I recently finished watching your video series on the Rogers-Ramanujan identities. Thank you for making these excellent video lessons. They are very good and your effort is appreciated. I plan to watch some your other videos too.

thank you so much

26:15 I thought the existence of a surjection is enough to show that the cantor set is uncountable because then the cardinality would be bigger than or equal to the cardinality of [0 ; 1] ?

Hi, where can I find the article by Andrew and Baxter about the Rogers-Ramanujan Identities?

Well presented.

Will there be "next few videos"🥲🥲 I appreciate your work for this series so much. I have watched and taken notes for every videos in this series. Thank you for making those videos though there are not many people watching it! It's really useful for students who is learning the same book in this ring-group-field order. 😍🥰

also I like

The strongest for real❣❣💯💯

Very sufficient and clear proof and explanations. I'm so pity that other students can't find this series or don't give it a look. Thank you for making those teaching videos!!

At 1:40, shouldn't the multiplies of 4 larger than 24 is 28,32,36 and so on? I think you might accidentally write 30.🙂

Thank you sooo much for making these series video!! You really explained it clearly, and all the proves are tidy and well-ordered.

Thank you for this quality video!

It is an interesting video. However, the content is old. I know a better solution, there is a 13-year-old boy has found the final solution for the integer partition problem. please check these links: kzbin.info/www/bejne/boqspJ-Ge7R1bpY ijcionline.com/paper/12/12423ijci06.pdf

Heyo, after i understood this i saw the beauty in it. Damn, this shits dope.

Took me 3 times, but I believe I understand this now.

Beautiful

Great

The Cantor function is NOT the "function that you see here".

Very nice job, thanks! I am currently working through the series and it helps a lot

omg thanks 🙏

Looking forward to your videos on inf-dim Lie algs.

Do you have some book references about abstract algebra with proofs included as done in this video ?

This was really helpful, thanks a lot👍

Great stuff 👌👌👌

Not clear

Thankyou so much for making this video. It helped me a lot! Please keep making these!

y

Thank you

Wave dampening is a surface phenomenon.

Don't you need stronger logic at 2:02? You've shown that this formula holds at any finite level of depth, but that doesn't necessarily imply that it should work at the infinite depth of the infinitely-nested root (because there's a large end value of f(n+k) that keeps getting replaced with new roots with every iteration). Is this really Ramanujan's original solution?

Could barely concentrate on the proof cause the music is so dope. Love it though. My new favorite way to see a proof for the first time

An alternative definition of a group is a structure (G, e, ~, *) on G, where e is a 0-ary function, e : {0} -> G, ~ is a 1-ary function, G -> G, and * is a 2-ary function, GxG -> G, such that, with all universal quantifiers implied, and no existential quantifiers, x*e = e*x = x, x*~x = ~x*x = e, x*(y*z) = (x*y)*z. A motivation for this alternative definition, aside from having a completely equational signature, is the fact that group homorphisms can now be defined more concisely in terms of compositions. In my opinion, this should have been taught earlier in the series, prior to rings. This is because, once you have been exposed to group theory, defining a ring axiomatically is extremely simple. A ring (R, 0, 1, -, +, ·) on R is a structure such that (R, 0, -, +) is an Abelian group, (R, 1, ·) is a monoid, and x·(y + z) = (x·y) + (x·z) and (y + z)·x = (y·z) + (y·x). I think there is motivation to define rings in this manner: it helps explain why the multiplicative monoid of a ring can never be a group: because it is not a cancellative monoid. In fact, generally speaking, there is a fundamental difference between a cancellable monoid, and a non-cancellable monoid. A cancellable monoid may not itself be a group, such as with the nonzero integers, but it is _embeddable_ into a group, the group of nonzero rational numbers. A non-cancellable monoid is _not embeddable_ into a group. Multiplicative monoids of a ring with zero divisors excluding 0 are examples of this. If x is a zero divisor, then there exists some a such a·x = 0, so y·(a·x) = 0 = z·(a·x) = (y·a)·x = (z·a)·x, and so, if a is cancellable, then x is not cancellable. So zero divisors satisfying a certain condition are never cancellable. In fact, this answers an important question: why can a ring not be extended in such way that 0 can attain a multiplicative inverse and allow 0·x = x·0 = 0 to be false? The reason is because 0 is not multiplicatively cancellable. Extending a ring to a field can allow every cancellable element to have an inverse, but not 0 or other non-cancellable nonzero divisors.

The music really makes these

What is the motivation? Just trial and error?

Nice vid. There was just one small issue, wiping out the content at 0:45 just one second after proving your key point “N is a natural number”. I was unable to process the final line this quickly, so I had to rewind and pause. Maybe next time let it sit for a few seconds before transitioning to the next “slide”, or scroll the content so people get a chance to follow the argument.

Awesome proof

Can you explain why N < 1/n is a contradiction? The way I see it, N is n! multiplied by the error of approximating e by the sum of the first n factorials. I can't see why this is necessarily a contradiction

nice ♥ do one for e is transcendental next😝

Why would anyone start with this function e^-it to begin with?

The Taylor Expansion one sticks in my head from school, but this one was fun 😊. Albeit, done with massive amounts of hindsight. Not always a bad thing!

This proof relies on the validity of the real rule for the derivative of e^(kx) when we set an imaginary k.

Cool!

Wonderful. Better than solving for y''=-y.

Wow, nice!! Very clever this idea of proving equality by showing that the division equals one :D

I don't know why I got this recommended, but I love it!

Ha, that is crazy. Nicely executed with the music and stuff

Wonderful! Thanks.

Amazing

Ok

nice one!