trigonometry like you've never seen it 

Hahahahahaha.

*click* Noice!

that actually made me giggle, haha

good one!🍹

Swine/coswine is the hamgent (a reach, I know. I'm sorry😅)

This blows my mind

That's a crazy fact, wow!

What if gcd=1?

​@@pelledanasten1615 Then the least common denominator is the product of the denominators, if i understand your question correctly.

@@samwalko I think he got lcm and gcd mixed up

CHARMANDER AGREES!!!

No arent you just thinking of squirtle???

@@user-en5vj6vr2u Yes. I was thinking exactly of Squirtle. However, Squircle sounds like what you'd get if you'd evolved a Squirtle alongside Porygon, which would be awesome

Ah yes! That's it!

Yes, it's Beta(1/4, 1/4). He got the numerical approximation correct tho. Python 3.12.3 (main, Apr 23 2024, 09:16:07) [GCC 13.2.1 20240417] on linux Type "help", "copyright", "credits" or "license" for more information. >>> from scipy.special import beta >>> 1/2 * beta(1/4, 1/4) 3.708149354602744

Doesn’t look like it. The beta function (described in the correct form in the video) would have to be evaluated at (5/4, 5/4) to look like the integral we have. Though the resulting denominator would be gamma(5/2), not gamma(1/2).

@@NStripleseven Michael made a mistake with the change of variables from u to x. Where he wrote 5/4 in the integrant, it should've been 1/4 instead. The resulting expression is what the original comment above said. Michael's final expression doesn't even evaluate to the correct answer.

Furthermore, since Gamma^2(1/4) = (2.π^1.5)/AGM(rt2/2,1), so rho = π/AGM(rt2/2,1), providing a nice relationship between rho and π: π/rho = AGM(rt2/2,1)

Where does this come from?

@@hybmnzz2658From “Modular Functions and Transcendence Questions” by Yuri Nesterenko, 1996.

@@hybmnzz2658 Modular forms of all places. I don't know the details, but apparently there is a theorem (Nesterenko's theorem) that for any z in the upper half plane, there is a set of three algebraically independent numbers among E2(z), E4(z), E6(z) (Eisenstein series) and q (=exp(2pi i z)). Setting z=i, we have E6(i)=0, so the other three must be algebraically independent. Their values are E2(i) = 3/pi E4(i) = 3Gamma(1/4)^8/(2pi)^6 and of course q(i) = exp(-2pi).

sup Queen's Uni 🍁

@@PillarArt ??

theres a whole book on the topic: "Squigonometry: The Study of Imperfect Circles"

@@lunaticluna9071 Yes I saw that. I didn’t know that before seeing this video, so I had a difficult time trying to figure it out on my own. I only managed to approximate the cosquine, as I’m not skilled enough to derive the formula myself

I also was looking for something like that to model my data. Now that I found it I am afraid of using it and having to explain to peiple 😂

Do you know any proof of the fact it converges to 4?

@@davidemasi__ That appears to be equivalent to 4 - ζ(2)/n² + ζ(3)/n³ - 9ζ(4)/(16n⁴) + k ζ(5)/n⁵ ∓ … You only have to prove this holds :-)

Yes

yes u r right

Well, the square roots only work for positive numbers (we don't want to have complex values for x and y), but then, the quircle is clearly symmetric, therefore the sign of the cosine times the square root of the absolute value of the cosine, and the sign of the sine times the square root of the absolute value of the sine should work quite fine, and indeed, every solution has to be some reparameterization of that. Indeed, I suspect the sign and absolute value make it non-analytic without appropriate reparameterization. Now we can try to calculate the derivatives of that, then we get (in the positive quadrant) x'(t) = -sin(t)/(2 sqrt(cos(t))) = -y^2/(2x), which looks quite different from the x'=-y^3 in the video. Note however that the differential equations in the video don't give a constant speed on the curve, since x'^2 + y'^2 = y^6 + y6, which clearly isn't constant. Therefore there is no reason to assume that the periodicity of the function tells us anything about the circumference of the curve. Now the fact that the number is between pi and 4 may be suggestive, but note that also the area of the unit circle is pi, and the area of the outer square is 4, so it could just as well be the area of the squircle. Or something completely different, of course.

What

I'm assuming it's a joke, as the last word would be roughly pronounced like "po-cue-pine" or, in other words, "porcupine", an animal with spikes on it's back. ​@@Durglass

Time is a flat squircle

is there a way of treating that smoothening such that it is like taking a limit?

Yes, and that results in rho = 0.3090..., which makes the arguement that it should be between pi and 4 problematic! Edit: Some folks pointed out the error was made at 22:43, where the exponents should be 1/4 -1.

No, the denominator is correct. The numerator should be (Gamma(1/4))².

Please make a video about how this is related to fractional derivatives

Wait, don't you mean sphere, not circle, since you are working with 3 variables here, instead of 2?

@@megauser8512 Cross section of a sq-cube with a plane is a subset of a plane that is a circle. For a regular cube that would be a hexagon. So, cross section of x^n + y^n + z^n = 1 is a circle not only for n = 2, but for n = 4 too, and it approaches a hexagon as n -> infty.

they're actually related , see "Squigonometry: The Study of Imperfect Circles"

Hahahahahahaha.

How did you come up with that???

@@neonlines1156 if you know that cos and sin satisfy cos^2+sin^2=1, then what satisfies (...)^4+(...)^4=1? An obvious choice is sqrt(cos) and sqrt(sin), at least in the interval of the argument between 0,pi/2. The absolute values and signs are just to make it parameterize the full unit squircle.

@@Kapomafioso very clever!

@@almostme80 thanks...I think what Michael was getting at (but didn't say explicitly) is, that while you can easily make up some parameterization, it might not be analytic (and mine, indeed, can't possibly work as analytic functions that would describe the whole squircle). You can see that immediately from the absolute values and signs, but also the series expansion for my y(t) features t^1/2 as the first term...

This parametrisation is not natural (in a sense of differential geometry) as t is not the arclength of the curve.

That would just be squine/cosquine

Or the sequant

​@@spacespark6922Squeecant? 😂

although, I did found recurrence relation, but couldn't solve it (because of m+1 term) f(n,m)=(m+1)f(n-1,m+1)+(2n-m+2)f(n-1,m-3) f(0,1)=1

Both z and w are 1/4 so the denominator is correct but the numerator should read gamma(1/4)^2. Final calc of 3.708 is correct. Typo

I don’t think there is any. Perhaps it’s helpful in the same way as circular trig, if for some reason you’re working with squircles a lot

it's interesting

Also the superhyperbolic

if I ever think about doing it i'll remind you

In contrast with the parametrization of the circle, the velocity (√((x')²+(y')²)) ain't constant for the squircle though.

You'd end up with a really nasty integral that can probably only be calculated numerically, unless there's some odd substitution that would clean it up.

Not really because vectors can start anywhere. The for example means you go over 1 on the x-axis, over 2 on the y-axis, and over 3 on the z-axis. But you can start it at any point you want because it doesn’t matter where a vector is, it's just some magnitude and direction being applied but it doesn't matter where it's being applied.

No because you have only two equations

I think you can't. You have a 9 variables vs 6 equations sistem. Infinite values satisfie that.