I just like maths and sharing it :) Video each Wednesday
A little bit about myself: I studied mathematics and physics and specialised on particle physics, writing lots of code and doing lots of integrals. :) Now I’m teaching mathematics and computer science.
Yes, this video contains a few minor errors, as others have noted before. If u = cos(2ϴ) (what had been defined before), then the bounds of the integral at the upper right at 11:30 should be ϴ₁ = arccos(u₁)/2 = arccos(1/2) / 2 = π/3/2 = π/6 ϴ₂ = arccos(u₂)/2 = arccos(1) / 2 = 0 And at 10:45 bottom right this should of course read cos(2ϴ) = 1 - 2sin²(ϴ) so that 1 - u = 2sin²(ϴ) 1 + u = 2cos²(ϴ) and du/dϴ = -2sin(2ϴ), rather than -2sin(ϴ) at 10:50.
Wikipedia says the conjecture has been tested out to something like 4E18. That seems like quite the induction step accomplished. So what is preventing the conjecture from being proven?
I thought you were a mathematician who'd made a breakthrough in the area. Big disappointment. 😓 The video is a great introduction to the topic though; I'd suggest writing the names with the portraits to better associate them together.
@@thepathintegratorum, Kepler laws are in class 10 and 11 in gravitation chapter, although mathematical proofs shown here are not available in our state books atleast.
hi, what source did you use for this proof? I would like to use this method for my assignment, but I can only do that if I have the real source, from an article, a text or something else :(
Interesting question! This would change the integral a bit and make it harder to solve. Also we would have to decide if it’s a negative or a positive curvature we are dealing with. Negative curvature probably would lead to a contraction of the universe and a smaller observable radius.
@@thepathintegrator If I remember correctly, it’s James Stewart’s “Calculus,” and the chapter with the derivations for Kepler’s 1st, 2nd, and 3rd laws should be under Chapter 13, “Vector Functions,” under the section “Kepler’s Laws of Planetary Motion.” I think there’s a PDF floating around online that has the entire chapter that you can view without having to purchase the textbook. :)
@@thepathintegrator In case you haven't had an opportunity to look it up: Chapter 13 of Stewart's book is now available on math.libretexts. Title: 'Calculus - Early Trancendentals' (The overview page still has a 'currently under construction' notification. In any case; Chapter 13 is there, including section 13.4 with the discussion of deriving Kepler's first law using vector calculus.)
Thanks, good to have that available on youtube! (Well, there's a lot of technical stuff in there; so, while the proof is 'clear', it still das not 'convince' someone 'obviously' of the relation between the 1/r²-law and the ellipse. Is there an easier way to realize this? Maybe starting with the special circular solution and disturbing it?
Thanks so much for your appreciation. I didn’t know this doesn’t already exist here on RU-vid. I think there are easier proofs of course but I liked that one specifically since it uses theoretical physics concepts and calculus.
It is very nice the information appears gradually, albeit a little bit fast, as it would be written by hand on a board. The nice animations being integrated with the mathematical derivation creates a wonderful presentation.
I am so happy to read your comment :) thank you so much for your appreciation, it’s absolutely amazing to see how you enjoyed it. I will also try in the future to slower my pace a bit. I’m always worried that you guys might get bored to easily because :)
*You emphatically do **_not_** need to compute integrals to determine the area of the ellipse.* You just need to know the general fact that Lebesgue measure changes of det(A) under a linear change of coordinates A. Then you apply this fact to the transformation A with x-> ax, y-> by, and det(A)=ab. The area of the circle is π, hence the area of the ellipse is πab. Easy peasy.
That’s the amazing thing there is always so many ways to do it. I personally just love integrals but going via change of coordinates is also an elegant way
cos(2φ)=2cos^2(φ)-1 so cos^2(φ)=((1+cos(2φ))/2 so Integ(cos^2(φ) dφ = =Integ(1/2)*dφ+1/2*Integ(cos2φ)dφ All Integrals under limits from φ=0 to φ=π/2. Fiinally A=4ab*(1/2*(φ))+1/2*(sin(2φ)) from φ=0 to φ=π/2 so Α=4ab*((1/2)*(π/2)+(1/2)*0)=4ab*π/4=πab.
Nice! Physics mostly deals with definite integrals, where the constants of integration cancel out during evaluation per the fundamental theorem of calculus. So they can be omitted from a side proof demonstrating the derivation of a pending formula substitution without lost of generality, as it were.
I have sent answer about the previous problem the ellipse that you asked. I have combined the Complex solution in one. Now i΄ll look for this solution of circle.
@@thepathintegrator Answer is in my comment to subject ''The solution is an ellipse''. Isay that it has a form of Complex Logaritm and you said ''Can you combine them to one?''. It was a good challenge and i did combine them. It was one of my very good works. I believe if you read it you''ll like it alot.
I don't really know how the algorithm works to recommend vids to people, so all I can say is that I already enjoy the videos a lot with how they currently are :D
