Hey KZread: Go follow my resurrected twitter: twitter.com/michaelpennmath tbh I don't really know how I will use it, but it seems like most "serious" youtube people are there... Any suggestions?

@goodplacetostop2973

Жыл бұрын

- Automatic tweet when a video or livestream is up - Retweet interesting math content like animations or proofs or facts - If you want to, tweet or retweet anything you want. You just listened a great song and you want to share it, you’re watching some great sport game, you want to talk about some breaking news, etc.

@goodplacetostop2973

Жыл бұрын

May I suggest some accounts I like on Twitter (besides math youtubers) : - @AlgebraFact - @math3ma - @kelseyahe - @LogicPratice - @TopologyFact - @Fermatslibrary

I thought using the generating functions to prove the relationship between the Leonardos and the Fibonaccis was really cool... thanks !

@DavesMathVideos

Жыл бұрын

I came here to remark this!

0:01 Catalan numbers are better in my book 0:02 Nothing will beat the 1987 version of Ninja Turtles 16:00 Good Place To Stop

@hybmnzz2658

Жыл бұрын

So quick

I suspect that the "Leonardo" in the name of these numbers is "Fibonacci" again, because his name was "Leonardo Pisano Fibonacci", which means "Leonard from Pisa, son of Bonaccio": "Fibonacci = Fi + Bonacci = Filius Bonacci". Perhaps the person who invented these numbers gave them such a name to underline that they are almost the same as the classical Fibonacci.

@Nikolas_Davis

Жыл бұрын

Nice! Since we're nitpicking, ;-) I would also point out that Leonardo the ninja turtle is named after yet another Leonardo, i.e. Leonardo da Vinci.

Volume is a bit low on your videos imo. Great content though!

@acborgia1344

Жыл бұрын

I agree

@howardcheung8304

Жыл бұрын

yeah

Thanks a million, Michael! I liked your derivation of the formulas that are new for me. All clear.

Fun. Thanks. I haven't seen generating functions for 35 years so I need to review them. And I wish that I had a more intuitive understanding of generating functions.

I thought another way to prove the second result could be simpler. First, just adding +1 to the left and right terms of the recursive definition equation of L, we see that L+1=M obeys to a Fibonacci-like recursive definition equation Mn=Mn-1+Mn-2. It is not Fn because of the initial terms of Mn, which are M(0)=M(1)=1. On the other hand we know (or can prove) than the suites obeying to this Fibonacci-like recursive definition form a vectorial space of dimension 2. What we just have now, to do, is find where Mn can be plotted in this vectorial space. The base vectors for this space are (phi^n) and ((-1/phi)^n) where phi is the golden ratio. Solving the equations based on the effective value of M(0) and M(1) we come to the same conclusion which is this linear relationship towards the Fibonacci suite.

I don't know if there's any causality but a couple of month ago I commented that there was a typo on the board which you didn't mentioned even when you noticed and needed to fix it, and since then I've seen in multiple videos you indeed pointed out the typos you'd fixed in cuts. So, as I mentioned, I don't know if there's any connection but after my earlier (and tbh a little grumpy) comment I feel I ought to mention that it seems to me you are making visible progress in this regard, and that's great.

I too like generating functions. A course would be greatly appreciated. Thank you, professor.

رائع كالعادة.استمر.أفضل أستاذ في افضل قناة

It is much faster to prove the relation by induction, but using generating functions is cool.

@tiripoulain

Жыл бұрын

Thing is, to prove something by induction you’d need to have noticed prior to that that such an identity seems to hold. Doing it directly with generating functions makes it so that we don’t need to know the result.

@SzanyiAtti

Жыл бұрын

That is true in general, however, in this particular example, it is pretty clear that the sequence is somehow related to the Fibonacci numbers, and it isn't too hard to find the relation by looking at the first few terms of both sequences.

@RabbidSloth

Жыл бұрын

@@SzanyiAtti I tried induction after reading your comment, but I got stuck. Would you be able to share your method?

@SzanyiAtti

Жыл бұрын

@@RabbidSloth Base case: n=0, L(0)=1=2*1-1=2F(1)-1 Lets suppose it holds for all natural numbers up to n. Then by definition, L(n+1)=L(n)+L(n-1)+1. We can replace L(n) and L(n-1), because we assumed the formula holds true for them. L(n+1)=2F(n+1)-1+2F(n)-1+1 Simplifying a bit, we get L(n+1)=2(F(n+1)+F(n))-1. By the definition of Fibonacci numbers, F(n)+F(n+1)=F(n+2), so we get L(n+1)=2F(n+2)-1, which is what we wanted. Our base case and inductive step both hold, therefore the relation is true for all natural numbers n.

@RabbidSloth

Жыл бұрын

@@SzanyiAtti Excellent. Thank you :)

And what of the Donatello Numbers?

Every time you say L(n) my brain thinks the natural logarithm.

Love it! A fun ride, even if you drive a little faster than I'm comfortable with....

This video just made me realize that generating functions are just the formal counterpart of the z-transform (with x = z^{-1}). How did I never think of this?

Generating functions always feel like magic to me. Never know how to work with them but they give very elegant proofs!

That’s basically it

awesome

do you ever compare the volume of your videos after upload to the ads that play next to them? because you probably should. if the volumes are more equal on average, that would be really cool

Dear Michael, thanks for super videos, could you please raise the audio level? On my phone you are considerably lower as anything else.

god tier thumbnail

Yay, respect for TMNT 2003

Hi Dr.! The volume is a little low.

For adding other integers than 1 L(n) = L(n-1) + L(n-2) + k = 2*L(n) - L(n-3) [ when L(2) = k+2 ] = (k+1)*F(n) - k where F(n) is the Fibonacci numbers.

@birdbeakbeardneck3617

Жыл бұрын

u prove this recursively(k)?

@schweinmachtbree1013

Жыл бұрын

I think the indexing is off slightly in L(n) = (k+1)*F(n) - k because this doesn't match the result in the video when k=1

You can also let K(n) = L(n) + 1, so that K(n) = K(n-1) + K(n-2), which is fibonacci recursion

@DeGuerre

Жыл бұрын

That was literally the first thing I thought of, although I used a different letter than K.

great

"Probably not Leonardo" so you're saying there's a chance .

I feel like the three step recursion is unnecessary (at least if all you want is to prove the generating function relationship).

cool

I really hate when "advanced" formulas are used to demonstrate simple recursive ones. It reminds me one of the first videos I watched where a simple geometric problem was resolved using a coordinate system and lots of calculations.... But overall, I like the choice of subjects and will continue watching even if it makes sometimes my hair stand on end

Its not Fibonacci numbers it's pingala numbers or hemachandra numbers. Fibonacci never claimed he had discovered these nos.

What does his t-shirt refer to?

Michael Penn, your claim that you proved those two facts is false. Nothing is proved until you knock a little coloured square on the blackboard.

@habermasnyc

Жыл бұрын

I think that only applies to lemmas.

Lit*

Too fast and loose with possible zero denominators for my taste ;)

👀

The fact that it is +1 and not -1 is very important: If it were -1, then the numbers are all 1!

Lim