the parabolic trig functions 

that's so egotistical

It's a nice idea, instead of i*f the solutions for 2i*f (just because they are shorter to write) are of the form C1*e^t*e^it+C2*e^(-t)*e^(-it) So you can see it's some combination of exponentials (so you could say of hyperbolic trig functions) and trigonometric functions (for f''=i*f it's the same expression but with t/sqrt(2) instead of t)

???? Bro what ​@@talastra

​​@@natepolidoro4565I think he was joking: if you read "i" as the english word the differenzial equation f"(x) = i*f(x) is an egoistic differential equation...

@@natepolidoro4565 think about i as a pronoun

(a*cost, b*sint) where a, b are semi-axis of ellipse. These give you well-known parametrization of ellipse

Sn and Cn? Jacobi functions? Or just a parameterization?

The interests on this channel are so...eccentric

The "unit" ellipse is the circule so you already hace elípticas trig functions

gotta catch 'em all!

notes are just sinusoidal waves, so it can happen

yup

In that form it’s still linear.

@@gamespotlive3673 but not fully generalized, as it's missing a constant term for what would correspond to the y-intercept

Right, but a second order diff eq needs two linearly independent solutions, hence the c1t and c2@@gamespotlive3673

I'd guess that if they did then someone would have invented them earlier.

Maybe they would have-maybe they have just been overlooked. Maybe a mathematical physicist could answer your question.

yes! well, at least, the rotational mathematics of parabolas, equivalent to that for hyperbolas and circles, does (i have yet to watch this video because of other obligations; it's just in my queue now); but: parabolic rotations can be used to represent translations as a type of rotation, and similarly to represent the "origin" for a positional space as a matter of arbitrary basis; the parabolic trigonometric functions i'm aware of are the components of these rotations edit: skimming quickly, the ones i'm aware of are much simpler than whatever looks like is going on, here; the ones i'm familiar with can be derived by just applying the exponential map to an arbitrary null-element (any element which squares to 0, but which need not be zero); for comparison, elements which square to positive and negative scalars will get you the hyperbolic and circular trigonometric functions, respectively

ok, i've watched the video now! ... unfortunately, i can only reiterate that there is a use for parabolic trigonometry, but this isn't it, because this was useless gibberish entirely divorced from its own motivation

The parabola seems like it should be important, but it is actually a degenerate conic produced when we cut the cone with a plane that parallels the edge of the cone. If we start with a plane parallel to the base of the cone, we get a circle, another degenerate case, but not quite as bad as the parabola. If we then tilt the base, we get an interesting ellipse, and if we tilt it more, we get an inspiring hyperbola. The lowly parabola is formed at the point where the ellipse transitions into a hyperbola. The axis of the ellipse becomes infinite at this point, and this kills off the y-squared term, leaving only a linear y term. Instead of resurrecting sinp, cosp, etc., let's enhance the trigonometric and hyperbolic functions to accommodate a linear term! -- e.g., a*x^2 + b*y^2 + c*y = 0! (Note that 0! = 1.)

How would that work? Like what equation would be the parameter

​@@advaitkamath8442like an ellipse, so similar to a circle but you scale one of the two axes by some additional parameter b, like x^2+y^2/b^2=1.

@@ojima510 but couldn't you also scale the other one or both, and the parameter can be any number that would look to make a lot of different scenarios and when you are doing it you would have to define the equation first for where the trig functions are being used

Are you guys sure? Don't these have anything to do with _elliptic functions_ ?

​@@advaitkamath8442scaling the other one is arbitrary (it'd be equivalent to swapping x and y as labels, which is just a conventional choice). And yes! That parameter b is a free parameter, so you actually end up with a class of functions sn(u,b) and cn(u,b) that simplify to just sine and cosine in the case where b = 1.

That's what i thought too. It leads to nicer functions with the usual symmetries: symmetric cosp and antisymmetric sinp. Btw the difference to the paper is just a shift: set t=0 at x=0 and swap x and y. Therefore the functions you get are the same as in the paper, but swapped and shifted to the left. The shift should be the arclength of x^2 from 0 to 1, which is ~1.47. But the formula says it's 4/3... idk why. Ok, now i know: The definition is not what I would use. In both circular and hyperbolic trig, the parameter is the arc length. But they used the angle. Unfortunately wolframalpha cannot give an explicit answer for the arc length dependend functions. We should just give them names 😅. Next i try and get the series expansions ... for more than a few terms i need to have a better technique. But it starts sinp(x) = x - 4/3! x^3 + 208/5! x^5 - O(x^7)

@@deinauge7894 hmmmm i’m not as stellar with advanced mathematics as i’d like but i may play around with this because it seems interesting 🧐

i was wrong at one crucial point: the parameter in hyperbolic trig is not the arc length. And if you scale an ellipse or a hyperbola until it becomes a parabola, you get the same "natural" parametrization for parabolic trig functions: sinp(t) = a t cosp(t) = b + a/2 t^2 what a surprise 😀 ps: a homogenious infinite scaling factor is divided out

"a new number system with a unit k where k^2 = 0" thats the dual numbers

These also satisfy the differential equation f''=0, which the one in the video seemingly do not, since they are not in the form ax+b which is the general solution. Yours also satisfy cosp(-t)=cosp(t), sinp(-t)=-sinp(t) for which I am not sure whether the funcitons in the video also satisfy them.

There is definitely something going wrong here. This is related I think, to the way you make the equation y''=ay, a!=0, to the case a=0. There is something more subtle than setting a=0. Usually, in a limit situation, you expect more symmetry than in the general case while here there is no symmetry left.

Agree, it seems that they do not satisfy the equation except for the trivial case 0cosp(x)+0sinp(t). I played with these functions in desmos and it seems that any linear combination of these is a non-linear function

I was thinking about exactly this and I think I have a resolution. The DE f''(x) = 0 specifies one criteria for f(x) to satisfy, but it doesn't specify everything for it. Namely, it says nothing about how f'(x) should behave. The traditional, linear solutions of y₁(x) = Ax and y₂(x) = B are the solutions that satisfy y₁'(x) = A and y₂'(x) = 0 (i.e. the solutions with constant velocity and zero velocity respectively). But what if we had non-constant velocity? That's where these guys come in. At 17:30 Michael shows the first order DE that these solutions satisfy, and it's definitely non-constant since it's a function of t (this makes total sense, the surface of a parabola does not have constant curvature) The takeaway is that a 2nd order DE needs 1st order information to fully determine it. The linear solutions are one basic solution, but they're not the only solution.

@@Zxv975I don’t quite get your point. For the ODE y’’=0, the most general solution is of the form y(t)=a t+ b, with a,b complex numbers. That’s it, there’s nothing magic in here.

showered and had a thought. so if we define exp(x) by its maclaurin series, then we can define cos, sine, cosh, sinh using the exponential and i and j as: cos(x)=(exp(ix)+exp(-ix))/2 sin(x)=(exp(ix)-exp(-ix))/(2i) cosh(x)=(exp(jx)+exp(-jx))/2 sinh(x)=(exp(jx)-exp(-jx))/(2j) where i^2=-1 and j^2=1 (where j≠-1,1). so, what is k^2=0 and k≠0 (the dual numbers), then we can define: cosp(x)=(exp(kx)+exp(-kx))/2 sinp(x)=(exp(kx)-exp(-kx)/(2k) working this out is actually kinda boring. since k^2=0, for all n>=2, k^n=0, so exp(kx)=1+kx, making cosp(x)=1 and sinp(x)=x they don't parameterize a parabola, but this is another answer for the parabolic cosine and sine i saw on quora somewhere, so it is *an* answer. it also satisfies the differential equations

this is a rabbit hole i LOVE SO MUCH. what happens when appending some kind of i-ish or j-ish object to the real/complex numbers? what if we choose some positive real number x and define k^2=x where k≠+/- sqrt(x)? what weird cosine-ish sine-ish functions do we get? what if our base system is the complex numbers and k^2=-1 but k≠+/- i? what about k^2=k and k≠0,1. what if k^2=i but k≠exp(tau/8i + tau n) for any integer n? this is such a deep rabbit hole

okay i've been having some fun. let c(x) and s(x) be the corresponding 'cosine' and 'sine' functions for k^2=(something) (not a formal def., but itll be clear from context) don't know if i mentioned it in another comment, but for k^2=0, we get c(x)=1 and s(x)=x, which parameterizes the line x=1, which i said was a 'boring' example, but i found something cool consider k^2=z for some z>0 (we'll break this later). after working it through, we can get: c(x)=cosh(sqrt(z) x) s(x)=sqrt(z)/z sinh(sqrt(z) x) these actually parameterize the right half of a hyperbola (clearly, since x=1 gives the split-complex numbers). this hyperbola is x^2-z y^2=1 now, if we break z>0 and let z=0, we get x^2=1, which we can think of as the right half of a hyperbola stretched vertically to infinity! so that helps understanding k^2=0 parameterizing the line x=1: its parameterizing a 'hyperbola' stretched out vertically to infinity now, if we consider k^2=-z, we get something very similar: c(x)=cos(sqrt(z) x) s(x)=sqrt(z)/z sin(sqrt(z) x) this parameterizes the ellipse x^2+z y^2=1, and, again, z=0 gives x^2=1, so k^2=0 ALSO parameterizes the right half of an ELLIPSE stretched vertically to infinity so cool. for k^2=0, c(x)=1 and s(x)=x is actually somewhat interesting now i'm thinking about what if k^2=k where k≠0,1. this seems kind of circular (really it just feels wrong), so we can consider it more as k(k-1)=0 where k≠0,1 this leads me to wondering about something like: let P(x) be a polynomial in x with real coefficients (and x is real) consider P(k)=0 where k is not a root of the polynomial (kind of contradictory definition, but i mean "root" here as specifically any complex number x where P(x)=0, and, since k is assumed to not be complex, it's not a root). also probably assume all the roots of P are real, because playing this game when P has a complex root would just give something isomorphic to the complex numbers right? i think

@@wyboo2019 Example: consider the matrix A = ((1 0) (0 0)) . A^2 = A but A ≠ 0 , 1 where 0 denote the 0 matrix and 1 denotes the identity matrix (I'll explain why I used this notation) So it is not wrong to imagine an element with unusual properties, much like complex numbers (i^2=-1), split-complex numbers (j^2=1 but j ≠ 1), and dual numbers (e^2=0 but e ≠ 0) We are now getting close to something. Something more general. We can imagine a set of elements together with some operation, a function that combine 2 elements into another element. This is what is discussed in abstract algebra, the concept of an algebraic structure. In general, 0 denotes the neutral element of addition and 1 denotes the neutral element of multiplication. And matrices together with the usual operations form a structure where weird properties like k^2=k but k≠0,1 or AB = 0 but A ≠ 0 and B ≠ 0, can happen.

The split-complex number k=1/2 (1+j) satisfies k^2=k. If you represent j as the Pauli matrix σ3, then the representation of k is just what the previous reply said.

also elliptical trig functions

sinp looks like a wibbly wobbly cot graph, and cosp looks like a wibbly wobbly -cosec graph

@@joeyhardin5903 Thanks! That's almost helpful! 🙂

Lol

This video is total gibberish. That would be ideal, but if cosp and sinp were at most linear, they wouldn't parametrize a parabola. What he states is that Acosp + Bsinp = c1t + c2 (with a typo), for some combination of constants, which isn't true either! The only thing that those functions satisfy is that random condition about the area under the curve.

Yes, you can make sinp_1(x)=1-sinp(x), and it should give y=x²

you may enjoy geometric algebra and 3D "Minkowski" spaces; regardless of how a plane is oriented relative to the null-cone, you can construct rotations within it; points on the cone will stay on the cone, moving within the intersection according to circular, hyperbolic, and parabolic rotations (but they won't look anything like what was shown here) you can use that behaviour to prove some properties of conic sections more easily than the traditional methods, too

No it's plus because the integral goes from cosh(t) to 1, but cosh(t) > 1. (It would be minus if you used the integral from 1 to cosh(t))

@@shirou9790 That means you are doing the right side under the integral, which is not in the shades area that is defined as t/2. t/2 = area of triangle with (cosh,sinh) as its corner, minus the part under the hyperbola

my problem is with the t/2 in the circle diagram. I think it should be "t". Checks out for t=pi/2.

I was bothered by that, but Penn's right. If you look carefully, you'll see that he indicates that t is the _angle_ in the circle diagram, not the area*. The area of a segment of a circle is half the angle it encloses (e.g. a full unit circ encloses and an angle of 2.Pi, and has area Pi). The expression on the left of the formula defining cosp t is the area of a segment enclosing angle t, which is correctly equated to t/2. *Penn confuses things by correctly labelling the _area_ of the "segment" in the hyperbolic diag as t/2 - effectively defining the hyperbolic "angle", by analogy with the circle case, as twice the area of the "segment" containing it. Ditto for the parabolic case@@petersiracusa5281

@@petersiracusa5281 the area is t/2. If you are Pi around the circle (half way) the area of half a unit circle is PI/2.

This episode was a little more theoretical than the usual videos. And it was getting even longer very quickly. It seemed clear to me. Perhaps you could get an undergraduate thesis out of exploring this further.

You said that just as though people "care what you like" Rather than whine, do a video of the parabolic trig function on your channel and see how you react to inevitable crybabies that are not being spoon fed up their fastidious standards.

yes you get elliptic functions

It’s actually the cosine of an arccosine

By “parabolic”, I meant straight-line-ic. And the version with the additional parameter is the Jacobi elliptic functions. I found out about them from the replies to another comment

Ha ha ha! Very clever!

barely if at all, lol to get ones that relate to exponentials, you'll need to use a non-zero element that squares to zero, and then, to draw parabolas with it, that element needs to be a null-bivector in a Minkowski space - none of the proper parabolas will share their plane with the origin

@@apteropith i'm not that far into math, but surely sin cos sinh cosh are defined by exponential functions

@@flavioxy yup, those ones are; as per Euler's equation: e^(i*t) = cos(t) + i*sin(t) for any i such that i^2 = -1 (assume t is scalar/"real") and this has a hyperbolic equivalent: e^(j*t) = cosh(t) + j*sinh(t) for any j such that j^2 = +1 (including j = 1) the parabolic version is similarly: e^(k*t) = 1 + k*t for any k such that k^2 = 0 which is a lot simpler

thanks!@@apteropith

You forgot to divide by 2 the whole way through.

@@megauser8512 No, I didn't forget that, but you can divide both sides by 2, if you want!

Before you ask: For that, you must divide all THREE sides (all 3 expressions) by 2: 1/(2 cos(t)) (sqrt(4 - 3 sin²(t)) - sin(t)) = 1/(2 cos(t)) (sqrt(1 + 3 cos²(t)) - sin(t)) = (sqrt(sec(t) + 3) - tan(t)) / 2

Ordinary circular cosine and sine are the elliptic cosine and sine. Just with different leading constants on each one.

Infinity exists infinity for all infinity?

@@carultch if you wish. I prefer infinities exists infinitely for all infinities either as an exclamation or a question. It does seem clunkier than ∞∀∞ tho

Let a be complex and then you don't need the - in front to get the circular/elliptic variants.

@@angeldude101 I was referring to the "heart" used as an apostrophe in "sol's" at the start of the video. I figured it was a reference to the upcoming dia de san valentin.

@@xizar0rg Oh. I completely didn't notice. I also somehow didn't notice the quotes around "love" suggesting that the word was important.

Putting t = 0, we get cosp 0 = {2 + √5)^(1/3) + {2 - √5)^(1/3) This looks wrong, but {2 + √5)^(1/3) + {2 - √5)^(1/3) = 1 (!) No, I didn't believe it either, but try it in your calculator Alternatively, show by direct calc. that x = {2 + √5)^(1/3) + {2 - √5)^(1/3) is a solution of x^3 + 3x - 4 = 0, which factorises over the reals as (x - 1).(x^2 + x +1) = 0; hence x = 1

This is just sin((sqrt(2)/2+i*sqrt(2)/2)x)

Just a product of "normal" and hyperbolic trig. functions.

Yes. Also, that function will be its own eighth derivative, and its negative fourth derivative, a solution of unity to the differential equation f(8) (x) = f(x). Also, you could get a function that is its own third derivative if the constant is equal to the prime cubic root of unity, or (-1+i*sqrt(3))/2.

@@bluelemon243 Yes. Also, that function will be its own eighth derivative, and its negative fourth derivative, a solution of unity to the differential equation f(8) (x) = f(x). Also, you could get a function that is its own third derivative if the constant is equal to the prime cubic root of unity, or (-1+i*sqrt(3))/2.