Відео

Yes the gudemannian function relates circular and hyperbolic trig functions, such that tan(θ)=sinh(φ) and sec(θ)=cosh(φ). Lol we're not there yet. I'm not claiming to come up with anything new other than simply this: It would be nice to have a treatment of hyperbolic trig that starts from right triangles and ratio definitions like we do for circular trig. Some interesting geometry, algebra, and intuitions fall out as a result. One benefit, for example, is that the forward-backward transforms that come up in calculus, such as arsinh(cosh(x)) have nice geometric methods that parallel those in circular trig, but I never see it taught that way.

paring down commitments. One byproduct is that I expect and hope to get back to this soooon

@@arithknot9740 I totally understand how 'real life' gets in the way of fun stuff like this. Good luck, and thanks for the video :)

In short, we can parameterize the hyperbola as (x,y) = (cosh φ, sinh φ). But we can define new coordinates along the asymptotes: (a, b) = (0.5 e^φ, 0.5 e^(-φ)). Geometrically, we can see this relationship in the unit hyperbola as we draw a perpendicular from the asymptote to our point. This will be the topic of the second and third videos.

There is, but it's not quite the same. In that method, you take a triangle at one fixed level of extension, and apply hyperbolic rotation to it. But we haven't discussed hyperbolic angle or hyperbolic rotation yet. So that's coming later.

@@arithknot9740where do the lines of triangle end up? Is the fixed length between the origin and the bottom of the hyperbola?

I study to answer my own questions. This stuff comes from my explorations into the unsolved question of the (ir)rationality of γ.

lol i dids not died. i just has many lots of responsabilitoeses

:( Haven't forgotten. Busy life though. Definitely want to get back to this still

From the other treatments I've seen, the asymptote corresponds to the speed of light. You might take a look at minutephysics' relativity series 😊

ua-cam.com/play/PLoaVOjvkzQtyjhV55wZcdicAz5KexgKvm.html&feature=shared

That's a non-square hyperbola. That would be like asking what the radius of an ellipse is. If a=b, then it's a square hyperbola. (x/a)^2 - (y/a)^2 = 1 x^2 - y^2 = a^2 So it has a fixed length of a. But yeah, if a doesn't equal b, you have an x-stretch that is different than the y-stretch, and you get a non-square hyperbola. The question I left hanging was how one might get a non-square hyperbola in the triangle context, or if that even makes sense.

@@arithknot9740 Thanks for the answer!!

Currently working on the second one

Hey, Look into my eyes, look into my eyes, not around the eyes, don't look around the eyes, look into my eyes. "Snaps 🤞" you're under. Now go and make another video immediately after reading this where you explain another math topic. Annnnnd you're out "🫰"

It's not forgotten. I'm just too terribly busy. I'm trying to pare down some commitments so that maybe I can get back to this soon. I've been so looking forward to it

Subscribed and awaiting the sequel!!

hes too busy grading my algebra 2 homework

@@jelly.timeeeThere's some truth to that 😅

@@arithknot9740 please come back 🙏🙏

Yeah, I agree! There’s simply not enough videos on hyperbolic trig when it describes so much, the perfect arch, a string held between two hands, and that’s just one function, the catenoid or hyperbolic cosine variant

That's exactly what I have in mind. I've done a just a little bit of hunting and haven't found anyone that has developed this topic this way, so I've been doing it from scratch. It took me about a month to come up with a nice geometric proof for hyperbolic angle addition. Hopefully I can get back to this hobby soon.

@@arithknot9740 Oh, sounds exciting! Please do it, I'll be waiting for it. My fascination for this came with the idea that e^(ix) can be thought of as (e^x)^i and e^x being coshx + sinhx, e^(ix) would be cosh(ix) + sinh(ix), which is cosx + isinx - Euler's formula So De Moivre's formula applies not only to e^(ix), but also to e^x. Just like e^(inx) = (cosx + isinx)^n = cos(nx) + isin(nx), We have e^(nx) = (coshx + sinhx)^n = cosh(nx) + sinh(nx) And so e^(ix) = (coshx + sinhx)^i = cosh(ix) + sinh(ix) = cosx + isinx

There's other fun connections to be had also. If you're that comfortable with the complex math, then you might appreciate something _far_ later I have planned in the series: the tie-in to split-complex numbers. If you've had any linear algebra, it is fairly easy to verify that the square of the matrix 0 -1 1 0 is -1 times the identity matrix. We can use this matrix as i in the field of 2x2 square matrices. If we instead use j = 0 1 1 0 this gives hyperbolic geometry when plotted on the real and j axes instead of the circular geometry that results from i above. j^2 = 1. e^jx = cosh x + jsinh x

@@arithknot9740 Fascinating. Let C = 0 -1 1 0 Let H = 0 1 1 0 Then, C^2 = -I and H^2 = I where I is the identity matrix. Then (complex) exponential of matrix A can be written as: e^(CA) = cos(A) + Csin(A) e^(A) = cosh(A) + Hsinh(A) Am I correct? Another interesting observation. The matrix C transforms a point (x,y) into (-y,x). For unit circle, y = sinθ and x = cosθ. So the point (cosθ, sinθ) is transformed to (-sinθ, cosθ). Interestingly, the rotation matrix in 2 dimensions is cosθ -sinθ sinθ cosθ Plug in θ = 90° and you get C. Goes on to show rotation by 90° just like i. H transforms (x,y) to (y,x). Kind of an identity matrix because it doesn't scale the vector. But also not because it changes the point.

Some greater care has to be taken when you take e to the power of a 2x2 matrix. It does work nicely, but it's worth going through the steps to see why it works as you expect. Let me just leave you with a few leads for now: 1) Search for split complex numbers 2) The rotation matrix by θ is cosθ -sinθ sinθ cosθ which can be decomposed into its I and C parts (since cosθ and sinθ are just scalars) I cosθ + C sinθ The same can be done for the hyperbolic rotation matrix, rotating by an arbitrary hyperbolic angle φ: coshφ sinhφ sinhφ coshφ Which decomposes into I and H components: I coshφ + H sinhφ

With four kids and a side job, it's difficult to find the time/money/energy to work on it. I've spent some time on it, but I'm hoping to make a lot of progress on it over the upcoming holiday breaks.

Thank you so much :)

Thank you so much. The software is mentioned at the very end. It's the manim python library originally developed by 3blue1brown's Grant Sanderson, which he released open source, called manim. I'm using the community edition.