Відео

🎉🎉🎉🎉🎉🎉 thank you very much for your beautiful work!!!!!!

Great video man. Thanks for being that organized, it has been really helpful for me :)

Can i get a link to the video for the result that you used!

Keep it up bro, this is very good. The way you explain is just different from every math's related youtubers

Haha that's a massive premise for an if -> then proof

To prove the fact that it's increasing, use proof by induction... Base case: x1 = 1, x2 = sqrt(2), so x2 > x1. Easy base case. Inductive case: Assume x(n) > x(n-1). Then 1 + x(n) > 1 + x(n-1). Therefore, sqrt(1 + x(n)) > sqrt(1 + x(n-1)). Therefore x(n+1) > x(n). Inductive case done.

Nice

I just love these! every singe step clearly explained! :))

:)

Also please prove the extreme value theorem.

Ok, this is going to be an adventure

can you show how to prove that the inverse of a bijective function is bijective

great resolution. hug from Brazil

Uraaaaaa ! Thank you so much for your detailed video analysis on this topic which is geometrically explained . I have thought that I will have to memorise this topic without understanding but you have come as a saviour before my engineering mechanics exam . Thanks again . A big cheers to your nice work .

idk whats going on but nice explaining man

I think that's the greatest title of a UA-cam video I've ever seen!

What is the point in still using an Epsilon Delta proof? Is there any instance in which it would be more efficient, because from what I’ve seen/know it only seems to complicate the proof.

Superb explanation brother. By the way,where do you live???? I am watching you from India 🇮🇳❤❤❤❤❤❤

How convenient that you got a |x-6|. If you had had, say |x-9| you would do |x - 6 -3| and apply triangle inequality to x - 6 and 3?

Getting cold?

It can be done more easily. The function f(x)=1/sqrt(3x+7) is obviously continuous in a neighbourhood of 6. But then lim_{x->6}f(x) = 1/sqrt(3(lim_{x->6})+7) = 1/sqrt(3*6+7)=1/5. The argument relies on the fact that lim_{x->y}f(x) = f(lim_{x->y}x) when f i contiuous at y.

haha nice. Just getting into proofs and i love your channel

Thank you

I am doing Calculus and hope I don't have to do this in the exam. Anyway it's really interesting

I think I'm gonna do another one of these exercises as my next video

support

I really like how you make sure these videos are easy to understand for everybody by explaining every step. Which means if it's a really hard problem I can trust that I will understand it with your explanations.

Can't you figure out the fact that it's a geometric series and say x_(n+1)/x_n=L where 1>L>0. Then you say L(x_n)=x_(n+1) and since L is between 1 and 0 that means the function of x_n is decreasing exponentially and is approaching 0 as n--->infinite.

Cool approach of induction on n for all n≥m≥1 with fixed m

Keep it up homie. I’m glad you’re so consistent bc it really helps me out sometimes, and it’s just cool stuff!

I believe this result is used in calculus for convergence of sequences all the time. Nice to see a proof of it.

I like how you outline the proof so that this sketch can be generalised. Love your videos!

Awesome ❤

I really needed this

Great content brother 🗿

This is a spectacularly well-explained and reasoned-through video. Bravo.

Nice

Great ❤

For the second part, you can use Bezout lemma and don't have to go back to the definition of gcd. Suppose gcd(a, b) = 1 and gcd(a, c) = 1. Step 1. a * p + b * q = 1 for some p, q in Z Step 2. a * r + c * s = 1 for some r, s in Z Want: a * x + (b * c) * y = 1 for some x, y in Z Step 3. a * c * p + b * c * q = c by multiplying step 1 by c Step 4. a * c * p * s + b * c * q * s = c * s by multiplying step 3 by s Step 5. a * c * p * s + b * c * q * s = 1 - a * r from steps 4 and 2 Step 6. a * (c * p * s + r) + (b * c) * (q * s) = 1 from step 5 Hence gcd(a, bc) = 1 by Bezout lemma

I am enjoying Your videos. Whenever I get time, I come here and just look into the statements and I try to prove myself. Helps me a lot in maintaining my form.

And for infinite sets?

I just want to say that this is so descriptive compared to other videos. Thank you!

sqrt((n)+sqrt(n))-sqrt(n)=(sqrt((n)+sqrt(n))-sqrt(n))*(sqrt((n)+sqrt(n))+sqrt(n))/(sqrt((n)+sqrt(n))+sqrt(n))=(n+sqrt(n)-n)/(sqrt((n)+sqrt(n))+sqrt(n))=sqrt(n)/(sqrt((n)+sqrt(n))+sqrt(n))=(sqrt(n)/sqrt(n))*1/(sqrt((1)+1/sqrt(n))+1)=1/(sqrt((1)+1/sqrt(n))+1) lim(1/(sqrt((1)+1/sqrt(n))+1)) x---> inf =1/(1+0+1)=1/2

I was dying for this solution 😢 THANKYOU SIR

amazing

Best explanation 😇

Another way to prove this would be to split the sequence to 2 cases: bounded and unbounded. If it is bounded, it would have a limit point and if it has a limit point, it would be easy to construct a monotone subsequence by adjusting epsilon appropriately. If it is unbounded on any one side, it is trivial to construct a monotone subsequence.

thank you

Something about this feels overly complicated. I'm a bit of an amateur (math-adjacent degree, but didn't get this far into formality). Let me see if I got this: If I have a sequence of *any* Real numbers (x_n), then I can pick out either a finite or (if available) an infinite number of elements of the sequence, in the order in which they appear in the original sequence, and in doing so, I can ensure that the elements that I choose have at least one of the following patterns: a) these still sequence-ordered elements are already in nonascending order, or b) these still sequence-ordered elements are already in nondescending order, or c) these still sequence-ordered elements are equal to each other (which counts as both nonascending and nondecreasing). Is that good common-wording understanding of this theorem? And if that's right, then it sounds like the Empty Sequence and single-element sequences (i.e.: single numbers) may be degenerate/trivial/vacuous cases, but then for any sequence of 2 or more elements, I can pick any 2 different-position elements, and either one must be greater than the other, or they must be equal to each other.