This is fantastic! While building a Desmos activity, I had to work out the trigonometric solutions myself from Cardano's formula. I knew the complex cube roots would make an equilateral triangle, but somehow I hadn't connected that with the geometric "curiosity", which I had seen mentioned in the Mathologer video. It was really neat to see it done that way, without even using complex numbers. As a geometry teacher myself, I find this especially satisfying.
So... Would complex analysis tell me I couldn't generalize this for any polygon and polynomial equation because... I lack degrees of freedom to exploit? Aka, rotating and expanding is not enough? I've never seen this... Method... Before, but I imagined it had connection with the polar representation, since x³=-1, for instance, gives us roots -1, 1/2+-(1/2)*√3i in an equilateral triangle and a coordinate transformation could probably connect this equation with a family of other ones. I must be missing something to not see why we aren't just solving all tame polynomial equations like this, then... I'll have to think about this I suppose... Btw, the algorithm loves you, because it just decided to show me this and I'll never feel like knowing the root formula for cubic equations ever again. It wasn't often, but I Know i tried memorizing it a few times...
for real values of s, the cbrt(s) = the real value for cbrt(s), (which can be found using a calculator or the above method he just showed, but if the other two roots are complex), you can get them by saying the 2nd root is cbrt(s)e^(2pi/3)i and cbrt(s)*e^2(2pi/3)i (thats why omega squared is another root of unity btw). Then, just use euler's formula to find cos(x) + isin(x) and youll land on exactly the complex roots of the number. This is true cause the complex roots always form an equilateral triangle when the value of s is real. (i tried modifying my formula to fit cube roots for any complex number, and it miserably broke, so im constraining it to the real numbers) Completely on a different note, the s-th root of c can be generalised to be x_(n+1) 1/s[(s-1)x_(n) + c/x_(n)^(s-1)] (hopefully its visible in that notation, its also an iterative process, just start with any value you can imagine for x_0 and iterate and it gets closer and closer to one of the s roots of c)
23:30 where the inscribed circle in the triangle is replaced with an inscribed ellipse in the triangle reminds me of inverting the elliptic integral which is then called an elliptic function is used in the parameterization of elliptic curves which are CUBIC curves. And the inscribed circle represents the Gaussian integers on the fundamental domain where complex(i) infinity is the cusp at infinity. Because the Gaussian integers is a square lattice representing the circle inscribed. And the ellipse inscribed representing for example the Eisenstein integers which is not a square lattice. So you go from a circle to an ellipse loosing one degree of symmetry. So this may give me a new way to look at elliptic curves.
as a fun fact, the circumradius is : a / sin A = 2R since it is an equilateral triangle, this becomes : s / ( sin 60 ) = 2R s / ( sqrt 3 / 2 ) = 2R 2*s / sqrt 3 = 2R so s / sqrt 3 = R and inradius, r is : r = A / s = ( 1 / 4 )*s^2 * sqrt 3 / (3s / 2 ) = ( 1 / 2 ) *s / sqrt 3 so that means that R = 2r QED
I’m not sure if this generalizes to the quartic, but it is also related to solving using cos3x’s expansion and it can work with cos4x for the quartic. Might be hard to find a nice geometric interpretation though.
I think I recall stumbling upon a hyperbolic solution for cubics with one real solution while researching for this video, might be interesting to look into.
I’m not sure that this geometric representation generalizes to the quartic or it might do so with a rectangle for example instead of a square since it’s not always symmetric. This triangle for a cubic is also related to the process of solving using the formula for cos3x and the same thing can be applied to the quartic with cos4x, could be a good starting point.
It also manages to work with 2 solutions, 2 vertices will be on top of each other but still can be solved. However it does not work for 1 solution and there is supposedly a hyperbolic solution for that instead
I can't believe so few people have seen this video with a SUCH an interesting fact in it! This is was really really good! Hope to see more!
This is fantastic! While building a Desmos activity, I had to work out the trigonometric solutions myself from Cardano's formula. I knew the complex cube roots would make an equilateral triangle, but somehow I hadn't connected that with the geometric "curiosity", which I had seen mentioned in the Mathologer video. It was really neat to see it done that way, without even using complex numbers. As a geometry teacher myself, I find this especially satisfying.
Welp, at least youtube is starting to recommend this great video
Very interesting material based on the relation of the triangle rotated and the cubic equation.
Once again youtube algorithm met me with a beautiful youtube channel
So... Would complex analysis tell me I couldn't generalize this for any polygon and polynomial equation because... I lack degrees of freedom to exploit? Aka, rotating and expanding is not enough?
I've never seen this... Method... Before, but I imagined it had connection with the polar representation, since x³=-1, for instance, gives us roots -1, 1/2+-(1/2)*√3i in an equilateral triangle and a coordinate transformation could probably connect this equation with a family of other ones.
I must be missing something to not see why we aren't just solving all tame polynomial equations like this, then... I'll have to think about this I suppose...
Btw, the algorithm loves you, because it just decided to show me this and I'll never feel like knowing the root formula for cubic equations ever again. It wasn't often, but I Know i tried memorizing it a few times...
for real values of s, the cbrt(s) = the real value for cbrt(s), (which can be found using a calculator or the above method he just showed, but if the other two roots are complex), you can get them by saying the 2nd root is cbrt(s)e^(2pi/3)i and cbrt(s)*e^2(2pi/3)i (thats why omega squared is another root of unity btw).
Then, just use euler's formula to find cos(x) + isin(x) and youll land on exactly the complex roots of the number. This is true cause the complex roots always form an equilateral triangle when the value of s is real. (i tried modifying my formula to fit cube roots for any complex number, and it miserably broke, so im constraining it to the real numbers)
Completely on a different note,
the s-th root of c can be generalised to be x_(n+1) 1/s[(s-1)x_(n) + c/x_(n)^(s-1)]
(hopefully its visible in that notation, its also an iterative process, just start with any value you can imagine for x_0 and iterate and it gets closer and closer to one of the s roots of c)
23:30 where the inscribed circle in the triangle is replaced with an inscribed ellipse in the triangle reminds me of inverting the elliptic integral which is then called an elliptic function is used in the parameterization of elliptic curves which are CUBIC curves. And the inscribed circle represents the Gaussian integers on the fundamental domain where complex(i) infinity is the cusp at infinity. Because the Gaussian integers is a square lattice representing the circle inscribed. And the ellipse inscribed representing for example the Eisenstein integers which is not a square lattice. So you go from a circle to an ellipse loosing one degree of symmetry. So this may give me a new way to look at elliptic curves.
Absolute legend!
as a fun fact, the circumradius is : a / sin A = 2R
since it is an equilateral triangle, this becomes : s / ( sin 60 ) = 2R
s / ( sqrt 3 / 2 ) = 2R
2*s / sqrt 3 = 2R
so s / sqrt 3 = R
and inradius, r is : r = A / s = ( 1 / 4 )*s^2 * sqrt 3 / (3s / 2 ) = ( 1 / 2 ) *s / sqrt 3
so that means that R = 2r
QED
Thank You, Sir
Is there something like this for a 4th degree polynomial? Not the Ferrari formula, but something similar to this method.
I’m not sure if this generalizes to the quartic, but it is also related to solving using cos3x’s expansion and it can work with cos4x for the quartic. Might be hard to find a nice geometric interpretation though.
Compliments! Interesting definitely 👍
Really good!
Me at 2 am: *scrolling tru youtube
UA-cam:
😂
Can you use hyperbolic solutions as well? Just asking....
I think I recall stumbling upon a hyperbolic solution for cubics with one real solution while researching for this video, might be interesting to look into.
out of curiosity, how would one apply this kind of theory to a quartic ?
I’m not sure that this geometric representation generalizes to the quartic or it might do so with a rectangle for example instead of a square since it’s not always symmetric. This triangle for a cubic is also related to the process of solving using the formula for cos3x and the same thing can be applied to the quartic with cos4x, could be a good starting point.
I feel like I saw some magic
Does this work in the case where the cubic has 1 or 2 solutions?
It also manages to work with 2 solutions, 2 vertices will be on top of each other but still can be solved. However it does not work for 1 solution and there is supposedly a hyperbolic solution for that instead
@@mihaiadamescu9861 wow, can you explain more or give me resource for 2 vertices ?
God bless you for this formula.
How to find the h value?
h is the x-coordinate of the inflection point. All cubics can only have one inflection point at x = -b/3a so h is always -b/3a.
i love you