Timeline for an identity between two elliptic integrals
Current License: CC BY-SA 4.0
28 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jan 9, 2020 at 5:53 | history | bounty ended | mark | ||
| S Jan 9, 2020 at 5:53 | history | notice removed | mark | ||
| Jan 5, 2020 at 15:08 | comment | added | mark | @Carlo Beenakker Yes, you are right. I miscopied the identity...$\theta$ should be equated to $2\arctan 3^{-\frac{1}{4}}$. Thank you. | |
| Jan 4, 2020 at 20:53 | comment | added | Carlo Beenakker | I think $\theta$ should be equated to $2\arctan 3^{-1/4}$, see the answer box where I have tried to work this out in some detail. | |
| Jan 4, 2020 at 19:52 | answer | added | Carlo Beenakker | timeline score: 3 | |
| Jan 4, 2020 at 16:31 | comment | added | mark | Two edits in the bounty offer: $theta$ should be replaced by $\theta$ and $\arctan(1/\sqrt(3)$ should be replaced by $\arctan(1/\sqrt(\sqrt(3))$ | |
| S Jan 4, 2020 at 16:26 | history | bounty started | mark | ||
| S Jan 4, 2020 at 16:26 | history | notice added | mark | Draw attention | |
| Dec 31, 2019 at 20:26 | comment | added | user145307 | @ToddTrimble, no problem! I agree with you that the way it was phrased in this particular example was non-optimal and I certainly don't know about past histories. | |
| Dec 31, 2019 at 19:09 | comment | added | Todd Trimble | @ElectricPenguin Thanks. Interesting point. I do question your assertion that there is more skill at Mathematics for solving such problems, but next time I see a case where this community finds a definite integral question on-topic and is defeated by it, I'll think about it. But based on a long history, it seems user64494 says the same thing just about every time he sees a definite integral problem for which Mathematica gives an answer (and, tbh, user's formulation sounds a bit rhetorical to me, as if asking "shouldn't you know better?", thus prompting my comment). | |
| Dec 31, 2019 at 16:56 | comment | added | user145307 | @ToddTrimble I don't know either the motivation or intention of the person who commented above, but it's pretty clear that there is a lot of expertise on MSE for computing every definite integral under the sun, certainly a lot more than has ever been demonstrated on MO. There's actually a case to be made for some definite integral questions that MSE is a better place because there are more experts there. I would ask you to reflect on whether any suggestion that MSE might be better is automatically "shaming" rather than actually directing the OP to a more useful site. | |
| Dec 31, 2019 at 14:07 | history | rollback | Carlo Beenakker |
Rollback to Revision 6
|
|
| Dec 30, 2019 at 2:05 | comment | added | Todd Trimble | @user64494 I agree with Yemon's response. Even worse, such comments serve to "shame" the OP for even asking here, when in fact the question is perfectly on-topic, for the reasons Yemon gave. Please reflect carefully on this. | |
| Dec 30, 2019 at 1:31 | comment | added | Yemon Choi | @user64494 Once again, not everything for which a CAS provides an answer (but no explanation!) deserves your "isn't MSE a right forum for such questions". Note that the OP desires a proof, presumably seeking understanding | |
| Dec 29, 2019 at 1:15 | review | Close votes | |||
| Dec 30, 2019 at 1:32 | |||||
| Dec 28, 2019 at 22:03 | history | edited | mark | CC BY-SA 4.0 |
added 2 characters in body
|
| Dec 28, 2019 at 20:32 | comment | added | user64494 | Mathematica reduces it to $$F\left(2 \cot ^{-1}\left(\sqrt[4]{3}\right)|\frac{1}{4} \left(\sqrt{3}+2\right)\right)=2 F\left(\tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt[4]{3}}\right)|\frac{1}{4} \left(\sqrt{3}+2\right)\right)$$ with reference.wolfram.com/language/ref/EllipticF.html | |
| Dec 28, 2019 at 20:21 | history | rollback | Carlo Beenakker |
Rollback to Revision 4
|
|
| Dec 28, 2019 at 20:18 | answer | added | Carlo Beenakker | timeline score: 1 | |
| Dec 28, 2019 at 20:15 | comment | added | user64494 | Isn't math.se a right forum for such questions? | |
| Dec 28, 2019 at 20:04 | history | edited | mark | CC BY-SA 4.0 |
deleted 9 characters in body
|
| Dec 28, 2019 at 17:45 | comment | added | Joe Silverman | Instead of writing $\sqrt{\sqrt3}$, you might use either $3^{1/4}$ or $\sqrt[4]{3}$, LaTeX for the latter is \sqrt[4]{3}. | |
| Dec 28, 2019 at 16:29 | comment | added | mark | @DimaPasechnik thank you, now corrected. | |
| Dec 28, 2019 at 16:29 | history | edited | mark | CC BY-SA 4.0 |
edited body
|
| Dec 28, 2019 at 16:27 | comment | added | Dima Pasechnik | your $d\phi$ are not where they should be. | |
| Dec 28, 2019 at 16:24 | history | edited | mark | CC BY-SA 4.0 |
Several changes in the integrand to bring out the fraction
|
| Dec 28, 2019 at 16:06 | history | edited | mark | CC BY-SA 4.0 |
I put parentheses around an upper limit of integration and put "2\phi" in place of "\2phi" in the second integral. I removed the equal sign after the final integral
|
| Dec 28, 2019 at 15:56 | history | asked | mark | CC BY-SA 4.0 |