits to articles that I cannot fully back myself up on, and I am currently very, very tired of the elliptic integral of the third kind, and already have my hands full trying to make sure I didn't make any mistakes with the elliptic integrals of the first and second kinds. Math Machine 4 (talk) 23:47, 29 July 2020 (UTC)

Okay, I've also found proof that the Carlson symmetric forms work for elliptic integrals.

First, construct the integral expressions for R_F, R_D, and R_J. Next, plug in the values for x, y, z, and p (cos²(θ), 1-ksin²(θ), 1, and 1-nsin²(θ) respectively). For the sake of avoiding absolute value bars, we're going to assume θ is between 0 and π/2 inclusive.

Next, apply the substitution u=t+1. Your integrals should now range from 1 to ∞. After that, apply another substitution v=1/u. After that step, you'll notice the integral is from 1 to 0, and has a negative sign. Feel free to swap the bounds and remove the negative sign.

Now, apply one more transformation, φ=arcsine(√(v)sin(θ)). After this, you should be able to simplify a lot of things. Remember your cosine and sine will both be positive.

Once all is said and done, you're basically done. F(θ|k) = sin(θ)*R_F(cos²(θ),1-ksin²(θ),1). When you plug in the integral expression, that'll simplify to F(θ|k)=∫1/√(1-ksin²(φ))dφ from 0 to π/2. E(θ|k) = sin(θ)*R_F(cos²(θ),1-ksin²(θ),1)-k*sin³(θ)/3*R_D(cos²(θ),1-ksin²(θ),1). When you plug in that integral expression, you'll get E(θ|k)=∫√(1-ksin²(φ))dφ from 0 to π/2. And if you do the same thing for Π(n;θ|k), you'll also get the integral definition. Math Machine 4 (talk) 18:41, 19 December 2021 (UTC)

Include R_G
I've added the definition of R_G, one of the standard symmetric integrals. I'm not quite sure why it was omitted. Simpler formulas involving R_G should be added in the following sections. cffk (talk) 13:52, 16 April 2024 (UTC)