User talk:Titus III

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And don't forget, the edit summary is your friend. :) Oleg Alexandrov (talk) 16:55, 8 January 2006 (UTC)

J-invariant Inverse

In February of 2014 you added some very useful methods, four actually, for inverting the J-invariant. They are very useful, but after searching the number theory literature for Ramanujan's alternative base theories for a few hours I have only found proofs of a few of them. Would you mind citing the reference you used to write down those formulae, and / or the original sources? Thanks. Jhhalverson (talk) 19:44, 5 January 2016 (UTC)

Proposed deletion of Sagan's Identity

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Unless the article has secondary sources to back it up, it violates WP:OR as "original research". Loadmaster (talk) 23:31, 7 September 2008 (UTC)

niobium

Hi. You added a list with the electronic configuration to niobium and I do not get the point why niobium is unique in this row? The chemistry of niobium and tantalum is so similar that separation is difficult and was impossible for severeal decades after their discovery. The atom orbital is mixed state of all undelying wave functions and therefore the lectronic notation does not help more than a drawing of the Bohr atom model. Can you provide a good source why this is necessary in the FA niobium? --Stone (talk) 21:15, 16 February 2009 (UTC)

Hello Stone. Well, I don't think the electron notation (as the number of electrons in shells K,L,M,...) is completely useless but can provide a convenient (though not absolute) heuristic. As you know, the outermost shell of the noble gases (excepting helium) has 8 electrons, those of the halogen group has 7, the oxygen group has 6, the nitrogen group has 5, etc. One may then hastily conclude that ALL groups have its member elements with the same number of electrons in the valence shell. As proven by niobium (and a few others), this is NOT necessarily the case, hence why I took the trouble of pointing it out so others may not make that conclusion.

P.S. As a sidenote, is it absolutely sure the electron configuration of niobium ends with 12,1? Pls refer to my question in the discussion page of lawrencium since there seems to be two contradictory forms for that element, one cited in the infobox and the other in the main text.Titus III (talk) 13:17, 17 February 2009 (UTC)

The group number and the number of electrons in the outer shell are a good way to understand main group chemistr, but for the transition metals the number is basically worthles. Also for hybridisation the number in each shell does not give you additional information either. The chemistry of tantalum and niobium is taht of oxidation state of 5 and 3 and they are very similar in chemical reactions. The 1 in the table suggests that with one valence electron the oxidation state 1 dominates the chemistry, which is in reality not the case. So what does the table benefit, than substituting one wrong assumption that all elements in the group have the same electron configuration by the wrong fact that niobium has one valence electron.--Stone (talk) 15:58, 17 February 2009 (UTC)

Hello Stone,

1. These tables (whether for the noble gases, halogen group, or any of the 18 groups) showing the electron configuration provide - in one glance - one easy way to see how member elements are related. For example, for the noble gases (other than helium) it is seen that the outermost shell contains 8 electrons which has implications for the chemical behavior of the family. For the transition metals, as you pointed out, the chemical implications may not be straightforward, but it turns out the sum of the number of electrons in the last two shells is a constant. Thus, with a list of all the transition metals showing just the number of electrons in the last two shells, even without knowing the atomic number, even with the element names replaced by arbitrary labels, one can sort them into the correct groups.

2. The information in these tables are found in the Infobox accompanying each article, spread over many webpages. For the case of niobium (and a few others), they were just put together for the convenience of seeing them in one place.

3. It is a statement of fact that niobium has a different configuration in the last two shells than the rest of its group. Note I took care not to write anything implying this has implications for its chemical behavior. Titus III (talk) 13:58, 18 February 2009 (UTC)

Tale of four constants

Titus, interesting stuff, but since it's your own unpublished work, not suitable for spamming on a bunch of WP articles. I reverted several. Dicklyon (talk) 06:08, 9 October 2010 (UTC)

Talkback

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Dougweller (talk) 17:48, 15 June 2012 (UTC)

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Titus To Titus

Hey Titus III,

I Don't Know you, but I made this for you:

Titus The III

A Signature! Hope You Like It! TitusFox 17:20, 26 February 2014 (UTC)

Some copy-editing

Hello.

Please notice the copy-editing I did on Ramanujan–Sato series and bear certain things in mind in future editing:

• I changed the title from Ramanujan-Sato series to Ramanujan–Sato series, and made the same punctuation change at each point in the article where that phrase occurs. I also changed McKay-Thompson to McKay–Thompson in several places, and Bailey-Borwein-Plouffe to Bailey–Borwein–Plouffe. That is is required by WP:MOS.
• I changed the links from other articles to links to the new title.
• I also found several ranges of pages in which I made the same punctuation change, e.g. 361-383 was changed to 361–383. Also in WP:MOS.
• I set the title phrase in bold at its first appearance. That is also required by WP:MOS.
• I italicized some subscripted "n"s. Generally variables in mathematical notation are italicized and digits, punctuation such as parentheses, and things like cos, det, max, log, gcd, etc., are not. That matches the style used in TeX, LaTeX, MathJax, etc.

Michael Hardy (talk) 17:03, 13 April 2014 (UTC)

Question on section Inverse function on the page j-invariant

Thanks for the formulas for the inverse of the j-invariant. They are quite useful. Could you add a reference to them? By the way, I found one more such formula:

tau = -i + 2 * i * 2F1([1/12, 5/12], [1], 1) * 2F1([1/12, 5/12], [1/2], 1-alpha) / 2F1([1/12, 5/12], [1], alpha)

where alpha = 1728/j. This formula fits in the same pattern as the other formulas, this time alpha being a solution of a linear equation. MvH (talk) 15:21, 17 June 2014 (UTC)MvH

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Euler's sum of powers conjecture

Hi Titus III, Yesterday you removed some information recently added to the article mentioned above. In your edit summary, your explanation involved (as far as I can tell) some factual claims and an external link not in the article. Perhaps you can add them in an appropriate place? Thanks! Best, JBL (talk) 15:43, 20 December 2014 (UTC)

Hello Joel. It is done. The early tables for (4,1,4) and (5,1,5) are found in the 1967 paper of Lander, Parkin, Selfridge. James Waldby extended the (5,1,5) table up to sums 10000^5 in 2009 (personal communication). I have now included his site in the "External Links" page. Titus III (talk) 17:31, 20 December 2014 (UTC)

That's great, thanks very much! --JBL (talk) 17:46, 20 December 2014 (UTC)

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Planet Nine "evidence"

Hi, since I've now apparently twice reverted your addition of "a" before "evidence" in Planet Nine, I just want to point out that "evidence" is usually an uncountable noun, so doesn't take an indefinite article. --Ørjan (talk) 02:41, 5 February 2016 (UTC)

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Manila tallest buildings

Hi Titus! Thanks for constantly updating the List of tallest buildings in Metro Manila. When you get the time, you may also want to review this Template:Manila Skyscrapers that i made that actually links to the individual skyscraper articles. Appreciate your contributions here!--RioHondo (talk) 13:32, 9 February 2017 (UTC)

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Ramanujan–Sato series

Hi Titus III

I don't agree with your cancellation. First of all this formula is a part of these kind of series as you will easy see when you compare this with that above from Ramanujan. Secondly as well 396^4k as 16^4k is a modul. Further Ramanujan made the enigmatic remark that there were "corresponding theories" ... and with this formular I found at least one new of them. Thus I reclaim this formula but moved this one paragraph deeper and explained the theory related a bit.

Sincerely yours

Ralf.steiner — Preceding unsigned comment added by Ralf.steiner (talk • contribs) 07:20, 23 March 2017 (UTC)

Hello Ralf. An example of a Ramanujan-Sato series is

${\frac {1}{\pi }}=\sum _{k=0}^{\infty }{\tbinom {2k}{k}}^{3}\,{\frac {Ak+B}{D^{k}}}$

There are infinitely many algebraic numbers $A,B,D$ which make the above equation true, and these three variables can be found using modular forms. Your equation has the similar form,

${\frac {1}{\pi }}=\sum _{k=0}^{\infty }C_{k}^{2}\,{\frac {Ak+B}{D^{k}}}$

where $C_{k}$ are the Catalan numbers. But unless you can rigorously prove that the single example of $A,B,D$ you found can be derived using modular forms or the Dedekind eta function, and can also be generalized to algebraic numbers using square roots, etc, then your formula's status as a Ramanujan-Sato series is still not established. However, as a compromise, we can create a separate section for it after the well-established levels. Titus III (talk) 13:39, 23 March 2017 (UTC)

Hello Titus III

I thank you very much for your above explanation and the compromise you find I fully agree with him. In the mean time at OEIS A001246 Mr. J. Arndt also had have a deeper look at my formula (and this link) and canceled my "it seems" regarding this question. Please further note that there are not only a single ONE formula (google for V_250116.pdf) ... there is a periodic of (4k+#) in the numerator of the sum to become a pure 1/Pi term each.

Sincerely yours,

Ralf.steiner — Preceding unsigned comment added by Ralf.steiner (talk • contribs) 17:10, 23 March 2017 (UTC)

Hello Titus III

Please note that I recently found a second periodic (in k) for the first formula and thus the proof that this formular is derived by modular forms and fulfils the requirements for this kind of series.

Sincerely yours

Ralf.steiner Ralf.steiner (talk) 16:20, 29 March 2017 (UTC)

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1999 Luzon earthquake

I think you should create this stub article and copy the two sentences you wrote to it. I would do it but I would prefer if you got the credit for creating the page. Then it can be listed in the "See also" section of the 2019 article. --- Coffeeandcrumbs 09:50, 23 April 2019 (UTC)

That's ok. You can create the article if you want. :) Titus III (talk) 15:26, 23 April 2019 (UTC)

Thanks. Done. --- Coffeeandcrumbs 16:23, 23 April 2019 (UTC)

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A barnstar for you!

The Original Barnstar

I could give you a Tireless Contributor Star instead of Original Barnstar for the work on COVID-19 pandemic, however because from what i've seen from edit summaries that is criticizing people from not doing their work numerous times i present to you Original Barnstar instead. (From what i have noticed, it appears that people is giving away more Tireless Contributor Barnstars and not Original Barnstars, so i consider Original Barnstars to be a greater honor) I used to work hard on COVID-19 pandemic pages in Wikipedia, but i've gone away because of burnout. Without you and LSGH i'm not sure if Wikipedia will stay up-to-date today. SMB​99​thx Email! 06:49, 20 August 2020 (UTC)

I can't handle takeover of Kazakhstan template anymore.

Hi Titus, i'm sorry with what happened on Iraq. I did not update Iraq template because i took some wikibreak on late July. At that time i was worried about not getting into my university. As such, i did not want to bother editing the Iraq template. I did not get hit by COVID-19, i just took some break.

Now, when i'm looking for a comeback, it seems that you completely took over the template i'm going back to editing again. As such, i'm back but only took over editing Guatemala medical cases chart. That was until an IP made major changes to the Kazakhstan medical cases chart. As such, i decided to take it over again. However, it got me into trouble with several IPs that was involved in maintaining the medical cases chart template. I learned a lesson to attribute any content copied within Wikipedia, and the situation is back to normal.

However, I had to face off with other IPs, again. I feel pain at handling the IPs again - I do not want to face further trouble and threaten my wikicareer. Kazakhstan is going to be bane of my problems if i keep it as it is. As such, i want to offer you a comeback in editing the template and maintaining it. Meanwhile, i'm going to remove Kazakhstan from my watchlist and adding back Iraq to the watchlist and updating Iraq to what happened in August 25. I do not want to face further problems at handling the Kazakhstan template again.

By the way, i did award you a barnstar some days ago. SMB99thx my edits 05:52, 26 August 2020 (UTC)

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Inanna

You should come join the fun on the talk page! Skyerise (talk) 14:40, 12 August 2021 (UTC)

Tempting. :) Titus III (talk) 15:19, 12 August 2021 (UTC)

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Koide formula

Hey,

the correction for the cubic equation actually makes it more difficult to see why a formula of koide type appears, so better I am reverting it and adding more explanation to the factorisation. Arivero (talk) 14:45, 18 June 2023 (UTC)

The explanation now makes more sense. But I suggest not adding the secondary variable $n'$ since it was not there in the original form, nor does the characteristic polynomial factorize if it is inserted. And I added "elementary symmetric polynomials" for more details. Titus III (talk) 04:31, 19 June 2023 (UTC)

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Solving the Bring Jerrard Quintic Equation with the Jacobi Theta Function

This is something I can answer you right away. I did find it out on the famous essay from Francesco Brioschi and Charles Hermite. I mean the work Sur la résolution de l'Équation du cinquiéme degré Comptes rendus that is equal to Sulla risoluzione delle equazioni del quinto grado by Hermite and Brioschi. To read that essay, click on that link: https://zenodo.org/records/2401804 And then read page 258 of that essay. There you see this formula:

${\theta }^{5}-{\frac {5}{2}}{\theta }^{4}-{\frac {(1-4k^{2}k'^{2})^{2}}{2k^{2}k'^{2}}}=0$

By multiplying by ${\textstyle 2\theta ^{-5},}$ a new quintic in terms of $\theta ^{-1}$ is:

${\frac {(1-4k^{2}k'^{2})^{2}}{k^{2}k'^{2}}}\theta ^{-5}+5\theta ^{-1}=2.$

Next, multiplying by the fourth root of the leading coefficient, we obtain:

${\biggl [}{\frac {(1-4k^{2}k'^{2})^{2}}{k^{2}k'^{2}}}{\biggr ]}^{\frac {5}{4}}\theta ^{-5}+5{\biggl [}{\frac {(1-4k^{2}k'^{2})^{2}}{k^{2}k'^{2}}}{\biggr ]}^{\frac {1}{4}}\theta ^{-1}=2{\biggl [}{\frac {(1-4k^{2}k'^{2})^{2}}{k^{2}k'^{2}}}{\biggr ]}^{\frac {1}{4}}$

This is a polynomial $x^{5}+5x=4c$ in Bring–Jerrard normal form (containing only terms of degree 5, 1, and 0) where

$:x={\biggl [}{\frac {(1-4k^{2}k'^{2})^{2}}{k^{2}k'^{2}}}{\biggr ]}^{\frac {1}{4}}\theta ^{-1},\qquad :c={\frac {1}{2}}{\biggl [}{\frac {(1-4k^{2}k'^{2})^{2}}{k^{2}k'^{2}}}{\biggr ]}^{\frac {1}{4}}:$

The elliptic modulus $k$ and the Pythagorean complementary modulus $k'$ satisfy $k^{2}+k'^{2}=1.$ The solutions for $k$ and $k'$ in terms of $c$ are:

$:k={\frac {{\sqrt {{\sqrt {c^{4}+1}}+1}}-c}{\sqrt {2c^{2}+2+2{\sqrt {c^{4}+1}}}}},\qquad :k'={\frac {{\sqrt {{\sqrt {c^{4}+1}}+1}}+c}{\sqrt {2c^{2}+2+2{\sqrt {c^{4}+1}}}}}.$

These two formulas can be presented in terms of the hyperbolic lemniscate functions:

${\begin{aligned}:k&=\operatorname {tlh} {\bigl [}{\tfrac {1}{2}}\operatorname {aclh} (c){\bigr ]}^{2},\\:k'&=\operatorname {ctlh} {\bigl [}{\tfrac {1}{2}}\operatorname {aclh} (c){\bigr ]}^{2}:\end{aligned}}$

That is the way it is!

And I already found it an even faster way. And I wrote it down in the under part of that article already:

https://hi.wikipedia.org/w/index.php?title=%E0%A4%A5%E0%A5%80%E0%A4%9F%E0%A4%BE_%E0%A4%AB%E0%A4%B2%E0%A4%A8

This is an article I wrote in the Hindi Wikipedia. Look at the under section of that article. There stands the foloowing pattern:

$x^{5}+5\,x=4\,c$

$Q=q{\bigl [}(2\,c^{2}+2+2{\sqrt {c^{4}+1}})^{-1/2}({\sqrt {{\sqrt {c^{4}+1}}+1}}+c){\bigr ]}$

$u={\frac {2\,\vartheta _{00}(Q^{5})\,\vartheta _{00}(Q^{1/5})-2\,\vartheta _{00}(Q)^{2}}{\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}}}$

$x={\frac {c\,u\,(5\,u^{2}-10\,u+4)}{5\,u^{2}-6\,u+2}}$

And in this Hindi Wikipedia article I wrote down those calculation examples:

Example one:

$x^{5}+5\,x=4$

$Q=q{\bigl [}(4+2{\sqrt {2}})^{-1/2}({\sqrt {{\sqrt {2}}+1}}+1){\bigr ]}\approx 0.1852028700803001414251518230736124606036037762504611138834393086\ldots$

$u={\frac {2\,\vartheta _{00}(Q^{5})\,\vartheta _{00}(Q^{1/5})-2\,\vartheta _{00}(Q)^{2}}{\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}}}\approx 0.3447357019439680107642844248203058218212701865144922007257031\ldots$

$x={\frac {u\,(5\,u^{2}-10\,u+4)}{5\,u^{2}-6\,u+2}}\approx 0.75192639869405948026865366345020738740978383913037835\ldots$

Example two:

$x^{5}+5\,x=8$

$Q=q{\bigl [}(10+2{\sqrt {17}})^{-1/2}({\sqrt {{\sqrt {17}}+1}}+2){\bigr ]}\approx 0.3063466544466074265361088194021326272090461143559097382981847\ldots$

$u={\frac {2\,\vartheta _{00}(Q^{5})\,\vartheta _{00}(Q^{1/5})-2\,\vartheta _{00}(Q)^{2}}{\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}}}\approx 0.26117326232214439979677125632439205703620052387333832254673599\ldots$

$x={\frac {2\,u\,(5\,u^{2}-10\,u+4)}{5\,u^{2}-6\,u+2}}\approx 1.16703618370164304731101943199639610129755211048801991\ldots$

Example three:

$x^{5}+5\,x=12$

$Q=q{\bigl [}(20+2{\sqrt {82}})^{-1/2}({\sqrt {{\sqrt {82}}+1}}+3){\bigr ]}\approx 0.370664951152024075624432522177568657151868089959747395750974\ldots$

$u={\frac {2\,\vartheta _{00}(Q^{5})\,\vartheta _{00}(Q^{1/5})-2\,\vartheta _{00}(Q)^{2}}{\vartheta _{00}(Q^{1/5})^{2}+5\,\vartheta _{00}(Q^{5})^{2}-4\,\vartheta _{00}(Q)^{2}}}\approx 0.2090460949758925319033273467025369651597551071050541509568731\ldots$

$x={\frac {3\,u\,(5\,u^{2}-10\,u+4)}{5\,u^{2}-6\,u+2}}\approx 1.38409179582314635924775512626713547488593506018067645\ldots$

And then I mentioned a more detailed formula that shall get secure that the elliptic nome really is used in the correct way:

$Q=q{\bigl [}(2\,c^{2}+2+2{\sqrt {c^{4}+1}})^{-1/2}({\sqrt {{\sqrt {c^{4}+1}}+1}}+c){\bigr ]}=$

$=\exp {\biggl \{}-\pi {\biggl [}\int _{0}^{1}2{\bigl (}z^{4}+2c{\sqrt {2{\sqrt {c^{4}+1}}-2c^{2}}}\,z^{2}+1{\bigr )}^{-1/2}\,\mathrm {d} z{\biggr ]}\div {\biggl [}\int _{0}^{1}2{\bigl (}z^{4}-2c{\sqrt {2{\sqrt {c^{4}+1}}-2c^{2}}}\,z^{2}+1{\bigr )}^{-1/2}\,\mathrm {d} z{\biggr ]}{\biggr \}}$

But I have to mention to you one special thing that can appear like a trap. When you enter the elliptic nome into the WolframAlpha computational knowledge engine and you want to have nome q(k), you have to write EllipticNomeQ(k^2) to get exactly that. I always took the modulus k with this value:

$k=(2\,c^{2}+2+2{\sqrt {c^{4}+1}})^{-1/2}({\sqrt {{\sqrt {c^{4}+1}}+1}}+c)$

And therefore:

$k^{2}={\frac {1}{2}}+{\frac {1}{2}}c\,{\sqrt {2{\sqrt {c^{4}+1}}-2c^{2}}}$

So if you want to get the right elliptic nome value for Q you always have to put this into WolframAlpha:

$\operatorname {EllipticNomeQ} {\bigl (}{\frac {1}{2}}+{\frac {1}{2}}c\,{\sqrt {2{\sqrt {c^{4}+1}}-2c^{2}}}{\bigr )}$

Then you get the correct Nome in this computational knowledge engine. Let me show you how I did this. I performed it in one of my own videos. Here it is:

https://www.youtube.com/watch?v=jX2s5xXGwSE

Yes! This is exactly the video that I created and in which I solved four different Bring Jerrard equations. And Yes! It really works. It always gives the correct solution of the Bring Jerrard equations. And by setting the imaginary fifth roots of one as a factor before the Q^(1/5) positions, you even get the correct four imaginary solutions of the Bring Jerrard equation. You can try it out. It works totally excellent!

I am glad and happy that you ask me this question. Because in this way my discoveries become known in the circles of the mathematicians. And that makes me extremely happy indeed. I say the truth. That gives me happiness because in this way I see that my research results become fertile.

And to answer your second question. There are indeed analogous formulas for theta_4 in a similar pattern as for theta_3 but in this case the sign of the coefficient of the linear term must be negative to the sign of the coefficient of the quintic term of the corresponding Bring Jerrard quintic equation. Then you have to create the elliptic modulus k after the pattern shown in this WolframMathworld article here:

https://mathworld.wolfram.com/QuinticEquation.html

Then you have to take the Jacobi theta quotient in the ground pattern I showed you, but this time you do not take theta_3 but theta_4 instead of that:

$t={\frac {2\,\vartheta _{01}(Q^{5})\,\vartheta _{01}(Q^{1/5})-2\,\vartheta _{01}(Q)^{2}}{\vartheta _{01}(Q^{1/5})^{2}+5\,\vartheta _{01}(Q^{5})^{2}-4\,\vartheta _{01}(Q)^{2}}}$

And then you enter it into the image of the mentioned broken rational cubic keylock. And that is it. You can see it performed in that video here:

https://www.youtube.com/watch?v=RYY67hd9Nzc

Hopefully you are successful in learning these algorithms and hopefully even many mathematically interested people more learn these algorithms too! Have a nice day and have a lot of success in solving the Quintic Bring Jerrard equations! Yours faithfully and sincerely!!

Lion Emil Jann Fiedler also known as Reformbenediktiner — Preceding unsigned comment added by Reformbenediktiner (talk • contribs) 18:57, 23 October 2024 (UTC)

Useful clue for your Update, 11/2/2012 and your case p = 7 and case p = 13

When I still had the access on the German Wikipedia platform I also discovered and entered some similar formulas also for the cases of Q = q^7 and Q = q^13 in a related way. If you want you can see them:

https://de.wikipedia.org/wiki/Ramanujansche_g-Funktion_und_G-Funktion

https://en.wikipedia.org/w/index.php?title=Modular_lambda_function&section=6 — Preceding unsigned comment added by Reformbenediktiner (talk • contribs) 20:23, 23 October 2024 (UTC)

It really is a related topic. But I am happy to see that you found out even these formulas. Have a lot of success with everything and have a wonderful nice time! Lion Emil Jann Fiedler also known as Reformbenediktiner — Preceding unsigned comment added by Reformbenediktiner (talk • contribs) 18:45, 23 October 2024 (UTC)

Wikipedia Article Adamchik Transformation

Very good! I am also very pleased and delighted about it. Now I have mentioned you in a Wikipedia article that I wrote at the moment. It is this one:

Adamchik transformation

Please read it carefully! And please tell me if you know even shorter coefficient formulas that set the coefficient of the different Tschirnhaus transformation forms in relation to each other. The Adamchik transformation uses a quartic key to transform the Principal Quintic form into the Bring Jerrard Quintic form. And there is also a way of transforming the Principal Quintic form into the Brioschi Quintic form by using a broken rational quadratic key at first and then take a second broken rational quadratic key to transform the Brioschi Quintic form into the Bring Jerrard Quintic form. But how do all these coefficients correlate to each other more accurately. Until now I could not shrink and shorten these coefficient relation formulas. Somehow the transformations to the Bring Jerrard Quintic must be able to be formulated even quicker and faster, or at least with shorter clue equations. But I do not know exactly how it is done.

https://www.oocities.org/titus_piezas/Tschirnhausen.pdf

https://www.oocities.org/titus_piezas/Brioschi.pdf

Please help me! I really want to find it out. And hopefully it is acceptable for you that I write this Wikipedia article and I also mention the formulas of your essays. My intention consists in becoming a master of solving all Quintic equations by using elliptic modular functions. I did already find out the elliptic solution of the Bring Jerrard form as you know exactly. Now I want to find out the elliptic solution of the General Quintic equation in the whole. Therefore I am glad that you talk to me. Therefore I am happy that both of us commit and entrust our formulas to help each other and to bring each other forward. Hopefully you have a further good idea how to compute General Quintics in its most regular case elliptically. Please tell me of your ideas how to make it! And please let me know! We are almost there!

Yours faithfully and sincerely! Lion Emil Jann Fiedler, Reformbenediktiner, Emil Jann Brahmeyer Reformbenediktiner (talk) 10:39, 25 October 2024 (UTC)

I want to mention two last things.

I mentioned the equation:

$x^{5}+5x=4c$

And I mentioned the corresponding elliptic modulus:

$k={\frac {{\sqrt {{\sqrt {c^{4}+1}}+1}}+c}{\sqrt {2c^{2}+2+2{\sqrt {c^{4}+1}}}}}=\operatorname {ctlh} {\bigl [}{\tfrac {1}{2}}\operatorname {aclh} (c){\bigr ]}^{2}$

But hopefully you know the Lemniscatic elliptic functions and especially the hyperoblic lemniscate cotangent and the hyperbolic lemniscate areacosine. I just wanted to mention this shortcut in creating the elliptic modulus for the equation.

And I have one more suggestion. We perhaps should collect some informations in the essays of Nikolaus Bagis from Macedonia in Greece or Soon Yi Kang from Kangwon in South Korea. Maybe you know these two mathematicians. Their essays could also be helpful for collecting even more information for the MSE essay. The idea of suggesting that just came into my mind. Hopefully everything is fine. And hopefully everything is proceeding excellent. Whenever you need help, and whenever you want to talk to me, you should just do it. I will always listen to you in a careful way. And I thank you so much for crediting the discovery to me. I am very thankful and grateful and appreciative for that. Reformbenediktiner (talk) 11:19, 25 October 2024 (UTC)

New article Brioschi quintic form

Alright, I made it! Now I wrote the Wikipedia article Brioschi quintic form and it looks very good. I took some of your essays as references. And this article does contain a further lagorithm to solve the Principal case of the quintic equations. More accurately this article shows how Principal equations are transformed into Brioschi equations via quadratic Tschirnhaus transformations. And then it shows a solution algorithm using the Hyperbolic tangent as a very important parameter. It really works too. But now I have a question to you. How are you progressing with writing the MSE essay? Is your MSE already completed? Or are you still in the editing phase? Do you eventually need my further help? Please tell me about your situation! Because in relation to this I really have curiosity! I am looking forward to the MSE. I wish you an excellent time. Yours faithfully and sincerely!!

Lion Emil Jann Fiedler, Reformbenediktiner, Emil Jann Brahmeyer Reformbenediktiner (talk) 9:05, 2 November 2024 (UTC)

Hi Emil. Sorry for the delay. I was busy trying to find a function with tetrahedral symmetry to solve the Bring quintic. Progress has been slow. And I've actually made a new post about your solutions to the Bring quintic using the Jacobi theta functions in this new MO post, credited to you. Kindly check. I will also update the MSE post with insights/comments from the new MO post.Titus III (talk) 13:06, 2 November 2024 (UTC)

Yes, I saw that and I also checked that. Did you already read the article about the Lemniscate elliptic functions and especially the section Combination and halving theorems Lemniscate elliptic functions#Combination and halving theorems? There I wrote a formula down that shows a special parametrization for the elliptic modulus k that is a useful clue for the relation between the coefficient of absolute term of the Bring Jerrard equation and the corresponding elliptic modulus. It is this one:

$k={\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}(f)]^{2}={\tfrac {1}{2}}{\sqrt {2+2f{\sqrt {2{\sqrt {f^{4}+1}}-2f^{2}}}}}={\bigl (}2f^{2}+2+2{\sqrt {f^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {f^{4}+1}}+1}}+f{\bigr )}$

$k'={\text{tlh}}[{\tfrac {1}{2}}{\text{aclh}}(f)]^{2}={\tfrac {1}{2}}{\sqrt {2-2f{\sqrt {2{\sqrt {f^{4}+1}}-2f^{2}}}}}={\bigl (}2f^{2}+2+2{\sqrt {f^{4}+1}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {f^{4}+1}}+1}}-f{\bigr )}$

In this way we can also say:

For the Quintic Bring Jerrard equation $rx^{5}+sx=t$ the corresponding modulus k is this:

$k={\text{ctlh}}{\bigl [}{\frac {1}{2}}{\text{aclh}}{\bigl (}{\frac {5t}{4s}}{\sqrt[{4}]{\frac {5r}{s}}}{\bigr )}{\bigr ]}^{2}$

This fulfills that formula:

$(3125k^{4}-3125k^{2})rt^{4}=(-256k^{8}+512k^{6}-384k^{4}+128k^{2}-16)s^{5}$

Then in the next step we use the formulas you already explained in your MSE article. But for this r-s-t-equation the mentioned fraction $(5u^{3}-5u^{2}+u)/(10u^{2}-6u+1)$ must be multiplied with $5t/s$ and then you get the x of the mentioned r-s-t-equation. This should be exactly that. You can test it out if you want. In the emergency case I can create another video by using that hyperbolic lemniscate function expression, but I am already totally happy about your MSE. You did great work and I am really glad and joyful about your production "Two new formulas to solve the Bring quintic using only Jacobi ϑ3(q) and ϑ4(q)"! Yes, I really am happy about your work! Very well done! Yours faithfully and sincerely!!

Lion Emil Jann Fiedler (Reformbenediktiner, Emil Jann Brahmeyer)

Reformbenediktiner (talk) 17:12, 2 November 2024 (UTC)

Hi Emil. Ah, so that's how. I made some edits to that "lemniscate section". Maybe we take this one step at a time. Since,

$k={\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}(f)]^{2}$

$k'={\text{tlh}}[{\tfrac {1}{2}}{\text{aclh}}(f)]^{2}$

Naturally,

$k^{2}+k'^{2}={\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}(f)]^{4}+{\text{tlh}}[{\tfrac {1}{2}}{\text{aclh}}(f)]^{4}=1$

Question 1. Is it always true that ${\text{ctlh}}[z]^{4}+{\text{tlh}}[z]^{4}=1$ for ANY $z$? Because this is would be the lemniscate version of the Jacobi theta function's $x^{4}+y^{4}=1$.

Question 2. What are easy formulas to calculate ${\text{ctlh}}[z]$ and ${\text{tlh}}[z]$ in Mathematica? There seems nothing in the article. Titus III (talk) 04:15, 3 November 2024 (UTC)

I want to answer your questions right away. The answer on question one is Yes! This works for every single real value of z indeed.

And now I want to answer the second question:

First of all you should know:

$\operatorname {aclh} (c)={\frac {1}{2}}\,\mathrm {F} {\bigl [}2\operatorname {arccot}(c),{\frac {1}{2}}{\sqrt {2}}\,{\bigr ]}$

And with the letter F of course I mean the incomplete elliptic integral of first kind in Legendre form.

For tlh and ctlh these formulas connecting to the Jacobi Amplitude functions have validity:

${\text{tlh}}(U)={\frac {{\text{sn}}(u;{\tfrac {1}{2}}{\sqrt {2}})}{\sqrt[{4}]{{\text{cd}}(U;{\tfrac {1}{2}}{\sqrt {2}})^{4}+{\text{sn}}(U;{\tfrac {1}{2}}{\sqrt {2}})^{4}}}}$

${\text{ctlh}}(U)={\frac {{\text{cd}}(U;{\tfrac {1}{2}}{\sqrt {2}})}{\sqrt[{4}]{{\text{cd}}(U;{\tfrac {1}{2}}{\sqrt {2}})^{4}+{\text{sn}}(U;{\tfrac {1}{2}}{\sqrt {2}})^{4}}}}$

The Jacobi amplitude sine of a value U with a modulus k you have to enter as JacobiSN(U,k^2) or sn(U,k^2) and the other function you have to enter as JacobiCD(U,k^2) or cd(U,k^2) you can also enter. Now you know how it is possible to enter terms for the hyperbolic lemniscate tangent and hyperbolic lemniscate cotangent function. But for the calculation example of the quintic Bring Jerrard equations the modulus always is computed by the square of the hyperbolic lemniscate cotangent. And for that I even can you give two further formulas:

$\operatorname {ctlh} (U)^{2}=\tan {\biggl \langle }{\frac {\pi }{4}}-\arctan {\bigl \{}\operatorname {cn} {\bigl [}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}-U;{\frac {1}{2}}{\sqrt {2}}{\bigr ]}^{4}{\bigr \}}{\biggr \rangle }$

And of course here is the counterpart:

$\operatorname {tlh} (U)^{2}=\tan {\biggl \langle }{\frac {\pi }{4}}-\arctan {\bigl \{}\operatorname {cn} {\bigl [}U;{\frac {1}{2}}{\sqrt {2}}{\bigr ]}^{4}{\bigr \}}{\biggr \rangle }$

And yes, the relation between the two now mentioned formulas fulfills the criteria $\operatorname {tlh} (U)^{4}+\operatorname {ctlh} (U)^{4}=1$ in an exact way!

And now watch even very carefully:

To create $k=\operatorname {ctlh} [{\tfrac {1}{2}}\operatorname {aclh} (c)]^{2}=({\sqrt {{\sqrt {c^{4}+1}}+1}}+c)\div {\sqrt {2c^{2}+2+2{\sqrt {c^{4}+1}}}}$ exactly this is what I enter into the WolframALpha machine:

k = tan(pi/4-arctan(cn(K(1/2)-1/4*elliptic F(2*arccot(c),1/2),1/2)^4))

Q = nome(tan(pi/4-arctan(cn(K(1/2)-1/4*elliptic F(2*arccot(c),1/2),1/2)^4))^2)

To get the nome Q for the mentioned equation $rx^{5}+sx=t$ please enter exactly this:

Q = nome(tan(pi/4-arctan(cn(K(1/2)-1/4*elliptic F(2*arccot((5*t)/(4*s)*(5*r/s)^(1/4)),1/2),1/2)^4))^2)

This is accurately what you have to enter into the WolframAlpha computational knowledge engine to get the nome you need!

Of course I enter it exactly this way because to get K(1/2*sqrt(2)) a person always has to enter it like K(1/2) into that WolframAlpha machine. And if you want to get the nome Q the person have to enter nome(k^2) into that machine. Hopefullly I did not forget any important information but really told you everything you must know. By the way, now I entered even more calculation examples into the article Brioschi quintic form that I created. At one position I even use the discovered solution algorithm and combined it with your essays. It also works in a wonderful way. I continue researching even more and I will definitely find out even more. Have a nice time!

Yours faithfully and sincerely!!

Lion Emil Jann Fiedler Reformbenediktiner (talk) 14:16, 3 November 2024 (UTC)

Hi Emil. Thanks for the response, I will do experiments on the formulas. Anyway, I see they have deleted/moved your Brioschi quintic article to the draft section. Just consider it a temporary setback. There is an art in creating Wikipedia articles and some of the guys here are strict.

Question 1: May I ask what is your first language? (Normally, the people in Wikipedia prefer extremely precise English.)

Question 2: More importantly, I suggest it would be more worthy of your talents if you focus on finding a new formula for the Bring quintic using the lemniscate elliptic functions. The trigonometric functions and $x^{2}+y^{2}=1$ can be used to solve the cubic. It would be extremely beautiful if the lemniscate elliptic functions and $x^{4}+y^{4}=1$ can be used to solve the quintic. (Hermite's method uses quotients of Jacobi theta functions so it is more $x^{8}+y^{8}=1$.) Do you think you can do it? Titus III (talk) 01:35, 4 November 2024 (UTC)

My first language is the German language. My original first Wikipedia account even was a German account. But on the German Wikipedia I got banned. The German Wikipedia users took away my rights of writing articles in the corresponding language. In other words, my German Wikipedia account was locked by the other users. For the German users I unfortunately was categorized as a dangerous user. In this way the English Wikipedia is one of my last chances to prove that I am really worthy of writing Wikipedia articles.

The names of mine that I entered into the articles, are my true names. My first forename is indeed Lion, my second forename is Emil, my third forename is Jann, and my official surname is Fiedler. So I always mentioned my real name. And yes, Reformbenediktiner was my Pseudonym already in the German Wikipedia. The formulation Emil Jann Brahmeyer is my Pseudonym on YouTube, or I could also say Emil Jann Brahmeyer is the name of my YouTube account and my YouTube channel. I have already linked some of the videos from my YouTube channel on your talk page here. When I find something out I often have the tendency to write Wikipedia articles about it. The latest two Wikipedia articles I wrote, the articles Adamchik transformation and Brioschi quintic form have many references that really fit to the topics. And some of those very important essays even come from exactly you. Therefore I am really thankful that you write these essays.

It is really a big honor for me to get to know you and to work together with you. It definitely is a very big honor for me. I am really totally happy about that. Sometimes I even wish that we could communicate even per E-Mail. My E-Mail address is very easy. Just write down my first forename, then the point, then my surname, then the at-symbol, then the gmx-letters, then again a point and at last the abbreviation de that is also typical for German URL-addresses. This is exactly my E-Mail-address. I let you decide if you want to communicate on E-Mail messages or on Wikipedia talk pages.

About your second question I have to mention one special aspect. You are totally right. The trigonometric and hyperbolic functions that often stand in Pythagorean relations, these functions are used to solve cubic and quartic equations in their regular forms. And yes, the elliptic functions are used to solve the regular case of quintic equations. Because the regular case of quintic equations can not be solved in elementary ways along with the Abel Ruffini theorem, as we all know. The equations of fifth degree, the quintic equations must be solved elliptically as long as hypergeometric terms are not used. The elliptic solution algorithms I took are direct modular functions of the elliptic nome q function as I showed you in my messages and pages to you. But there are indeed further ways to solve the quintic equations elliptically and without elliptic nome, but instead of this using the Jacobi amplitude functions, especially in their so called incomplete forms. In one special case I even can show you how. So please watch this video in which I explain exactly this algorithm:

https://www.youtube.com/watch?v=R-STgFaPgNM

You can see clearly that I use this code in the WolframAlpha engine:

r*x^5 + s*x = t

k = tan(pi/4-arctan(cn(K(1/2)-1/4*elliptic F(2*arccot((5*t)/(4*s)*(5*r/s)^(1/4)),1/2),1/2)^4))

w = tan(arctan(cn(3/5*K(k^2),k^2)/cn(1/5*K(k^2),k^2))+arctan(cn(3/5*K(1-k^2),1-k^2)/cn(1/5*K(1-k^2),1-k^2))-1/2*arctan(2))

x = 5*t/s*w*(1+w-w^2)/(1+3*w)/(1+w^2)

This generates the real solution x for this quintic Bring Jerrard equation. It is a very beautiful way to solve the equation only by using an easy Jacobi Amplitude function at four positions in the formulation of the key. And this way of solving really is related to the new algorithm I showed you already.

And of course you noticed the fact correctly that there is also a further Jacobi elliptic analogon for solving the quintic equations in their completely general form. If we take the elliptic analogon to the trigonometric and hyperbolic functions, we indeed at first take an inverted Jacobi Amplitude function and divide the value by five. And then we enter it into the same Jacobi Amplitude function again and then construct a quintic polynome of it. And this quintic polynome interestingly solves another quintic polynome that can not be solved in an elementary way. I want to demonstrate you a lemniscatic elliptic calculation example for that. This calculation example is about the Hyperbolic lemniscate sine and its inverse, the Hyperbolic lemniscate areasine as a fundament. So watch carefully:

Most probably you know this already in a very accurate way:

$v={\frac {\operatorname {sn} (u;k)}{\operatorname {cd} (u;k)}}=\tan {\bigl [}{\frac {1}{2}}\operatorname {am} (2u;k){\bigr ]}$

$\operatorname {slh} (u)={\frac {\operatorname {sn} (u;{\tfrac {1}{2}}{\sqrt {2}})}{\operatorname {cd} (u;{\tfrac {1}{2}}{\sqrt {2}})}}=\tan {\bigl [}{\frac {1}{2}}\operatorname {am} (2u;{\tfrac {1}{2}}{\sqrt {2}}){\bigr ]}$

And most probably you also know very accurately, that this is the inverted function in relation to t of that:

$u={\frac {1}{2}}\mathrm {F} {\bigl [}2\arctan(v);k{\bigr ]}=\int _{0}^{1}{\frac {v}{\sqrt {(v^{2}\varphi ^{2}+1)^{2}-4k^{2}v^{2}\varphi ^{2}}}}\,\mathrm {d} \varphi$

$\operatorname {aslh} (v)={\frac {1}{2}}\mathrm {F} {\bigl [}2\arctan(v);{\tfrac {1}{2}}{\sqrt {2}}{\bigr ]}=\int _{0}^{1}{\frac {v}{\sqrt {v^{4}\varphi ^{4}+1}}}\,\mathrm {d} \varphi$

And therefore:

${\frac {\mathrm {d} u}{\mathrm {d} v}}={\frac {1}{\sqrt {(v^{2}+1)^{2}-4k^{2}v^{2}}}}$

${\frac {\mathrm {d} }{\mathrm {d} v}}\operatorname {aslh} (v)={\frac {1}{\sqrt {v^{4}+1}}}$

This so called incomplete elliptic integral must be used!

The following part is completely new!

Now I give you a special quintic equation pattern as an example that can be solved in following way:

${\frac {x^{5}-5{\sqrt[{4}]{5}}(7-3{\sqrt {5}})x^{3}+5({\sqrt {5}}-2)x}{5({\sqrt {5}}-2)x^{4}-5{\sqrt[{4}]{5}}(7-3{\sqrt {5}})x^{2}+1}}=z$

For every single real value z you can create the solution of this equation using following algorithm:

$y_{0}=\operatorname {slh} {\bigl [}{\frac {1}{5}}\operatorname {aslh} (z){\bigr ]}=\tan {\biggl \langle }{\frac {1}{2}}\operatorname {am} {\bigl \{}{\frac {1}{5}}\mathrm {F} {\bigl [}2\arctan(z);{\frac {1}{2}}{\sqrt {2}}{\bigr ]};{\frac {1}{2}}{\sqrt {2}}{\bigr \}}{\biggr \rangle }$

$y_{1}=\operatorname {slh} {\bigl [}{\frac {1}{5}}\operatorname {aslh} (z)+{\frac {2}{5}}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\bigr ]}=\tan {\biggl \langle }{\frac {1}{2}}\operatorname {am} {\bigl \{}{\frac {1}{5}}\mathrm {F} {\bigl [}2\arctan(z);{\frac {1}{2}}{\sqrt {2}}{\bigr ]}+{\frac {4}{5}}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )};{\frac {1}{2}}{\sqrt {2}}{\bigr \}}{\biggr \rangle }$

$y_{2}=\operatorname {slh} {\bigl [}{\frac {1}{5}}\operatorname {aslh} (z)+{\frac {4}{5}}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\bigr ]}=\tan {\biggl \langle }{\frac {1}{2}}\operatorname {am} {\bigl \{}{\frac {1}{5}}\mathrm {F} {\bigl [}2\arctan(z);{\frac {1}{2}}{\sqrt {2}}{\bigr ]}+{\frac {8}{5}}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )};{\frac {1}{2}}{\sqrt {2}}{\bigr \}}{\biggr \rangle }$

$y_{3}=\operatorname {slh} {\bigl [}{\frac {1}{5}}\operatorname {aslh} (z)-{\frac {4}{5}}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\bigr ]}=\tan {\biggl \langle }{\frac {1}{2}}\operatorname {am} {\bigl \{}{\frac {1}{5}}\mathrm {F} {\bigl [}2\arctan(z);{\frac {1}{2}}{\sqrt {2}}{\bigr ]}-{\frac {8}{5}}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )};{\frac {1}{2}}{\sqrt {2}}{\bigr \}}{\biggr \rangle }$

$y_{4}=\operatorname {slh} {\bigl [}{\frac {1}{5}}\operatorname {aslh} (z)-{\frac {2}{5}}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )}{\bigr ]}=\tan {\biggl \langle }{\frac {1}{2}}\operatorname {am} {\bigl \{}{\frac {1}{5}}\mathrm {F} {\bigl [}2\arctan(z);{\frac {1}{2}}{\sqrt {2}}{\bigr ]}-{\frac {4}{5}}K{\bigl (}{\frac {1}{2}}{\sqrt {2}}{\bigr )};{\frac {1}{2}}{\sqrt {2}}{\bigr \}}{\biggr \rangle }$

In WolframAlpha you have to enter this for all these y-values:

tan(1/2*am(1/5*elliptic F(2*arctan(z),1/2),1/2))

tan(1/2*am(1/5*elliptic F(2*arctan(z),1/2)+4/5*K(1/2),1/2))

tan(1/2*am(1/5*elliptic F(2*arctan(z),1/2)+8/5*K(1/2),1/2))

tan(1/2*am(1/5*elliptic F(2*arctan(z),1/2)-8/5*K(1/2),1/2))

tan(1/2*am(1/5*elliptic F(2*arctan(z),1/2)-4/5*K(1/2),1/2))

By entering y into that quintic polynome you get the solution for the quintic equation shown above:

$x_{01234}={\frac {y_{01234}^{5}+{\sqrt[{4}]{5}}(3+{\sqrt {5}})y_{01234}^{3}+({\sqrt {5}}+2)y_{01234}}{({\sqrt {5}}+2)y_{01234}^{4}+{\sqrt[{4}]{5}}(3+{\sqrt {5}})y_{01234}^{2}+1}}$

In WolframAlpha you have to write following expression:

(y^5+5^(1/4)*(3+sqrt(5))*y^3+(sqrt(5)+2)*y)/((sqrt(5)+2)*y^4+5^(1/4)*(3+sqrt(5))*y^2+1)

This creates your five real x-solutions for the mentioned equation indeed!

And then you can check the correctness by entering your five x solutions:

(x^5-5*5^(1/4)*(7-3*sqrt(5))*x^3+5*(sqrt(5)-2)*x)/(5*(sqrt(5)-2)*x^4-5*5^(1/4)*(7-3*sqrt(5))*x^2+1) = z

That is the way it always works!

You can try every value z on this and it always works!

It always gives you the five real solutions on the formulated equation pattern!

If you want, you can make experiments with that.

It looks so fascinating.

This is what I am researching right at the moment. And yes, there are analogies that use other modulus values than k = 1/2*sqrt(2) but they have to be created in a special way I will explain you next time. If I find out how to use this algorithm for solving the general case of quintic equations I even straight away will tell you this too. But I still have to research how this is really made in the singular steps. So I need my definitely much longer time of researching until I really got the solution. It is my next big puzzle. Hopefully I will be successful. My current situation looks comparatively good. I continue my researches consequently. Have a nice time and have a lot of success in your career!

Yours faithfully and sincerely!!

Lion Emil Jann Fiedler Reformbenediktiner (talk) 10:30, 4 November 2024 (UTC)

Hi Emil Thank you for the reply. Give me time to digest all these information. I am at home with the Jacobi theta function, the Dedekind eta function, and many others, but I've somehow neglected the lemniscate elliptic functions so it will take me time to be familiar with it. However, I've found two new ways to solve the Bring quintic using Ramanujan's g-and G-function and Hermite's method. I believe it is connected to your two methods using $\vartheta _{3}(q)$ and $\vartheta _{4}(q)$ since the same octic and quartic in $k$ appear, but I don't yet know how to prove it. This is the MO link. Titus III (talk) 05:49, 6 November 2024 (UTC)

Of course I give you time to digest all information. I definitely do not want to practice a so called feeding up. But I nonetheless will give you one further thing. It is a link about the Ramanujan g-function and G-function indeed. I do not want to make your talk page to full. But I myself discovered something that I wrote down immediately about the Ramanujan formulas about the reciproke of the circle number 1/π and it is standing in some of my good pages from the German Wikipedia. Therefore I decide to give you the link to the German page of exactly these two functions g and G and two further essays that do also give you the solution clue for the famous Ramanujan 1/π sum series. Do me a favor and click on these three links:

https://de.wikipedia.org/wiki/Ramanujansche_g-Funktion_und_G-Funktion

https://de.wikipedia.org/wiki/Jacobische_Thetafunktion

https://de.wikipedia.org/wiki/Elliptische_Integrale

And please read every single Wikipedia article of those three especially where the word Kreiszahlformeln appears. All three of them! You can make the strg-f trick and enter the word Kreiszahlformeln if you want. So you can find it right away in these three articles. But read the corresponding sections of these three articles very carefully! The formulas you can find at these sections are correct. I checked it myself. It took me a while to find them out. But you can learn them all. I always want to be a helping hand for you. We definitely work together indeed. I do not want to make your talk page to full. I want to give you time for digesting all my information to you. I do not want to jostle and I do not want to hustle either. Therefore I end this comment here and wait for your next answer. I wish you a lot of success in digesting everything.

Yours faithfully and sincerely!

Lion Emil Jann Fiedler Reformbenediktiner (talk) 15:09, 6 November 2024 (UTC)

Calculation examples of the Brioschi form

I think that I also should give you this:

https://www.youtube.com/watch?v=0vTce-1iTVY

This video shows how I solve the Principal Quintic equations $x^{5}-x^{2}-1=0$ and $x^{5}+x^{2}+x-1=0$ accurately. I use these input in the WolframAlpha engine:

x^5 - x^2 - 1 = 0

q = nome(cos(1/2*arccot(sqrt(3)*coth(1/3*arcoth(1/24*sqrt(a/3)))))^2) where a = 1/2*(31757250331 - 249778425*sqrt(16165))

f = (2*theta(3,0,q^5)*theta(3,0,q^(1/5))-2*theta(3,0,q)^2)/(theta(3,0,q^(1/5))^2+5*theta(3,0,q^5)^2-4*theta(3,0,q)^2)

z = f*(5*f^2-10*f+4)/(5*f^2-6*f+2)

(4-4*sqrt(a/3)*tanh(1/3*arcoth(1/24*sqrt(a/3)))*y)/(a*y^2 - 3) = z where a = 1/2*(31757250331 - 249778425*sqrt(16165))

x = (1/2*(398625 - 3133\sqrt(16165))y - 1/2*(sqrt(16165) - 125))/(1/2*(31757250331 - 249778425*sqrt(16165))*y^2 - 3)

x^5 + x^2 + x - 1 = 0

q = nome(cos(1/2*arccot(sqrt(3)*coth(1/3*arcoth(1/24*sqrt(a/3)))))^2) where a = 1/20511149*(983008525*sqrt(8870)+67467103347)

f = (2*theta(3,0,q^5)*theta(3,0,q^(1/5))-2*theta(3,0,q)^2)/(theta(3,0,q^(1/5))^2+5*theta(3,0,q^5)^2-4*theta(3,0,q)^2)

z = f*(5*f^2-10*f+4)/(5*f^2-6*f+2)

(4-4*sqrt(a/3)*tanh(1/3*arcoth(1/24*sqrt(a/3)))*y)/(a*y^2 - 3) = z where a = 1/20511149*(983008525*sqrt(8870)+67467103347)

x = (1/24389*(2110353+1312*sqrt(8870))*y-1/29*(sqrt(8870)+82))/(3 - 1/20511149*(983008525*sqrt(8870)+67467103347)*y^2)

This concept really works! I am so glad that I already found these solution ways out. I continue the solving way of quintic equations by Lemniscatic elliptic functions. And I decide to make a Tschirnhaus transformation on the quintic pattern for the lemniscate function and then create a Brioschi form in order to be able to solve the Brioschi form by Lemniscatic function expressions. Hopefully everything will be excellent. Have a nice time!