Great explanation! I think it doesn't rule out the possibility of a quintic formula for real coefficients only. The key argument is continuity of the formula within the complex numbers. Am I right?
a/b+c + b/a+c + c/a+b = n a,b,c positive integers. sequence of n: 4, 178, 896, ... What are some higher numbers and how to find.? Is the sequence infinite?
I feel the video is good but there needs to be some corrections. The cubic formula he showed, I don't even know what it is, and me being someone whos actually solved equations using cardano's formula, yes it does give you quite an ugly but exact answer. It's called Cardano's formula and it can be used to solve for 1 root, another seperate derived much easier formula for 2 roots, and 3 roots are pretty hard but possible with Cardano's. 4th degree, you use Ferrari's method to solve for 1 root, and get a left over cubic equation which you do the exact same thing like I said. So the so called quintic equation should reduce it to a 4th degree, not give all 5 answers.
Can we get an example where a nested root changes under a commutator? What make nonsense to me is that, if we can prove sqrt{B} get back to its origin under a commutator, why nested root sqrt{A+sqrt{B}} might move to another value, because sqrt{B} return to its origin and A returns too. Does it mean, under a commutator, when sqrt{B} returns to its origin, sqrt{B} might has gone a circle around -A, so an outer sqrt root might change sqrt{A+sqrt{B}} from x1 to x2? Since the prove that "sqrt{B} returns to its origin under commutator" assume the 0 (which is the origin of the complex plane) as center of the circle and compute the argument of the complex number, it does not guarantee the path of sqrt{B} could form another circle around -A, so the conclusion might evade if we have nested root?
You should show the infinite series formula for all roots of quintic or any polynomial. I like Bernd Sturmfels 2000 paper in Journal of Combinatorics the best, but I am sure others exist.
You mentioned commutators in the context of loops around the origin. Is there any connection between permutations of roots and homotopy classes? Can the complex plane be seen as the universal cover of some space whose fundemental group is related to these permutations?
I had always found all material on Galois theory extremely confusing, and I had already accepted that I would never understand why there's no quintic formula. This video changed everything. Very nice.
A good explanation, but I wanted to point out a little something extra. That pleasant music in the background was very beneficial. It gave the feeling that the explanation was one long smooth continual line of reasoning instead of a bunch of disconnected equations and ideas thrown together. Our attention was able to remain engaged throughout, like listening to a long song. Though I must admit, I had to set the video on a slower speed to be able to absorb it all.
I don't understand one thing: You proved that there is no cubic/quartic/... formula with one nth root by contradiction, but if we take the simple cubic equation: X^3-a=0, then we get the formula x=cubic root(a) which gives the roots. So there are special cases in which one root is enough. How can you explain this ?
There are two cube roots in the solution formula. Both of those cube roots can be interpreted in 3 ways, because you can put 3 different arguments to them. If you choose the arguments of the two cube roots independently, it looks like the formula actually gives 9 solutions. It turns out that 6 of them are wrong, and 3 of them are right. So the solution formula doesn't really work unless you know how to choose the arguments of the cube roots correctly.
Theres no way to use a finite succession to solve, using complex analysis like cauchy integrals You can solve it, which confused me at the start. Good day
This is spectacularly good (it’s literally a great spectacle)! As a retired topologist it makes me wish I could give one more course in complex analysis. (Never thought I’d say that!)
I don't know if others felt similarly but having the variable in the equation be z and having the solutions be z sub n was honestly a little confusing, it would be better if it was for example your variable stayed z and the solutions were y sub n
I am really surprised by the enthousiastic comments. I found the video somewhat interesting but a lot of points are confused or at least not strait mathematically. In fact it is difficult to promise a demonstration with no Galois theory and immediately jump into root permutations as Galois theory is exactly the theorisation of these permutations ! Thanks for serving us scrambled eggs while arguing they where made with no eggs at all !
20:39 First of all, wouldn't it make sense to rewrite the theorem like this: "The quadratic equation cannot be solved with a continuous single-valued function", because by convention the root function is also continuous and single-valued and hence is not essential to the argument. The argument would still be valid. However, the argument only works for certain values of the coefficients a, b and c. In your example, you're looking at the quadratic equation H(z; theta) = z^2 - 4 exp(2 i theta) = (z - 2 exp(i theta)) (z + 2 exp(i theta)) so that H(z; 0) = H(z; Pi). In this particular case the argument is valid. However, for other quadratic functions H(z; theta) with H(z; 0) = H(z; Pi), this is not true. For instance H(z; theta) = z^2 + 3 z - 4 exp(2 i theta), the functions z1(theta) and z2(theta) do move along a closed path, albeit a discontinuous one, while for H(z; theta) = z^2 + 5 z - 4 exp(2 i theta), z1 and z2 return to their starting point even on a continuous path. So, essentially we have three categories of quadratic functions defined in terms of the continuity and closedness of the z1 and z2 paths. Could it be then, that a similar categorization is also possible for equations of degree 5 and that for some of these categories there do exist explicit algebraic solutions?
Is there a proof that it is always possible to take some path around the origin to swap two of the roots and leave the others unchanged? Wouldn't it always just cycle through all the roots?
39:11 It’s really quite remarkable that there exists arbitrarily nested commutators that genuinely swap around 5 points, when there are only 5! = 120 possible orderings of those 5 points; quite far removed from ∞ 😮.
@not all wrong 21:00 Isn’t that kind of circular reasoning, though; since the 2 solutions *_COME FROM_* the formula involving a square root? Then, the premise kind of already assumes/presupposes the presence of a square root, in the formula. So, it’s like: ”OK; so, we have 2 solutions, because we have a square root; and we need the square root, because we have 2 solutions.”, right? 🤔
11:53 At least, in the factorized form; the expression is not the same, though. The value is the same; as are the coefficients, in the polynomial form; but the expression is different, because the order of some of the factors has been flipped. Just like, for example, the expressions: ”5*2” and: ”2*5” are different; although, their values and coefficients are the same. The expressions look different, since they feature a permutation; meaning: The two expressions don’t have the same numbers, *_AT THE SAME PLACES, IN THE SAME ORDER;_* hence, the expressions are different. They’re 2 different expressions of the same product. For example, for quaternions; the value of the products wouldn’t even be the same: ijk ≠ ikj ≠ jik ≠ kij ≠ kji ≠ jki (that’s, actually, what the commutators, in Rubik’s cubes are based on: The order matters).
14:27 So, in this case, you’d multiply z² by (-1)², resulting in z²*(-1)² = (z*(-1))² = (-z)² = w*(-1)² = w*1 = w; but, instead of just multiplying w by 1, straight away, you multiplied it by (-1)², to make sure that w completes a full revolution around the origin, instead of just staying put; and, to put it algebraically: You’ve multiplied z² by (-1)², instead of just 1, to make sure that the extra factor, ”added” to z, is really -1, instead of +1. 💡 😃
I guess all cyclic permutations commute with each other; because they’re, in a way, ”1-dimensional” permutations, there’s only 2 directions (clockwise & anti-clockwise), 2 options to permute the elements, for the permutation to still be cyclic. There’s not enough ”space to maneuver” / ”freedom to move around” for a non-trivial commutator. 🤔
