the Fibonacci sequence is a trig function??
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“Anything is a trig function… if you try hard enough.”
- Fourier
Euler: real and imaginary parts
Binet: rational and irrational parts
Just wondering if there are similar relation involving other metalic ratios. I know what I am going to do this weekend! Great video as always.
"Hey Ferb, I know what we're gonna do today!"
I think it would be any extension of one field with an object that does not exist in that field, but is the square root of something that does.
For example, i is non-real, but i^2 is real, so every even power lands you in the reals.
Similarly, sqrt(5) is non-rational, but squared becomes rational, so again every even power lands you in the smaller field.
Please post a series on Functional Analysis/Linear Algebra too, sometime, if you can, sir. Decent lectures on graduate level subjects are hard to find. The ones on Lie Algebras and Differential Forms are super useful. Complex Analysis too.
This video is fascinating. Love watching these.
What sort of cost free world do you actually live in? How many weeks do you think it would take to prepare the lessons for these subjects; plus he has to work for a living as a Prof. Be grateful for what he does prepare for us and all at his own expense in time & effort. Why not ask your own Profs to do this and see the reaction!
Michael does have a second channel MathMajor. That one is more oriented for stuff you'd see in a classroom lecture. Maybe what you want has already been covered over there.
Bro 💀💀💀@@ChuffingNorah
@@ChuffingNorah ... Which is why I asked. I wasn't demanding it. :P Besides, he said himself earlier in another video that he wants to gradually build the list of playlists until he covers every topic he can possibly think of/that is of interest to him and given that Functional Analysis is used pretty much everywhere, I figured he might be into it. 🤷🏻♀️ I've no idea why you're being so unnecessarily aggressive about this, but it sounds like you have your own issues, so whatever, I guess. Peace out.
@@bsmith6276 Oooh, I love that one, I binged his Abstract Algebra series there in two days, they're fantastic. And he's a wonderful teacher. :) Shared them with some of my friends too.
After playing around for a few hours, I think that the better correspondence is to the hyperbolic trig functions. Via the OEIS:
"Suppose psi=log(phi). We get the representation F(n)=(2/sqrt(5))*sinh(n*psi) if n is even; F(n)=(2/sqrt(5))*cosh(n*psi) if n is odd."
"Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd."
It screams hyperbolic sine and hyperbolic cosine being derivatives of each other, as well as a comparison of the (continuous) differential operator to the (discrete) difference operator.
The sooner you learn that all mathematics is interrelated, the sooner you can move on to the truly difficult stuff.
17:39
Super interesting! Thanks so much for sharing!
@ 17:13 The sums in the expressions for L_mn and F_mn should be multiplied by 1/2^m, not 1/2^(m-1).
I think this is to say that exponentiation works similarly in the "augmented" fields Z(sqrt(-1)) and Z(sqrt(5)), right?
I'm also thinking about ℤ[i] and ℤ[√5]. I think it's related to the automorphisms f(i) = -i and g(√5) = -√5.
@@user-oe5eg5qx4c Oh, nice mathy characters. I like that!
I made a mistake. I originally meant to say Q(sqrt(5)), but 's just a power of two away from Z(sqrt(5)) in this case really. That's why Penn has those 2^m scaling factors to deal with.
We have x^2 = -1 and x^2 = 5 being the relevant polynomials for characterizing the adjoint fields. You need to pay attention here because the D is in the details. Honestly, I know too little about this topic so I'm just going to take your comment about automorphisms on faith. Trust and verify!
So, if I understand correctly, there is an similarity between R[sqrt(5)] and C, so that there is a parallel between complex exponentials and powers of phi? Would that even make sense given that Fibonacci and Lucas numbers are discrete? Clearly there is more to this but I am not sure my group theory is strong enough to get to the bottom of it.
At minimum, the fact that sqrt(5) is linearly independent of 1 in Z[sqrt(5)] like how i is linearly independent of 1 in C means that some identities can be translated between the two. It just so happens that L_n and F_n act nearly identically in the former ring to how cos and sin do in the latter.
EDIT: when talking about linear independence, I meant to treat Z[sqrt(5)] as a vector space on Z and C as one on R. Also, correcting R[sqrt(5)] to Z[sqrt(5)].
@@minamagdy4126technically Z[sqrt(5)] would not be a vector space, but a module, as Z is not a field but a ring. That's said in this case we're probably thinking more of Q[sqrt(5)] which is a vector space over the field Q.
@@shirou9790 technically, terminological technicalities are pretty much irrelevant. Specially in this case, with Z, not just a commutative ring, but a integral domain, that may be immersed in its field of fractions. The modules over Z, meaning the abelian groups, all have vector space structure over Q. So ... just let the guy or the girl, I don't even know, show his or her knowledge in peace.
Well ... I don't think so. The similarity between R[sqrt(5)] and C is just that C = R[i], sqrt(5) and i are algebraic numbers (they are roots of polynomials with integer/rational coefficients) and ... that's it, essentially. I think the "correspondence" established is as good as saying "there is a correspondence between cos, sin and f(x)=x, f(x)=1/x. Which correspondence? They are both pairs of functions". Nothing REALLY special was shown between cos and sine functions versus Fibonacci and Lucas numbers.
The Fibonacci and Lucas numbers are more closely related to the sinh and cosh fuinctions. L_n = i^n * cosh(n[a - i * pi / 2]) / sinh a where a = ln(phi) and F_n = i^n * sinh(n[a - i * pi]) / cosh a. Note the F_n and L_n are the same but with cosh and sinh swapped, and sinh a = 1/2 and cosh a = sqrt(5) / 2.
Now THIS is a guy who knows mathematics.
10:37 16:08 The odd parts should be sums of k=0 to floor( (m-1)/2 ). e.g. For m=4, k= 0,1 produces 2k+1 = 1, 3.
It's still fine though because m choose m+1 is usually defined as 0. (In fact, m choose k is always equal to 0 when km, for m>=0 that is.)
@@shirou9790Interesting. Thanks. I was wondering the same thing.
Oh, then -infinity to infinity is still fine....
I was gonna say that
Awesome - just awesome.
Amazing never saw this before.
this is super cool
Amazing!
That is something new to me
Well, I have never anything quite like that. But very fun.
what would be the equivelent identity for cos(ix) = cosh(x)? would it be like L_(sqrt5 * n) / 2 = some other sequence?
You would need to use a continuous extension of L for that to even begin to make sense. That's possible through Benet of course, but it would be interesting to see what pops up. It feels like nothing obvious to me as an irrational exponent isn't likely to produce nice results even in terms of square roots.
Great video! Will you do a follow-up with the continuation to arbitrary, non-integer exponents using a phase factor exp(i*pi*x) in the Binet formula? I've poked around with that before and found very nice Fibonacci double-angle formulas.
Specifically, e.g. F(z)=(phi^z - exp(-i*pi*z)*phi^(-z)) / sqrt(5)
and similarly for Lucas function
I found that the "trigonometric" identities seem to not generalize with the "standard" continuation with cos(pi*z)
Do you have a formula for Pi^n and EulerGamma^n too?
@12:25 how can we just take two sums to the floor of m/2?
What if m isn’t even, then we’re losing out the last term of the sum here?
Pretty cool!
One now wonders what sequences would play similar roles for cosh and sinh...
Well, that's not even a well established correspondence, is it? For any sequence Cn, (complex) constants k,E≠0 and a, we can define
Sn := (aEⁿ - Cn)/k
It will satisfy
Cn + kSn = aEⁿ
Also, your question doesn't make much sense. Because he didn't show that Fibonacci and Lucas numbers "play similar roles for cosine and sine".
@@samueldeandrade8535Prof. Penn claims to have shown exactly that (see 17:18). Are you asserting that he did not in fact show what he claims to have shown?
@@jrbrown1989 I am asserting that there is nothing sufficiently PRECISE about the statement "there is a corresponde between cosine and sine functions and Fibonacci and Lucas numbers". It is a vague statement. The "correspondence" is vague. Hahajajajaha I just realised your "Proof" at the start of your comment. I guess I already did. What's the criteria to establish that there is a correspondence between cosine and sine functions and a pair of sequences? IF the criteria is to exist a relation
Cn + kSn = aEⁿ
then ANY sequence admits another sequence such that the pais is in correspondence with cosine and sine. Is that relevant? I can't see why! What part didn't you get?
@@samueldeandrade8535 The part where an extension to the integers, like √5 here, gives rise to a set of sequences mirroring the odd/even behavior of sine/cosine. Do you consider the question to be trivial or unsolvable? I don't consider it to be either of those.
@@jounik you have no idea about what you are talking, do you? Odd/even behavior of sine and cosine? Actually, you don't give a sh1t, right? "Do you consider the question to be trivial or unsolvable?" This question of yours make no sense. Trivial? Unsolvable? Man, you know what? Believe anything that you want. About your first question, consider that the hyperbolic sine and cosine are just
sinh(x) = -isin(ix)
cosh(x) = cos(ix)
So, make what you can or want with that.
I really like your videos, but the volume is so low I just can't hear them :(
This is the first time I see the Fibonnaci numbers starting with 0. I mean, the first term equals 0.
Many people do just talk about the "first two terms", i.e. F_1 and F_2, being 1,1. It is perfectly reasonable (and not uncommon) to talk about a zeroth term and is easy to see that the value must be 0 for the recursion to hold. Indeed, you can even extend into F_-1 and F_-2 etc. You will find you get the alternating sign version of the same sequence.
at 1:28 for Fibonacci sequence f_0=1 and f_1=1 am I true?
No, F_1 and F_2 are 1 and 1. Try plugging in n=0,1,2 into Benet for yourself. (1/sqrt(5)*[phi^n - phiconjugate^n])
And it's a good time to take out quotient rings and generalize it.
Hehe. Sir Penn, you and other math tubers I follow must be inspired. Thanks for sharing your knowledge, and all else. My signature: ALL life is chosen, and the Earth belongs to [G_D|DESS]. And math affirms this point, just like everything else, if a person is honest. Gracias. :) (Edit: I also like the "Sybermath" channel, I imagine he's also inspired (I say imagine to not be presumptuous but I'm sure he is, and likewise I imagine you are inspired. Thus)
I also like the "Sybermath" channel, I imagine he's also inspired (I say imagine to not be presumptuous but I'm sure he is, and likewise I imagine you are inspired. Thus)
We’re always here
This to me is what mathematics is all about. Borderline mystical.
🔻
So, you got your Fib#s programmed on your screen on a slider, you move it around, and all the other transcendentals adjust themselves accordingly? In fact, the whole universe is now at your finger tips?
I'm just going to shut my mouth. I think there might be $$$ in this. I will say this looks like something Polar Pi did. You getting real close to the bigger picture though.
I truly question what America offers to attract 'top talent' when excellent videos like this are available anywhere in the world. Perhaps it's my intersection that leads me to this perplexion.