Talk:Elliptic integral

message: I would like to see more history, for example that many
mathmaticians tried to find a closed form (a formula) for them
until it was discovered that there is no closed form. 
(User Meni Taub) 
:Well, is that so interesting? Even if it's true, which it probably is. Probably the most important single case was the equation of the pendulum. Charles Matthews 19:27, 25 Sep 2004 (UTC)
thread_title:  More history 

message: The articles about complete elliptic integrals are rather short, I think they could be included in this article. It seems they have been here before, and were splitted later, but I could not find any discussion supporting that. In my opinion the coverage about complete elliptic integrals does not require a separate article for each one and the potential reader would be interested either on elliptic integrals as a whole, or a particular kind of elliptic integrals. Both requirements may be fulfilled in Elliptic integral. Rjgodoy 07:20, 26 May 2007 (UTC)
replies: text: I have transcluded these articles, so the change can be easily reverted if necessary. Rjgodoy 07:25, 26 May 2007 (UTC)

text: Note: Ellipse links Elliptic_integral#Complete_elliptic_integral_of_the_second_kind instead of Complete elliptic integral of the second kind. Rjgodoy 19:01, 9 July 2007 (UTC)

text: This page has been merged from Complete elliptic integral of the first kind and Complete elliptic integral of the second kind. Agreement by silence. Rjgodoy 19:01, 9 July 2007 (UTC)
thread_title:  Merge 

message: Since there is no closed form for elliptic integrals, and there are times when you need to calculate them outside the confines of a tool such as mathematica, I think it would be nice to open a discussion on the standard numerical methods used to calculate them.
— Preceding unsigned comment added by Abm2 at Wustl (talk • contribs) 17:34, 15 August 2007 (UTC)
thread_title:  Numerical methods 

message: I think that K(1)=ln(4) instead of K(1)=\infty. See Abramowitz and Stegun page 591 equation 17.3.26  —Preceding unsigned comment added by 67.172.132.138 (talk) 04:08, 31 March 2010 (UTC)
thread_title: Inccorect Expression for K(1)

message: The formula listed for E(1/4*(sqrt(6)-sqrt(2))) is erroneous.
The correct value of E(1/4*(sqrt(6)-sqrt(2))) = 1.5441504969146733661864210210267 not 1.2969170687813684714021678185866. The other values for E are correct
however. Given the Legendre relation one readily computes the correct expression for E(1/4*(sqrt(6)-sqrt(2))) as follows:
1/216*2^(1/3)*3^(3/4)*(27*GAMMA(2/3)^6+12*2^(1/3)*Pi^3+4*2^(1/3)*Pi^3*sqrt(3))/Pi/GAMMA(2/3)^3
or equivalently one has 
1/8/Pi*2^(1/3)*3^(3/4)*GAMMA(2/3)^3+1/18*2^(2/3)*3^(1/4)*Pi^2/GAMMA(2/3)^3*(1+sqrt(3))
This last result is based on an identity of Legendre as it appears on page 527 Section 22.81 of "A Course of Modern Analysis -- Whittaker &
Watson -- 4th Edition".
Orbtax 20:22, 29 June 2007 (UTC)
replies: text: The original formula for E(1/4*(sqrt(6)-sqrt(2))) has been edited to it's correct form in terms of GAMMA(1/3) consistent with the  companion formula for E(1/4*(sqrt(6)+sqrt(2))). Orbtax 22:44, 29 June 2007 (UTC)

text: Please post, at least in here all the forms of this so I may review them. Just dump all that you have on this of relating the gamma function with the elliptical integral of the second kind. I am most interested in it. Thank you. Numbertruth (talk) 00:48, 26 December 2012 (UTC)
thread_title: Incorrect expression for E(1/4*(sqrt(6)-sqrt(2)))

message: I've enjoyed copy editing math related articles and have learned a lot by doing so.  I hope that I have not bothered any of your work.  This page seems to be getting jumbled with section titles and equations.  I might do some re-organizing for mere aesthetic reasons as long as there are no objections. --Kenneth M Burke 19:50, 12 August 2007 (UTC)
replies: text: In overlooking the organization of the page, I think the section titles will probably be fine with some explanatory text.  I could undertake writing some paragraphs for the sections. I would have to find sources, unless there is someone who can suggest some. --Kenneth M Burke 17:46, 15 August 2007 (UTC)
thread_title:  Organization 

message: Hi, The article states that there is a relation between Schwarz-Christoffel transformations and elliptic integrals, but it doesn't say how. I've encountered no elliptic integrals in the few schwarz-christoffel transformations I've done so far, so what's the connection? Greetings,
Roger.  —Preceding unsigned comment added by 86.80.203.194 (talk) 17:03, 27 September 2007 (UTC)
thread_title:  Schwarz-Christoffel transformations 

message: This article is not geometric! There are no references to actual ellipses, and the relation of those angles to the angles in an ellipse (like true anomaly or eccentric anomaly) or to the ellipse's eccentricity. Albmont (talk) 17:16, 27 February 2009 (UTC)
thread_title:  Geometry 

message: When wiki is ready, they may post my new findings. I am coming up with more identities.
E(e^2)=(pi*(e-1)*e*(e+1)*(d/dx)[agm(1,(1-x=e)/(1+x=e))]-pi*(e-1)*agm(1,(1-e)/(1+e)))/(2*(e+1)*agm(1,(1-e)/(1+e))^2)
E(e^2)=2e^2(1-e^2)*d/dk(K(k=e^2))+(1-e^2)*pi/(2*AGM(1,sqrt(1-e^2)))
These are the best closed form there is for computation, but being I am not a PhD in math it will take years or decades until the world would get to see these readily...some day people will be assessed on their merit that goes beyond sitting in classrooms and spending money to obtain credentials.  — Preceding unsigned comment added by Numbertruth (talk • contribs) 08:02, 22 December 2012 (UTC)
thread_title:  New findings 

message: If one expands E(k) about k=±1 one finds that the coefficient of the k2 term in the series is infinite. As a consequence the series does not converge well for these values of the modulus which might be pointed out in the article. --Jbergquist (talk) 05:08, 8 August 2013 (UTC)
replies: text: If one tries to expand about k=1 one gets E(k)=1+(1-k)\int_0^{\frac{\pi}{2}}\frac{sin^2\phi}{cos\phi}d\phi-\frac{(1-k)^2}{2}\int_0^{\frac{\pi}{2}}\frac{sin^2\phi}{cos^3\phi}d\phi+\cdots with the integrals infinite. If one does the expansion using ε2=1-k2 about ε=0 one gets the the coefficient of the linear term equal to zero and that of the ε2 term infinite. --Jbergquist (talk) 07:29, 8 August 2013 (UTC)
thread_title:  Convergence of the series expansion of E(k) 

message: The lead discusses integrals of the form f(x)=\int _^R\left(t,{\sqrt {P(t)}}\right)\,dt, where r is a rational function and P(t) is a polynomial, yet in the rest of the article in concentrates on integrals with trigonometric functions. Would it make sense to discuss the trigonometric form in the lead, as this is the form most people will be familiar with?--Salix alba (talk): 21:48, 18 February 2015 (UTC)
thread_title:  Difference between lead and main text 

message: I think something is messed up in the display, symbols seem to be missing, making parts of the article hard to read. (See Argument Notation bullets)  Also, as a personal note, I found the equivalence of the complete integral 1st kind to a agm function to be very useful and I would love to see the 2nd kind get the same treatment. RDXelectric (talk) 14:20, 30 October 2016 (UTC)
thread_title:  Missing symbols 

message: In the section Incomplete elliptic integral of the first kind, the second equation reads,
This is the trigonometric form of the integral; substituting  and , one obtains Jacobi's form:
:  F(x ; k) = \int_{0}^{x} \frac{dt}{\sqrt{\left(1 - t^2\right)\left(1 - k^2 t^2\right)}}.
Should it not be 
:  F(x ; k) = \int_{0}^{arcsin x} \frac{dt}{\sqrt{\left(1 - t^2\right)\left(1 - k^2 t^2\right)}}.
since the limits have changed?
— Preceding unsigned comment added by 139.80.80.8 (talk) 02:12, 10 October 2017 (UTC)
thread_title:  Incomplete elliptic integral of the first kind 

message: The section Continued fraction is as follows in its entirety:
:A Continued fraction expansion is:
::\frac{K(k)}{2\pi}=-\frac{1}{4}+\sum^{\infty}_{n=0}\frac{q^n}{1+q^{2n}}=-\frac{1}{4}+\frac{1}{1-q}+\frac{(1-q)^2}{1-q^3}+\frac{q(1-q^2)^2}{1-q^5}+\frac{q^2(1-q^3)^2}{1-q^7}+\frac{q^3(1-q^4)^2}{1-q^9}+\ldots,
:where the nome is .
What is the connection to a continued fraction? As stated, there is no continued fraction mentioned. 2601:200:C000:1A0:5D0:73D9:68BF:7D30 (talk) 17:59, 16 March 2021 (UTC)
thread_title:  What does this have to do with a "continued fraction"? 

message: There's a near-equation in this section, viz.: 
K\left(k\right)\approx\frac{\pi}{2}+\frac{\pi}{8}\frac{k^2}{1-k^2}-\frac{\pi}{16}\frac{k^4}{1-k^2}
Shouldn't the last denominator be fixed to read:
K\left(k\right)\approx\frac{\pi}{2}+\frac{\pi}{8}\frac{k^2}{1-k^2}-\frac{\pi}{16}\frac{k^4}{1-k^4}
? k just looks wrong. Of course since this is just an approximation anyway, it's arguable that k and k both work here.
In fact, I'd like to see the derivation of this approximation. Is it available anywhere? This entire section could use a few citations... JohnH~enwiki (talk) 20:27, 9 August 2023 (UTC)
thread_title:  Typo in "Asymptotic expressions"?