Ok so guys I'm in 6th grade and this video popped up in my reccomendations and i don't understand anything so please remind me of this channel when i start college
@stewartzayat7526
4 жыл бұрын
Ok, see you in a few years
@suhaibalkhaldi
3 жыл бұрын
Ok
@kabascoolr
3 жыл бұрын
Will do!
@froyocrew
Жыл бұрын
Almost there!
@iduggix8604
7 ай бұрын
A few years still left
@tomdeneckere
4 жыл бұрын
Only recently I started noticing that people (in educational contexts) go to great lengths to prove boundedness on both sides in a proof using the monotone sequence theorem, like you do in this video. When I studied maths, this theorem was formulated as two cases: 1. an increasing sequence that is bounded from above is convergent; 2. a decreasing sequence that is bounded from below converges. Granted that it’s concise to merge the two parts by saying: “a monotone sequence that is bounded converges”. But this has (to my taste) the strange effect that people go on and insist on proving that the sequence is bounded from both sides. Not only is a monotone sequence trivially bounded by its first term (from below or from above, depending on the case), but moreover this trivial upper or lower bound contributes in nothing to the sequence being convergent. So, again to my taste, this is a strange, confusing, counterintuitive practise and since I only saw it appear recently, I wonder if this is due to a new tendency in mathematical didactics (kind of like the introduction of “negative one” vs. “minus one”.) No criticism though, and only praise for your nice videos. Just curious actually.
@suhaibalkhaldi
3 жыл бұрын
Yes , i totally agree with you But this is how student learnt to prove the convergence so we have to stick with that at least on the exam
@antananarywa
Жыл бұрын
I study economics, and we get to know a theorem which states that if a sequence is monotonous, for a big enough n, and it is bounded, then such a sequence is convergent. So, it would be enough if my sequence started with a 3 and I proved it is decreasing with an inf=2 to say that the limit of this sequence is 2? No need to look for an upper bound, is that correct?
@thesecondderivative8967
Жыл бұрын
I agree. I also didn't think it was necessary to show that it was bounded below.
@stronkman9157
3 ай бұрын
@@thesecondderivative8967 Pretty sure that to make the proof as rigorous as possible you have to show that it is monotonous + bounded to show it converges.
@goodplacetostop2973
4 жыл бұрын
8:41
@sumukhhegde6677
4 жыл бұрын
🙌🙌
@DrDailbo
4 жыл бұрын
This problem can also be done from an elementary algebra standpoint by setting the nested root 2's equal to x. Squaring both sides gives the same quadratic, x^2 = 2 + x.
@snnwstt
3 жыл бұрын
@@angelmendez-rivera351 True, but "if" it converges, you get the limit. And if the limit is equivalent to a_n+1 = a_n = 0, or similar, it is probably because it does not converge for anything else that a_n == 0. In fact, it is a way to get the forking points of some well known chaotic recurence formula.
@mathsfermattest6494
4 жыл бұрын
Hello Michael I hope you will read this comment I have an formula about inverse sin and inverse cos and *Pi* in a form of infinite series. And I am going to share a special case of it a formula for Pi Pi = lim x->infinity (2^x *×* sqrt( *2* -sqrt(2+sqrt(2+sqrt(2+... x times))))) if you take x=11 it is correct upto 10 or 9 decimal places While counting 2s you need to start it from first 2 that is bold see up I am gonna mail you. Please make a video on it
@calcul8er205
4 жыл бұрын
Khushank Verma sounds like Vieta’s product for π
@mathsfermattest6494
4 жыл бұрын
@@calcul8er205 yes looks similar but note that this is not the actual formula I want to show you. It is a special case of it. There would be possibly a way to show both are same
@mathsfermattest6494
4 жыл бұрын
@@calcul8er205 you would probably see the formula that came to me in future videos of Michael. I wish he will make a video on it. I had mailed him the formula I want to show all the proof of it.
@mathsfermattest6494
4 жыл бұрын
@Adam Romanov I know that the formula that I showed is already discovered I am just telling that this formula is a special case of another formula that came to me. I have not shown you the actual formula I am talking about.
@Joomarcks
4 жыл бұрын
Is it correct to do it this way? When n -> infinity, we have: L = sqrt(2 + sqrt(2 + sqrt(2 + ... ))) L = sqrt(2 + L) L^2 = 2 + L L = 2 or L = -1 So L = 2
@johannesoertel5764
4 жыл бұрын
You must first know whether or not the sequence ( aₙ) ₙ is convergent because you're assuming that by using the fixed point equation.
@rileyabel5387
3 жыл бұрын
you probably dont give a shit but does someone know of a method to get back into an instagram account?? I somehow forgot the login password. I love any help you can give me!
@tylercairo193
3 жыл бұрын
@Riley Abel Instablaster :)
@rileyabel5387
3 жыл бұрын
@Tyler Cairo I really appreciate your reply. I got to the site through google and I'm in the hacking process now. Takes a while so I will get back to you later when my account password hopefully is recovered.
@rileyabel5387
3 жыл бұрын
@Tyler Cairo it worked and I actually got access to my account again. I am so happy:D Thank you so much you saved my account :D
@SuperSilver316
4 жыл бұрын
Ahh this take me place to those early analysis days
@WhiterockFTP
4 жыл бұрын
i feel it too. good old times ...
@amircanto5416
4 жыл бұрын
I had a very similar exercise left in a class and the difference made it more interesting, it only changed the a_1 and give a suggestion(dont remember it now), and now a_1>0 (or a_1>=0 I'm not so sure), but when you change that, you could either take cases when a_1< or > sqrt2 where in each one the sequence is monotone, but how could you prove that?, since in my case I got it from a graphic calculator. The solution was given to me by another student in mathstackexchange, who could prove what was suggested and the limit wouldn't change.
@elgourmetdotcom
4 жыл бұрын
Hi Michael! Excellent videos as usual! I have a related problem I couldn’t solve perhaps you find it interesting! It says: prove that there is no real number K such that a_n > 0 for all n given that a_1=K and a_n= -1 + a_(n-1) . n / (n-1)
@mathsfermattest6494
4 жыл бұрын
I think I don't understand the problem. a_1 can be any real number then how a_n can be greater than 0 for all n a_1 can be any number
@elgourmetdotcom
4 жыл бұрын
That’s exactly what the textbook says. I understand that no matter the real K you pick, you’ll never guarantee the whole sequence is positive
@mathsfermattest6494
4 жыл бұрын
@@elgourmetdotcom I understood it thanks I just read the last part. So thanks a lot.
@mathsfermattest6494
4 жыл бұрын
I got a proof. I am really sorry that I can't write it here. Since I am writing everything from my phone and it would take time but I have a hint make a closed formula a_n and here is what I found a_n = - 1 + n( k - (1/2 + 1/3 +1/4... 1/(n-1)) It's clear that harmonic series will diverge and at some point it will be greater than so n(k - harmonic) will be negative since n is positive adding negative 1 will never make it positive
@elgourmetdotcom
4 жыл бұрын
Khushank Verma got it! It worked, thanks!! I couldn’t see the pattern at first. Great :)
@davidmoss9926
4 жыл бұрын
The proof by induction felt like a magic trick! Let's just add 2 to both sides ! @ 3:18
@electrikshock2950
4 жыл бұрын
i find it particularly interesting that if we use sqrt(1+an) instead we get phi for a(infinity)
@electrikshock2950
4 жыл бұрын
I know it's obvious in the final form but it's super cool that it would be the case for first form as well
@shafrazameer631
4 жыл бұрын
sir its awesome wow ................................
@yesil1026
4 жыл бұрын
I remember I solved this during a holiday. an is actually 2cos(pi/2^(n+1)) a1 is 2cos(pi/4) and using the fact that 2cos(2a) + 2 = 4cos^2(a) we can reach this result.
@rahimeozsoy4244
4 жыл бұрын
Karekökle ne alakası var
@suniltshegaonkar7809
4 жыл бұрын
Everything is okay, but I am astonished to find the solution of the limit to be exactly 2, no more or no less. I was expecting some rational number between 1.5 to 2.2. Sq root 2 = 1.41, but if we add sq rt of 1.41 and continue adding roots, you will reach at 2. Hard to digest but mathematics has shown that it is a natural number, not a rational one.
@punditgi
3 жыл бұрын
Very slick!
@voxclamantisinextermo
4 жыл бұрын
Suppose we have a geometric progression c(1)=1, c(2)=1/2,c(3)=1/4, c(n)=1/(q^(n-1)),(where q=2) e.t.c; Sum of this progression equal Sc=c(1)*(1-(1/q^n)/(1-q); For this sum right Sc1=1, Sc2=3/2, e.t.c, i.e Sc always less than a(n), Sc
@joudalturki9781
2 жыл бұрын
thank you so much
@yahav897
4 жыл бұрын
a1 = sqrt 2 a2 = sqrt (2 + sqrt 2) ... An+1 = sqrt (2 + An) = sqrt (2 + sqrt (2 + sqrt...)) = x as x approaches infinity, the term sqrt (2 + sqrt...) is x so x = sqrt (2 + x) x^2 - x - 2 = 0 x = 1+- sqrt (1+8) / 2 x = (1+-3)/2 x = 2, - 1 Apologize if I'm wrong, havent had enough time to watch the video and I'm not very good with these type of questions, lol
@suhaibalkhaldi
3 жыл бұрын
So we don't have to prove the monotone by induction ? I mean is the method you used work of all problems?
@thesecondderivative8967
Жыл бұрын
6:56 Does anyone know how to perform the exercise in a formal way?
@Joseph2302
2 жыл бұрын
Surely if a1 is 2, then a2 is also 2 (as 2+2)^0)5 = 2. And so an is 2 for all n
@hjdbr1094
4 жыл бұрын
You can put the limite inside the square root because sqrt(x) is continuous at [1,2]?
@josephturbo7606
4 жыл бұрын
Continuous on [1,4], because 1
@hjdbr1094
4 жыл бұрын
@@josephturbo7606 thx
@demenion3521
4 жыл бұрын
can you use such a nice method for limits also for homogeneous recurrence relations? The obvous strategy of replacing the sequence elements by their limit doesn't work in this case unless the limit is 0. I'm asking because I've recently seen the problem of calculating the limit of a sequence where each element is the mean of the previous two and the first two elements are 1 and 2. The best way, I could come up with was calculating the exact generating function and from there the explicit formula for the elements and take the limit of that. But I feel like there has to be a better way to get to the limit of a recursively defined sequence
@demenion3521
4 жыл бұрын
@VeryEvilPettingZoo thanks for the very informative answer. that's a nice trick that you used to make the relations inhomogeneous :)
@hectorkim440
4 жыл бұрын
let an=2cosθn
@connorbredall3112
4 жыл бұрын
I challenge you to find the closed form of this sequence.
Пікірлер: 62