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Thx I was very confused before this

This channel is going to be huge one day fr

Is there any book recommed for set theory..?

You can't use n in your inductive hypothesis, because the number you are assuming the statement holds for is somewhere between the smallest number in your set of inputs and n, not n itself.

I love you

Great explanation!

or you could simply use the ratio convergence test and see that lim k->+inf [ (x^(k+1) * k!)/(x^k * (k+1)!) ] = lim k->+inf [ x/(k+1) ] = 0 for any fixed x ∈ ⁠ℝ.

So the series absolutly converges for all real Numbers x?

Great great video...you are the one! Tanks for your beautiful work

any clue what the proof would look like for any (positive integer) power n beyond just 2?

maybe i wont take real analysis lmao

Just wanted to say i appreciate how good these proofs videos are, they are so clear that I can follow and enjoy them in real time. I wasn’t used to enjoy math like this before.

Great video! The grounded style felt accessible and I appreciate your concise yet digestible pace. Thanks for the useful example too! Its been valuable seeing a proof in terms of this structure, especially your methods of indexing. Inspirational, great work!

I m watching this theoram from india mp gwalior.😅😅😅

Excellent explanation!

Trivial

Excellent video! I was wondering, do you have any book recommendations for formal logic? Especially books that actually prove everything, and not just tell you the theorems. Thanks in advance!

But does this mean that a+b=b+a?

I want to prove the (⇐) direction by the contra positive: Given (1) "x_n not convergent" means: ∀x∈ℝ ∃ε>0 ∀N∈ℤ⁺ ∃n≥N (|x_n-x|≥ ε). Want to show (2) "x_n not Cauchy" i.e: ∃ ε >0 ∀N∈ℤ⁺ ∃n,m≥N (|x_n-x_m|≥ ε). Pf: SPS "x_n not convergent", for N∈ℤ⁺ arbitrary, take x=x_k for some k≥N (since (1) holds for all x∈ℝ, it holds for each term in the sequence x_k). For that choice of x_k, choose ε >0 and n∈ℤ⁺ s.t n≥k (in (1), N≤k) and |x_n-x_k|≥ ε. (Note: n>k) With that ε>0 In (2), if for all N∈ℤ⁺ we take n,m∈ℤ⁺ such that m=k≥N, n>k (same n from previous), then we have |x_n-x_m|≥ ε, thus "x_n not Cauchy"

Bro went from 0 to 100 in two seconds. Great video. I love your from ground up approach to proofs.

i like such content but i wonder to know from where you get these theorems isnt like a pdf book or smthg..?

For the base case you can have a=0 and b=0 which makes the expression have an undefined 0^0.

Great video!

I expected the difference of same power factoring but this 100% works.

Thm 2.1.9 says it's an exercise to prove 0 <= a <= e , but I don't see how it's any different from this. Solution manual says choose e = a>0 deriving the "contradiction" 0<e<=a but that's not a contradiction if e=a Look into it pls thx

I love your channel

bro this is in the as level maths textbook as well 💀

sir you helped alot

WAIT THIS IS TRUE

Real analysis is tough and I'm not a math major, but I feel like you missed proving that S is not empty.

I think you made a mistake, you cant suppose inductively that Xn<(1+√5)/2 and prove X(n+1)<(1+√5)/2 because if you do so then you can also inductively suppose that for n>1 Xn>(1+√5)/2 and then prove that X(n+1)>(1+√5)/2 by your own same steps shown from the time stamp 6:20

Pfft! It's as easy as solving a quadratic equation; just need to form a recursive relationship. Let, √(1+√(1+√(1+...))) = x. So, x^2 = 1 + √(1+√(1+√(1+...))). Thus, x^2 = 1 + x [∵ √(1+√(1+√(1+...))) = x] Solving this quadratic equation leads us to the Golden ratio. ∴ x = (1±√5)/2 = ϕ QED.

Couldn't we just prove it using Peano axioms?

Thank you, that’s a great review💙

nice.

x = sqrt(1+sqrt(1+sqrt(1+...))) x = sqrt(1+x) x^2 - x - 1 = 0 so x = phi :)

The continued fraction of phi is 1+1/1+1/1+.... then there is a relegion of these two formulas .

In my head, if you denest one it's the same expression so say it equals x. We square both sides and minus one to get the same expression, which is x, equals x^2-1. Just move it over and famous phi formula

Or just solve x = sqrt(1+x)

Very short proof. Can you please put up a link to your shortest video ever?

🎉🎉🎉🎉🎉🎉 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.