The most advanced definition of sine and cosine? 

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The next step is to design a trig course with this as the first lesson.

An alternative method would be, from the time we have s²+c²=1, to plot the trajectory in the (c,s) space. We proved the curve is confined to the unit circle. Its velocity is (c,-s), making it go counter-clockwise at a fixed speed of 1! So it will run the full circle in a time equal to the perimeter of the circle: 2pi

ah that joke about cats in the sponsored segment was hilarious, you know every sponsored segment should be like that XD

An alternative to the "cheat" of the definition of π/2 is to define 2π as the length of the unit circle, and parametrise the circle with t -> (c(t), s(t)). The interval [0, 4\theta] covers the circle exactly once (otherwise 4\theta would not be the smallest period). Now for any curve t -> (f(t), g(t)) parametrised uniquely by t in [a,b] we can compute its length as ∫_a^b √((df)^2 + (dg)^2) = ∫_a^b √(f'(t)^2 + g'(t)^2) dt Therefore the length of the circle gives 2π = ∫_0^{4\theta} √(c'(t)^2 + s'(t)^2)) dt = ∫_0^{4\theta} √(-s(t)^2 + c(t)^2) dt = ∫_0^{4\theta} dt = 4\theta,

I love how the angle sum and difference formulas appeared.

No, I’m not going to forgive you 24:00. I watched the whole video to see where you would pull pi from and you pulled it from the air.

Homework 16:15: Because it's the first zero after x=0, and c(0)=1, the derivative at the zero is negative. But, this derivative is -s(x), which means s(x) is positive.

I remember doing this in college. Thank you, professor.

That reminds me of a functional equation we discussed at the math power lessons in late grammar school: We have two differentiable functions s and c with s(x1-x2) = s(x1)*c(x2) - c(x1)*s(x2) and s'(0) = 1. We proved a couple of things about their properties until we could prove that there's exactly one solution and determine the function. That's over 35 years ago, but I found it so amazing back then that I still remember. Unfortunately, I didn't keep my notes about it.

1:42 - I was blowing at my monitor. Thought it was a tiny flybug 🤣 Btw. great video as usual 🙂

you said "fetch quest" which sent me down the rabbit hole of tvtropes, so thanks for that, in addition to this fascinating video

This video is very cool. I love looking at different ways of defining basic concepts.

at 21:18, c(x+theta) should be equal to -s(x) not -s(theta)

Honestly a very nice video. Kinda hate that some "extreme" theorems have to be used, like Taylor's series and that last definition of π/2, but that's just maths sometimes.

This video reminds me of something. I took an unusual Complex Analysis 101 course a few years ago. In one of the earlier lectures the exponential function was defined apriori as the inverse of the antiderivative of 1/x. Starting from this definition alone all the proprties you can think of were proven, including that e^(ix) is periodic. Which led to defining pi as half its period. It was a wild course, the professor had a very interesting approach.

10:02 "Radius of convergence of McLaurin series of any bounded and smooth function is infinite" - is that right? How about 1/(1+x^2) wich is bounded and smooth but series converges only when -1

the alternative definition of pi/2 is a bit of circular reasoning

There should be a supplementary "rabbit hole" video, as that does feel like a huge cheat.

15:44 At this point in the video Michael uses the Intermediate Value Theorem to show there are values of θ such that c(θ) = 0. But then he says "let's take the smallest such θ". I think he maybe jumped a bit with that assumption. It's not actually immediately obvious at face value the zeroes of c must have a minimum. For instance, it could be that c() "wiggles" around faster and faster between (1/4, 3/4) hitting 0 an infinite number of times in that interval and the sequence of those zeroes has no minimum value. The minimum does, I think, exist in this case but it's maybe something that actually should be proven. I think you can demonstrate it exists from c() being continuous on (0,2) and showing that, if there is an infimum of the zeroes on that interval, c()'s continuity at that infimum implies the limit of its values there must equal c() at that point as well.

I was really looking forward to how pi pops out of that, so just offering the alternate definition felt like I was being cheated. That’s definitely one rabbit hole I’d love to see explored.

I also see a path to figuring out the Pythagorean identity without using our prior knowledge. s'' + s = 0 2s's" + 2ss' = 0 2s"s' = -2ss' ((s')²)' = (-s²)' (c²)' = (-s²)' c² = -s² + C c² + s² = C

21:36 on the right : =-s(x) ?

13:30 "since it is an alternate series, if we truncate it before a minus, we end up with something larger, if we truncate it before a plus, we end up with something smaller". Is this true in general? I'd think this is only true if you know that the coefficients are decreasing in absolute value.

@3:37, when you use the dofferential equation to extract the boundary condition seems incorrect. You should texhnically justify why the OdE holds at x=0. The function has value or slope specified on the boundary so technically the ODE need not hold. Thus establishing c'(0)=s''(0)=-s(0) using y"=-y is not trivial.

Cool!! It would be so great to see a video about that second definition of π/2 :)

A classic derivation every advanced math student should see.

i guess you didn't want to go there, but without using sine and cosine you can just solve the equation by its characteristic polynomial as a superposition of the exponential functions exp(ix) and exp(-ix). using the initial values we get the desired solution in the exponential version of sine

Thanks Mr Penn

A better thing to do would be to multiply the original equation by y'(x) and integrate up to get the relavent constant. simply differentiating s^2(x)+c^2(x) seems too forced, and not really natural to me.

In general, truncating an alternating series before a positive sign need not lead to something smaller. It's true if the absolute value of the terms is decreasing.

You can also directly apply the 2π translation operator e^(2π d/dx).

This very idea was brought up in a footnote in my diff eq textbook this year! Very cool

Best rarely seen derivations. Prof. Penn always wows.

For those who might appreciate a deeper look into this kind of technique, Steam Algebras using coinductive reasoning have been suggested as an alternative formalization of calculus, I suggest you look at the work of Rutten and Escardo.

Why can we take the smallest theta? How do you know it exists? What if all rationals in (0,2) work?

You could start using the integral as a definition of the inverse sine. This was done in the first edition of Alfors Complex Analysis book.

I would directly integrate the equation given and find out that the equation resulting is the equation of a circle/ellipse. Proceeding further, if I am at a position of that unit circle then I go 2*pi radians to meet on the same position, then the solution is periodic, along the phase space plane. Only thing is to evaluate an expression for theta in relationship to y and y’. Which I think that’s what Michael was doing but maybe I don’t follow😂😂

How do we know there isn’t a smaller number that can be the period. We showed that it is periodic in 2pi but why is that the smallest period.

How do you know there is a smallest theta in (0,2) - if theta was satisfied by every 1/n there would not be any smallest

Please, correct me if I am wrong, but here your proof requires the assumption that this arbitrary s solution is n-th differentiable for every n. Not saying that the idea is incorrect, but it lacks some rigor in this sense, IMO.(It is quite easy to prove this differentiability, but this step shouldn't be forgotten) Still, I liked the idea used!

How do we know that min root exists for s(x) =0? What if set of roots is inething like {1/2, 1/3,...}. Which one is minimal?

If there can be infinite θ with c(θ) = 0, how do you know there is a smallest one??

Very cool!

10:02 This line of argumentation is invalid. Arctangent is bounded and analytic but its MacLaurin series doesn't have infinite radius of convergence. There are also smooth functions which disagree with their MacLaurin series in every positive radius. Consider the piecewise function f(x)=exp(-1/x) when x>0 and f(x)=0 otherwise. It can be proven that f(x) is infinitely differentiable everywhere, critically at x=0 it's taylor series agrees with the zero function, but f(x) is nonzero for all x>0, so the function disagrees with its maclaurin series. The way you could salvage this argument is by proving that: 1. The described taylor series has infinite radius of convergence 2. The described taylor series satisfies the described initial value difeq y''=-y. 3. The solutions to the initial value difeq are unique. Those are actually not too hard to prove, and would let you infer that C(x) and S(x) agree with the expected MacLaurin series. You may have meant to say that we're assuming C and S are analytic, in which case the above arguments aren't necessary. This is different from assuming that they are smooth, which simply means that they are infinitely differentiable at all points. We get smoothness for free: since we assume y''=-y, then we must assume that y is twice differentiable, and thus y is in C^1. In general, if y is in C^k then -y is also in C^k, but since -y=y'' then y'' is in C^k and hence y is in C^(k+2). Continuing inductively, we see that y is in C^k for all k, hence y is smooth.

Hmmmm one slight detail: why is theta well defined? In other words, why do we neccesarily have a "smallest" such root on that interval?

11:23 the powers of c reminded me of the powers of the i with a small change in their order

Michael, this is my favorite video of yours. I love the cleverness of it all. I do have a question though. How do you know the smallest solution to c(x)=0 on [0,2] exists? You could have an infinite set of zeroes, but no minimum. (The function cos(π/(2x)) has zeroes at {1/(2n+1) | n is in Z}, yet this set has no minimal number.)

Can you use the function which finds the remainder (mod) to create a sin wave? I can make it do a zig zag up and down and feel like that can be tweaked

How did we assume that c'' exists? Are we assuming automatically that s is infinitely differentiable. What could we prove if we assumed that s'' was not differentiable?

Some cool intuition: if you think of the start point as where sin = 0 this problem is like an object being thrown up into and mashed between two gravitational fields

Overcomplicated. First take s the solution and c=s' then c' = - s. So if you define g(x) = - s(-x) then g'' + g = 0, g(0) = 0 and g'(0) = 1 so by uniqueness we must have s=g. Hence s is an odd function, so c is an even function. Now if you set e(x) = c(x) + i s(x) then e' = i e and e(0) = 1 so it follows that e(x) = exp(ix), which prove that exp(ix) = c(x) + i s(x) and by even/odd-ness we have exp(-ix) = c(x) - i s(x). Substract the two equalities to conclude that s(x) = ( exp(ix) - exp(-ix) )/2i is a sum of 2pi-periodic functions.

WOW ! Awesome ideas !!!

Muy bien aplicado el teorema de existencia y uniciadad de soluciones de ODE en s(x+t)=s(x), saludos

14:45 is it valid to use the IVT since you don't know if the function is continuous on [0,2] or is that assumed for solutions to DE's?

whats sine and cosine?????

wonderful!

I feel violated! Deus ex machina in the final step. In fact he only proved s(x+4θ)=s(x) where c(θ)=0 for the first time. Ok, I see it now…

Very nice. So we get the definition of pi as the root of a power series. Might be nice to relate this derived constant to pi from other contexts. Is there a way to derive a series for pi from this root of a series?

What’s up with that tiny ghost floating around the room?

Hello 👋🏻

this video was messssy girl

Do you even realize how ridiculous "...most ADVANCED definition..." sounds?

مدهل

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