But the variable should approach infinity

I wish you we around 10 years ago when I was first tackling these kinds of problems!

That is so beautiful. Thanks for sharring.

Excellent problem and solution. Good choice of numbers too.

"Recreational Mathematician" I did not know there was a name for us 🧠

6+1+7+4=18 1+8=9 I am not surprising it breaks down to 9 the most divine in Sacred Geometry

love from Bangladesh 🖤

Just tell it no one loves it, no need to overcomplicate things.

What is the practical application of this? Like if if you quickly arrive at a number more then the number of atoms in the known universe what the hell are you even counting?

Dressed up really nicely Very sophisticated

What would be a good linear algebra book for self study that has the Cayley-Hamiltonian and problems such as finding square roots of matrixes?

Much too complicated. Let's do it differently: Let A as described in the video. Then, let X^2 = A, with X = matrix of x1...x4. Then, a1 = x1^2 + x2*x3; a2 = x1*x2 + x2*x4; a3 = x1*x3 + x2*x4; a4 = x2*x3 + x4^2. Four equations, four unknowns. Solve. Done.

I sent question in ur mail but no response yet from you

goat.

Nice exercise!

4:03 😂😂😂 I'm dead. I havent laughed that hard in a math video in a long time. Hahahaha 😆 😂 😆 But for real... I hate this problem... sometimes I wish math was easier.

plzzz prove Cayley-Hamilton theorem

You need to show your preferred strategy 🙂for the integral...

So interesting! Thanks.

x^2-2x+1=(x-1)•(x-1)=(x-1)^2 ∫ 5 -> -1 (x-1)^2•dx <=> ∫ (x-1)^3/3•dx [(5-1)^3/3- (-1-1)^3/3)•dx] 5 -> -1 [21+1/3 + 2+2/3• dx] = ∫ [24•dx]5 -> -1 By using the definition, we can calculate the summary as follows: (x^2-2x+1)=0 <=> p=2; q=1 -p/2± [(p/2)^2-q] <=> x=1 ∫ [24•dx]5-> -1=lim n->5 Σ i=1f(24i)•6•24 => Σ=24^2•6•1 = 3.456

Sir, why one of those cube root of 1 is named omega? Is there a reason for it? 🤔

I have another idea about the proof. Let √A = k ( A + p I ) Then A = k^2 ( A + p I )^2 = k^2 (A^2 + 2pA + p^2 I) Then use Cayley-Hamilton theorem to reduce A^2 in terms of A and I, and then comparing the coefficients of A and I on both sides and then solve for k and p.

Imma dance at outro

Amazing sir this video is really help me

Well said stop learning stop living

Well done

Is there a formula for the nth root of the 2 x 2 matrix A?

Prime newtons you sound like Richard Mofe Damijo and I imagine you are a Nigerian

Bravo !!!!! Very good, Sir.

As pointed out by others, the proof near the end of the video is incorrect, and completely lacking in rigor. However, I do not blame you for this mistake, because as it happens, this Diophantine equation is actually an unsolved open problem in number theory, and there have been many dozens of research papers published on this question, but with no success in solving it. The problem here is that it is possible that a = b. To discard the possibility of other solutions besides (3, 3, 4) existing, one must discard a = b for a > 3. This enables proving that a! = sqrt(c! + 1) + 1, so for c > 4, c! + 1 must be a perfect square. This leads to the auxiliary Diophantine equation c! + 1 = d^2, known as Brocard's problem, which you can search for yourself online and see that this is indeed an unsolved problem, as there has never been a proof published to correctly demonstrate that there are no other solutions besides c = 4, c = 5, and c = 7. c = 5 and c = 7 do not lead to new solutions for a!, but as it has not been proven that there do not exist any other such c!, there is also no proof that there are no other a! satisfying a! = sqrt(c! + 1) + 1. These comments in the comment section do attempt to prove that there exist no other such a!, but they are all incorrect and make mistakes which have mostly been pointed out by those who replied to said comments.

You missed a trick. Use the Cayley-Hamiltonian theorem again, X^2-tr(X)X+det(X)I_2=0. Note that taking the trace is a LINEAR operation. Take the trace to obtain: tr(X^2)-(tr(X))^2+2det(X)=0. Note that X^2=A, and det(X)=sqrt(det(A)) and rearrange to get: (tr(X))^2=tr(A)+2sqrt(det(A)), take square roots to get tr(X)=sqrt(tr(A)+2sqrt(det(A))). I think that this is slicker.

Thanks!

It's apparent that A can't be singular for this to work, right? You can't have a square root of a singular matrix, is that right?

Congrats my friend. Like the blackboard and working with chalk. You 're a brilliant example for all the teachers...!!

Note that tr(A) and dat(A) are invariants of the matrix. So I suspect that there is a topological derivation of this result which is quite simple in application.

16

Why can't you use your own name rather than Newton's?

you can say n2^n + 1 is a square so by (a+b)^2 we can say a^2 + 2a = n2^n let a be 2^x n2^n = 2^2x + 2*2^x n2^n = 2^2x + 2^(x+1) n2^n = 2^(x+1) [ 2^(x-1) + 1] n2^n = [ 2^(x-1) + 1] * 2 ^ (x+1) so here if we assume x + 1 = n then 2^(x-1) +1 should also be equal to n x+1 = 2^(x-1) + 1 x = 2 ^ (x - 1) 2x = 2 ^ x how here you can either make an educated guess that x = 1 and x = 2 satisfies solution or you can solve for x with some rearrangement and lambert w function so if x = 1 then n = x +1 aka n = 2 and for x = 2 ; n = 3 or you can use other equation also for finding out n If any one is reading this please let me know is there any problem with my solution as i am not from maths background

Tetration is literally- so large, I can see why he chose a base of 2 for the example. ²2 equals to 4, but ³3 already equals to 7,625,597,484,987.

Very good professor!

Thank you for teaching me this one, now imma trick my classmates hehe😅

Superb! For linear algebra I recommend Prime Newtons.

Muy buenos videos amigo, saludos

Hi, I'm from Mexico, and I´m studying computing engeneer, and this kind of exercises caught my attention, this formula or this topic I've never seen on my Linear Algebra course, and I would like to know how can I find this theme or if this is particularly on a Lineal Algebra Course, Very nice video i learned something new. Thanks

Вы хорошо объясняете и всё понятно кроме одного как вычислить W? поскольку у меня на калькуляторе есть кнопка [ln], но нет кнопки [W]

Outstanding. Best maths tutor by far.

The squareroot of a matrix is a SPINOR!!!!!

You have made my work easier sir 😁😉

Thank you so much. God bless you 🙏❤ You're a man of your word. Thank you for the likes 😊