the parabolic trig functions

[@Kram1032]

some interesting weirdness about these: - neither of them are symmetric around the y-axis - they instead have symmetries around x=1 - cosp(x) is *antisymmetric* - sinp(x) is *symmetric* All of that seems weird to me. They also have two intersection points: 2/3 (2 - sinh(3 log(1/4 (-1 ± sqrt(5) + sqrt(22 ∓ 2 sqrt(5)))) at 0.63661 and 4.3634

[@BikeArea]

7:56 c1*t+c2*t should read c1*t+c2, shouldn't it?

[@wargreymon2024]

yup

[@gamespotlive3673]

In that form it’s still linear.

[@BetaDude40]

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

[@quarantined_buritto9109]

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

[@MatthewDickau]

Changing up the defining property results in a simpler (albeit somewhat boring) definition: In complex numbers with the usual imaginary unit i where i^2 = -1, exp(it) = cos(t) + i sin(t) In split-complex numbers with a unit j where j^2 = 1, exp(jt) = cosh(t) + j sinh(t) By analogy, in a new number system with a unit k where k^2 = 0, we could define exp(kt) = cosp(t) + k sinp(t) Which results in cosp(t) = 1 and sinp(t) = t.

[@papetoast]

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

[@Sinthoras-zo7cy]

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.

[@akshaymurthy4955]

@@Sinthoras-zo7cy These functions do not parametrize a parabola however. Perhaps complex units do not play the same role in parabolic trigonometry as they do in circular or hyperbolic?

[@Albeit_Jordan]

totally read this as 'pentatonic trig function' and thought you were taking mathcore metal to a whole new level

[@photinodecay]

notes are just sinusoidal waves, so it can happen

[@thelocalsage]

is using the relation y = 1-x^2 the best way to construct parabolic trigonometry? i’m thinking about the conic sections geometrically and i feel like the concavity of the parabola should go in the same direction as the hyperbola. some other elegant features like all the foci staying on the x-axis would make sense to me too if you think of smoothly moving the cross-section from circle to hyperbola.

[@deinauge7894]

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)

[@thelocalsage]

@@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 🧐

[@deinauge7894]

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

[@joysanghavi13]

The focus of this unit parabola is also on the x-axis

[@thelocalsage]

@@joysanghavi13 uhhh nope, the coordinate of the focus is (0, 3/4). Not sure where you’re getting that.

[@sinecurve9999]

Also have elliptic trig functions as well sn(u), cn(u), and dn(u).

[@advaitkamath8442]

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

[@ojima510]

​@@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.

[@advaitkamath8442]

@@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

[@rv706]

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

[@ojima510]

​@@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.

[@akshaymurthy4955]

Correction at 22:00; if my understanding is correct, those formulas are presented to propose a means of shifting phi being the "parabolic angle" into the circular angle represented by theta. The role of the input for parabolic and circular functions are different, and they are expressed through the simple relation tan(Theta) = tanp(phi) where tanp is the parabolic tangent function.

[@benhbr]

So what's the connection with the DE f''(x) = 0? Shouldn't sinp ans cosp satisfy it, i. e. be linear functions?

[@rgicquaud]

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.

[@Eye-vp5de]

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

[@Zxv975]

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.

[@pyrotas]

@@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.

[@TacoDude314]

17:57 and 21:57, the second derivatives of these are not 0.

[@sophiophile]

The hyperbolic trig functions have really useful applications in relativity. Are there any places where parabolic trig simplify physics (from a math or explanatory standpoint)?

[@enricobianchi4499]

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

[@edwardlulofs444]

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

[@apteropith]

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

[@apteropith]

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

[@r.w.emersonii3501]

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.)

[@officiallyaninja]

Now i just need the elliptical trig functions to complete the set!

[@timothytiberius487]

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

[@Noam_.Menashe]

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

[@caseyglick5957]

The interests on this channel are so...eccentric

[@xoppoy]

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

[@MichaelGrantPhD]

gotta catch 'em all!

[@eliyahzayin5469]

I've done something similar but with the parabola defined by y^2=1-x, as it preserves cos as an even function and sine as an odd function. In terms of elegance, though, the best way to define a set of parabolic trig functions would be to create a general conic set paramaterized by the ecentricity and the focci position such that the regular and hyperbolic functions are special/limited cases.

[@maestrobrutalizador9605]

Every conic can define a set of two trigonometric functions using the Minkowski space and the cone of light together with a plane at an angle that intersect it to define a conic. Then the projections on this plane of a parametrization by arc length of the conic (respect to some chosen initial point where you are measuring the length) can define trigonometric functions. Using this I get cosp t = t^2/2 and sinp t = t using as the reference point where the parabola is symmetric.

[@wyboo2019]

we know that in the complex numbers, where i^2=-1, we get: exp(ix)=cos(x)+i sin(x) in the split-complex numbers, where j^2=1, we get: exp(jx)=cosh(x)+j sinh(x) for what nonreal k do we get: exp(kx)=cosp(x)+k sinp(x) going to try and work out later today

[@wyboo2019]

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

[@wyboo2019]

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

[@wyboo2019]

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

[@YouTube_username_not_found]

@@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.

[@decare696]

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.

[@matiasmardijs8675]

Correct me if I'm wrong. But this video has a serious mistake. You only asked for the area condition to be true. You never asked the Acosp(t) + Bsinp(t)=c1t +c2 to be true. So the second derivative is not 0 as you pointed. Besides, for me it would have been interesting if cosp(t) and sinp(t) independently had 0 as their second derivative, and formed a base for the solutions to f''=0. But that would force cosp and sinp to be at most linear. So (cosp(t), sinp(t)) wouldn't parametrize a parabola. So all of this stuff turns out to be completely useless.

[@mathisnotforthefaintofheart]

Ok, silly question perhaps: Why with hyperbolics is there a plus in front of the integral sign (3:30), why not minus? And 8:04, C2 also has a t? Why?

[@Jordan-zk2wd]

Damn I really wanted to see a plot of the parabolic trig functions tbh ;-; edit: the equations provided let you plot them easily though!

[@physnoct]

At 13:09, I understand the integral term means the region hashed in green, but I don't get where the terms 't/2' and '½cosp t sinp t' come from. From there on, he's speaking klingon.

[@nahuelcaruso]

I think that the region is the "wedge" with vertices: (0,0), (cosp(t), sinp(t)), and (1,0)

[@physnoct]

After thinking about it, the graph axis should be labelled as usual: horizontal = x, vertical = y. The u and du in the integral may be renamed x and dx to make it more obvious. The term ½cosp t sinp t represent the area formed by the triangle defined by these points (0,0); (cosp t, 0); (cosp t, sinp t). The integral is the area between the triangle and the parabola (same principle for circle and hyperbola). However, why the area would give t/2 is not obvious yet.

[@nahuelcaruso]

​@@physnoctt/2 is the same as before, in the classic and hyperbolic trigonometric functions definition...

[@physnoct]

@@nahuelcaruso "t/2 is the same as before," I didn't learn about that definition in my trig courses, so I would need some demonstration or some video explaining this.

[@goodplacetostop2973]

23:00

[@lapiscarrot]

Lol

[@charleyhoward4594]

you realize that he NEVER ANS. the ques. (at 8:00) what exactly c1t + c2t was actually , dont you ?

[@tomholroyd7519]

That so amazing! Three types of conic sections. I've studied hyperbolic, and circular; parabolic is unknown to me this is so cool

[@gigantopithecus8254]

also elliptical trig functions

[@edwardlulofs444]

Thanks. This is a super explanation of a math paper that I likely would not have understood on my own. I found this very interesting. A good change of pace from the normal problem solving. If I had had you for a math prof perhaps I would have become a mathematician instead of a physicist. I liked your notation more than the papers notation but I understand why a pro would want to change the notation for publication. 😊

[@Demo-critus]

What do the graphs of cosp and sinp look like?

[@joeyhardin5903]

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

[@Demo-critus]

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

[@stephenhamer8192]

Can we do "logarithmic" trigonometry by taking the area of the "segment" enclosed by the curve y = In x, the x-axis, and the line-segment joining the origin to P = (x, In x) = (cosl t, sinl t) to be half the logarithmic "angle" t (definition of the "log" angle and the associated "log"-trig function cosl and sinl)? Then cosl t satisfies: ½cosl t.sinl t + Int [coslt to 1] In u du = ½t We also have: sinl t = In cosl t cosl 0 = 1

[@ingiford175]

Should the hyperbolic function definition for cosh(t) be 1/2 cosh(t) sinh(t) MINUS? and not plus?

[@shirou9790]

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))

[@ingiford175]

@@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

[@petersiracusa5281]

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

[@stephenhamer8192]

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

[@ingiford175]

@@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.

[@ramziabbyad8816]

I think the lower bound of integration in the first chart where cos(t) is defined should be cos(t) instead of t. This is because x varies from the adjacent side of the triangle to the point where radius = 1.

[@tommyyuan4054]

Why t/2? For hyperbolic and parabolic cases

[@JJ_TheGreat]

16:02 Could you please remind me how you used the Fundamental Theorem of Calculus to differentiate the integral? Thanks.

[@orisphera]

I've thought of this a long time ago. I haven't watched most of the video yet. Here are the formulas I thought of: sinp x = tgp x = x cosp x = secp x = 1 ctgp x = cosecp x = 1/x I also have an idea for trig functions with a parameter. I'm not sure on how exactly to denote them, but: sink x = (sin kx)/k cosk x = cos kx tgk x = (tg kx)/k ctgk x = k ctg kx seck x = sec kx coseck x = k cosec kx The functions would actually be denoted by putting k² as a lower index after the short function name. Putting -k² there would result in the same except with hyperbolic functions. (Alternatively, it can be the other way around: positive for hyperbolic and negative for elliptic.) If it's 0, parabolic functions would be used instead

[@orisphera]

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

[@iabervon]

If you try expressing the result of the cubic formula with functions R->R, you end up replacing the complex cube root with the cosine of an arctan (the real part of the complex number whose argument is 1/3 the argument of the complex number whose real and imaginary parts came from different parts of the formula). This seems like a good way to connect cosp with circular trig functions.

[@mossy8419]

It’s actually the cosine of an arccosine

[@ultrametric9317]

This would be interesting to present from the perspective of projective geometry and groups of motions. Essentially it must be a choice of metric in the Cayley-Klein sense. Also I smell line geometry :) Very fun!! Never had heard of this before.

[@AKA-f7p]

cosp(x)=(12ax)^(1/3) sinp(x)=1-(12ax)^(2/3) a="an arbitrary constants that follows those equation"{i took a= 1 because i derived this from that area relation}

[@Sinthoras-zo7cy]

If the parabolic trig functions are some functions involving cube roots and other things, then they can't satisfy the differential equation f''=0 because the general solution of this is ax+b, or can they?

[@matiasmardijs8675]

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.

[@GlenMacDonald]

Cool! I wasn't aware that these existed, but it's a perfectly natural extension. Thanks!

[@tomholroyd7519]

cosp(0)=1 defines this upside down unit parabola ... could you still do it with y=x^2?

[@Eye-vp5de]

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

[@michaelaxton5005]

Shouldn't the definition of Cosh have a minus sign in front of the integral sign? These changes would makes sense if the integral gives the area under the hyperbola.

[@AlbertTheGamer-gk7sn]

Also, because of that we have 3 forms of polar coordinates in 2 dimensions: Euclidean polar coordinates: The polar coordinates we are used to. x^2 + y^2 = r^2, x = r*cos(θ), y = r*sin(θ), Jacobian determinant: r. Setting r constant gives a circle with radius of r, setting θ constant gives a line of slope arctan(θ). Hyperbolic polar coordinates: Same as Euclidean polar coordinates, but with a transformation through the imaginary world: x^2 - y^2 = r^2, x = r*cosh(θ) = r*(cos(θi)), y = r*sinh(θ) = r*sin(θi)/i, Jacobian determinant: r. Setting r constant gives a branch of a hyperbola with transverse axis 2r, setting θ constant gives a line of slope 1. Parabolic polar coordinates: x^2 + y = r^2, x = r*cosp(θ), y = r^2*sinp(θ), Jacobian determinant: 2r^2(1-1/(cosp^2(θ)+1) + cosp(θ)*sinp(θ)/(cosp^2(θ)+1)) = r. Setting r constant gives a parabola of roots +-r, setting θ constant gives a line.

[@Altair705]

It's a shame I've never heard about this until now. Thanks a lot for the video. This could probably help in astrodynamics, when dealing with the 2-body problem. In this context, you have 3 classes of trajectories: ellipses, parabolas and hyperbolas. In particular, the Kepler equation has 3 different forms. The elliptic and hyperbolic forms have obvious similarities (replace "sin" by "sinh" and so on...), while the parabolic form is very singular in comparison (though much simpler overall). I wonder if parabolic trigonometry would allow to write the parabolic variant under a similar form, thus allowing some sort of unification.

[@alexedurojas2]

How is the formula for minute 12:42 deduced?

[@trueriver1950]

There is a solid geometry hint that parabolic functions exist between the "circular" ones and the hyperbolic ones. Slicing a cone with a plane parallel to the base gives a circle. If the plane is tilted parallel to a plane that's a tangent to the curved surface we get a parabola. Tilt the plane further and we get a hyperbola. Unfortunately, a smaller tile gives you an ellipse, which is based on the circular trig functions (with a scale factor in one axis). There are elliptical functions but as far as I know they are not planar sections of a cone... enlighten me if I'm wrong on that 😊

[@trueriver1950]

Michael: if something's not positive or negative it must be equal to zero Something that when squared gives -1 has just entered the chat

[@DanielSimanek]

Is there a generic "conic" trigonometry? Like define your unit cone as y²+x²=z, and then your conic trig function needs parameters to define a plane that intersects the conic section, plus one more for the angle, just like normal trig. I'm trying to imagine what would happen to the sin function (for example) as you sweep the angle of the intersecting plane ...

[@apteropith]

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

[@tommyproductions891]

when do we get the sinusoidal trig functions (is this too far)

[@colinpitrat8639]

I would have loved for this video to finish with "and that's why we don't use parabolic trigonometry" 😂

[@ArthurvanH0udt]

maybe ALSO interesting idenitiy to remark: cosp(X+Y)-cosp(X-Y) = 2sinpXsinpY

[@omargaber3122]

Please explain Selberg sieve and bounded Gap between prime numbers

[@gp-ht7ug]

Don’t like when you skip everything. It would have been interesting to see a graph of these parabolic trig functions

[@edwardlulofs444]

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.

[@MyOneFiftiethOfADollar]

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.

[@andrewhone3346]

All smooth conics (genus zero curves) are rational, so they can be parametrized by rational functions.

[@ilyafoskin]

What about the exponential representations?

[@roberttelarket4934]

In your Brilliant ad you reveal Mike that you were a physics major.

[@alexeyl22]

I'd like to see a rigorous explanation of hyperbolic trig functions and how it was related to solving Special Theory of Relativity cases. Just to get intuition into why STR is so heavily relying on it and demonstrates it as "proof" of what essentially is just a theory. 😊

[@flavioxy]

how do they relate to exponential functions?

[@apteropith]

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

[@flavioxy]

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

[@apteropith]

@@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

[@flavioxy]

thanks!@@apteropith

[@conando025]

Am I the only one that really adores the look of the chalk?

[@punditgi]

Excellent video

[@robshaw2639]

ill bet Randolph college is getting very strong applicants to their math program because of this channel….

[@eartphoze]

The dog can't eat homework, zombie can eat dog, WW3 zombieland

[@JourneyThroughMath]

Is trigonometry defined differently if we are looking at an ellipse that is not a circle?

[@gigantopithecus8254]

yes you get elliptic functions

[@richardchapman1592]

Can you have a continuous function that converts derivatives of the hyperbolic to regular si and cos.

[@michakaras7925]

so tanp(x) = tan(x)?

[@rainerzufall42]

1/cos(t) (sqrt(4 - 3 sin²(t)) - sin(t)) = 1/cos(t) (sqrt(1 + 3 cos²(t)) - sin(t)) = sqrt(sec(t) + 3) - tan(t) by the way!

[@megauser8512]

You forgot to divide by 2 the whole way through.

[@rainerzufall42]

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

[@rainerzufall42]

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

[@pedroalonso7606]

That's really cool!🤯

[@semyon422]

interesting video! it would be interesting to watch about a complex case f''(t) = (a + b*i) * f(t)

[@Alan-zf2tt]

Incredible! Not only does it intrigue the interested it also identifies the motivated. So in this exposition mathematical and pedagogical excellence is identified and encouraged along with third party resources. What is not to like? Oops: new signature ∞∃∞∀∞

[@carultch]

Infinity exists infinity for all infinity?

[@Alan-zf2tt]

@@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

[@pnadk]

Next do diabolic trig functions 🙂

[@roberttelarket4934]

Ha ha ha! Very clever!

[@ArthurvanH0udt]

VERY interesting!!

[@boanergesaraujo1501]

Do we get a new trig family if we use "i" as the constant, on f"=i.f?

[@bluelemon243]

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

[@pedroalonso7606]

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

[@AlbertTheGamer-gk7sn]

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.

[@AlbertTheGamer-gk7sn]

@@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.

[@talastra]

I enjoy listening even when I don't understand.

[@MessedUpSystem]

I'm salty because I was thiniking about this some time ago and I was so close to actually finishing it hahaha

[@xizar0rg]

Clearly general solutions "love" that f''(t)= -a^2*f(t).

[@angeldude101]

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

[@xizar0rg]

@@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.

[@angeldude101]

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

[@apteropith]

this is an interesting exercise, but the entire approach is incorrect and overwrought: you cannot produce the rotational geometry of parabolas, and hence their trigonometry, in a purely two-dimensional context; to produce a parabola by rotation at all you need the "vertex's" offset from the origin to be independent from parabola's own plane - without this you only get trivial lines associated with skew transformations incidentally, by a rotational approach the parabolic cosine and sine are merely 1 and the argument respectively, which exactly parametrizes the unit skew-line - the Pythagorean quadratic form here is (cosp t)^2 + 0*(sinp t)^2 = 1, which is exactly between the circular and hyperbolic versions the structure of a parabola doesn't emerge until you introduce a null-cone (as found in Minkowski space of 3 or more dimensions), and its rotational properties have many unique quirks (for example, in the 3D Minkowski space, the "axis of rotation" for a parabola is parallel to its plane, very much unlike for a circle or hyperbola) ... also, even ignoring what I said above, everything about that diagram (12:00) of an area-definition of a parabolic angle is wrong. everything. this has no comparison with the equivalent definitions for circles and hyperbolas, and no business being so incomparable - start at the vertex! sweep out area from the origin! maybe the authors of the paper(?) justify this in some way, but I've become convinced this approach is hopelessly misled regardless

[@joeeeee8738]

And the elliptic cos and sin?

[@carultch]

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

[@mihaleben6051]

DEAR GOD.

[@tomholroyd7519]

these are not simple looking functions is there something else going on here? Because I expected the parabolic case to be degenerate ...

[@WilliamHesse]

I agree - If the curve used for the parabolic case is x^2 + 0 y ^2 = 1, or x = 1, the area definition gives cosp = 1, sinp = t. This fits the other identities better, and is more obviously the degenerate limit between the circular and hyperbolic functions.

[@comdo777]

asnwer=1/3 isit

[@leewilliam3417]

Great 😊

[@stephenhamer8192]

Am I right in thinking: cosp t = { -½(3t - 4) + [ ¼(3t - 4)^2 + 1]^½ }^(1/3) + { -½(3t - 4) - [ ¼(3t - 4)^2 + 1]^½ }^(1/3) = the single real root of the depressed cubic x^3 + 3x + 3t - 4 =0? Quel horreur !

[@stephenhamer8192]

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

[@Kiba0169]

Well, I was interested in how the function actually look, but when I plot the parabolic trig functions using the normal trig functions and then the hyperbolic ones, I get different results.

[@aram5642]

2:00 and I got already lost. Time for me to go.

[@peterjones7886]

few typos

[@Axacqk]

Where did f"(t)=0 go?

[@owensthethird]

Now I we just need n-gonal trig and that's all the shapes

[@NStripleseven]

Pine, copine, pangent, copangent, pecant, copecant.

[@lumina_]

cool!

[@vicar86]

More like diabolic trig functions 😏

[@sepdronseptadron]

That ending is such a stretch that function does not relate well at all

[@p0gr]

whats all this weird area business? didnt you calculate a different area not starting at 0? dont we want cosp to be even and sinp to be odd?

[@skeleton_craftGaming]

Okay [I think] but why?

[@richardchapman1592]

Can you have a co

[@menjolno]

0 citations hahahahahaha

[@zenverak]

simp(t), cosplayer(t)

[@dreznik]

absolutely useless

[@orisphera]

I don't understand this video. Here's my understanding of it: First, you give the differential equation they should satisfy. The solutions to it are clearly linear (affine) functions. But you give a formula that gives some, but not all, of them. In particular, it doesn't give f(x) = 1. I can rewrite it as f(x) = (c1+c2)x. Then, you pretend you can't just plug in the starting conditions (s(0) = c'(0) = 0, s'(0) = c(0) = 1) and try to derive them using the other way. So, you define a unit parabola in a rather arbitrary way. (I'd try to define it as 0y² = 1-x². It wouldn't be a parabola, but it would work analogously to the circle and hyperbola.) Then, you solve for parabolic functions using both definitions at once. You don't prove that what you get satisfies either. I'm sure it dissatisfies at least one of them

[@boanergesaraujo1501]

Do we get a new trig family if we use "i" as the constant, on f"=i.f?

[@talastra]

that's so egotistical

[@luckycandy4823]

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)

[@natepolidoro4565]

???? Bro what ​@@talastra

[@ivandebiasi6657]

​​@@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...

[@talastra]

@@natepolidoro4565 think about i as a pronoun