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 Razonar #101 Posted : 04 November 2022 20:25:39(UTC) Rank: Advanced MemberGroups: Registered Joined: 28/08/2014(UTC)Posts: 1,374Was thanked: 834 time(s) in 528 post(s) Originally Posted by: алексей_алексей It turned out that Draghilev's method is capable of solving Diophantine equations.Can be viewed here.Could that be related to one of the seven millennium problems: Birch and Swinnerton-Dyer conjecture? (In Spanish, hope that can add automatic translations). Can someone tell Grigori Perelman?Best regards.Alvaro.
 алексей_алексей #102 Posted : 04 November 2022 21:49:32(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Originally Posted by: Razonar Could that be related to one of the seven millennium problems: Birch and Swinnerton-Dyer conjecture? This question is not for me, I just suggested a simple way to solve equations in integers. Such theories are too complex for me to understand.
 Razonar #103 Posted : 05 November 2022 19:53:35(UTC) Rank: Advanced MemberGroups: Registered Joined: 28/08/2014(UTC)Posts: 1,374Was thanked: 834 time(s) in 528 post(s) Originally Posted by: алексей_алексей ... Such theories are too complex for me to understand.For me too, that's why I use an outreach video aimed at a general public and in my mother tongue.Best regards.Alvaro.
 алексей_алексей #104 Posted : 17 December 2023 03:44:54(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Rolling without slipping. Once again. There, small pictures are links to full-fledged animations. The equation of the rolling surface is displayed on the graph. And just a few words in Russian.1. 2.
 Jean Giraud #105 Posted : 17 December 2023 11:12:34(UTC) Rank: GuestGroups: Registered Joined: 04/07/2015(UTC)Posts: 6,866Was thanked: 981 time(s) in 809 post(s) Originally Posted by: Razonar Originally Posted by: Ð°Ð»ÐµÐºÑÐµÐ¹_Ð°Ð»ÐµÐºÑÐµÐ¹ [/url]It turned out that Draghilev's method is capable of solving Diophantine equations.[url=https://www.mapleprimes.com/posts/218880-Diophantine-Equations]Can be viewed here.We have Diophantine in Smath 30405, works superb from MCD style.Jean
 grelkin2 #106 Posted : 18 December 2023 20:41:36(UTC) Rank: Advanced MemberGroups: Registered Joined: 12/12/2020(UTC)Posts: 35Was thanked: 3 time(s) in 3 post(s) Originally Posted by: алексей_алексей Rolling without slipping. Once again. There, small pictures are links to full-fledged animations. The equation of the rolling surface is displayed on the graph. And just a few words in Russian.1. Можешь выложить анимацию или уравнения с начальными условиями сюда?. Качество очень низкое в превью.
 алексей_алексей #107 Posted : 19 December 2023 02:59:42(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Можешь выложить анимацию или уравнения с начальными условиями сюда?. Качество очень низкое в превью.Там картинки являются ссылками. Вот тема на киберфоруме, где все эти рисунки присутствуют. Они в конце. Вот тема на MaplePrimes, где есть основной текст. Он сильно корявенький, но и я не программист.Edited by user 19 December 2023 03:01:46(UTC)  | Reason: Not specified
 алексей_алексей #108 Posted : 19 December 2023 09:02:54(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s)
 алексей_алексей #109 Posted : 30 December 2023 03:40:21(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Mobius strip.Edited by user 03 January 2024 23:21:39(UTC)  | Reason: Not specified
 алексей_алексей #110 Posted : 30 December 2023 03:51:04(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s)
 алексей_алексей #111 Posted : 07 January 2024 21:21:12(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) On a stationary surface x1^4 + x2^4 + x3^4 - 1 = 0, its reduced copy rolls without slipping, the equation of which at the origin is x1^4 + x2^4 + x3^4 - 0.01 = 0. It rolls by "sideways ". The equation corresponding to the current position is displayed on the graph.Rolling without slipping
 алексей_алексей #112 Posted : 10 January 2024 02:17:43(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Rolling without slipping. Another example, but with a very simplified and detailed graph.
 grelkin2 #113 Posted : 25 January 2024 06:17:06(UTC) Rank: Advanced MemberGroups: Registered Joined: 12/12/2020(UTC)Posts: 35Was thanked: 3 time(s) in 3 post(s) [RUS]Рассмотрим движение окружности по параболе без проскальзывания. Центр окружности двигается по эквидистантной кривой к параболе. Для того чтобы движение окружности было без проскальзывания, необходимо чтобы скорость вращательного движения была равна скорости центра окружности, и в точке касания к параболе эти скорости должны быть противоположными по направлению. При выполнении этих условий скорость в точке касания равна нулю и мы наблюдаем "зубья". Полученные кривые являются циклоидными.[/RUS][ENG]Consider the motion of a circle along a parabola without slip. The center of the circle moves along an equidistant curve to the parabola. In order for the motion of the circle to be slip-free, it is necessary that the velocity of the rotational motion be equal to the velocity of the center of the circle, and at the point of tangency to the parabola these velocities must be opposite in direction. If these conditions are met, the velocity at the point of tangency is zero and we observe "teeth". The obtained curves are cycloidal.[/ENG] kol1.avi (1,815kb) downloaded 2 time(s). [RUS]В случае самопересечения эквидистантной кривой картина движения усложняется, тем не менее все условия продолжают выполнятся. В данном случае окружность выходит за границы параболы. Если необходимо, чтобы границы соблюдались, то выполняется "перекат" окружности с одной ветки параболы на другую. Перекатом я называю скачкообразное изменение точки касания окружности к параболе. На рисунке ниже показано "перекат" из точки А в точку В.[/RUS][ENG]In the case of self-intersection of the equidistant curve the picture of motion becomes more complicated, nevertheless all conditions continue to be fulfilled. In this case, the circle goes beyond the boundaries of the parabola. If it is necessary that the boundaries are observed, the circle is "rolled" from one branch of the parabola to another. I call a roll a jump-like change of the point of tangency of the circle to the parabola. The figure below shows a "roll" from point A to point B.[/ENG] kol2.avi (1,640kb) downloaded 2 time(s). kol2a.avi (1,196kb) downloaded 4 time(s).
 алексей_алексей #114 Posted : 29 January 2024 10:01:35(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Draghilev's method also allows us to construct equidistant surfaces.https://en.smath.info/forum/resource.ashx?image=1703https://en.smath.info/forum/resource.ashx?image=1702https://en.smath.info/forum/resource.ashx?image=1701https://en.smath.info/forum/resource.ashx?image=1700Edited by user 30 January 2024 01:02:40(UTC)  | Reason: Not specified
 grelkin2 #115 Posted : 30 January 2024 00:46:51(UTC) Rank: Advanced MemberGroups: Registered Joined: 12/12/2020(UTC)Posts: 35Was thanked: 3 time(s) in 3 post(s) Originally Posted by: алексей_алексей Draghilev's method also allows us to construct equidistant surfaces.https://en.smath.info/forum/resource.ashx?image=1703https://en.smath.info/forum/resource.ashx?image=1702https://en.smath.info/forum/resource.ashx?image=1701https://en.smath.info/forum/resource.ashx?image=1700Не открывается, поправь пожалуйста.
 алексей_алексей #116 Posted : 05 February 2024 08:40:39(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Originally Posted by: алексей_алексей Draghilev's method also allows us to construct equidistant surfaces.A little more on near this subject. Link to one Russian-language forum.
 grelkin2 #117 Posted : 10 February 2024 11:20:45(UTC) Rank: Advanced MemberGroups: Registered Joined: 12/12/2020(UTC)Posts: 35Was thanked: 3 time(s) in 3 post(s) Originally Posted by: алексей_алексей Originally Posted by: алексей_алексей Draghilev's method also allows us to construct equidistant surfaces.A little more on near this subject. Link to one Russian-language forum.[RUS]Поверхность можно получить разными способами. Один из способов получения - решение нелинейных уравнений при помощи метода Драгилева. Полученные решения представляют собой набор замкнутых кривых, составляющие поверхность. При помощи полученных данных можно построить полигональную поверхность.[/RUS][ENG]The surface can be obtained in different ways. One way is to solve nonlinear equations using Dragilev's method. The obtained solutions are a set of closed curves that make up the surface. Using the obtained data, a polygonal surface can be constructed.[/ENG] pov1.png (592kb) downloaded 4 time(s). pov1.mov (3,241kb) downloaded 3 time(s).Edited by user 10 February 2024 13:14:36(UTC)  | Reason: Not specified
 алексей_алексей #118 Posted : 01 March 2024 03:32:47(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Find all solutions to a system of equations z^2 - 0.1*x^4 - 0.05*y^4 +1 = 0x^3 + y^3 + 0.05*z^3 - 1 = 0-2*cos(3*x) + 2*cos(3*y) - 2*cos(3*z) + 1 = 0
 Razonar #119 Posted : 02 March 2024 09:14:50(UTC) Rank: Advanced MemberGroups: Registered Joined: 28/08/2014(UTC)Posts: 1,374Was thanked: 834 time(s) in 528 post(s) Originally Posted by: алексей_алексей Find all solutions to a system of equations z^2 - 0.1*x^4 - 0.05*y^4 +1 = 0x^3 + y^3 + 0.05*z^3 - 1 = 0-2*cos(3*x) + 2*cos(3*y) - 2*cos(3*z) + 1 = 0Hi Alexey. Using Draghilev's method for the Solve Block from this post, with this setup, 40 solutions can be found. By varying the parameters you can find more. DraghilevSolveBlock - alexey example.sm (274kb) downloaded 4 time(s).Best regards.Alvaro. 1 user thanked Razonar for this useful post. on 02/03/2024(UTC)
 алексей_алексей #120 Posted : 02 March 2024 10:42:52(UTC) Rank: Advanced MemberGroups: Registered Joined: 30/09/2012(UTC)Posts: 61Was thanked: 13 time(s) in 8 post(s) Originally Posted by: Razonar 40 solutions can be foundHello Alvaro! Yes, thanks for the work done. This was, one might say, a test example for the Method. This system of equations has 116 solutions. The first two equations describe a closed curve; moving along it, we track the change in the sign of the last equation. There are 116 solutions in total.Edited by user 02 March 2024 21:13:40(UTC)  | Reason: Not specified
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