Integrating the impossible

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[@Keithfert490]

I believe it should be x'=y^2 and y'=-x^2, instead of cubed

[@krisbrandenberger544]

Yes

[@aniruddhvasishta8334]

Glad this comment exists, I was about to say that

[@blqbfish2643]

This same antiderivative can also be expressed in terms of the Gaussian hypergeometric function: the antiderivative is x*2F1(-1/3, 1/3; 4/3; -x^3)+C

[@disgruntledtoons]

I'd do the engineer approach: Work out the Taylor series until the margin of error is good enough, and call it a day.

[@Demo-critus]

"Underscore 3" should be "subscript 3" 🙂

[@tomholroyd7519]

some people think in LaTeX

[@hiddebalk98]

When parametrizing the curve |x|^3+|y|^3-1 you could see it as the solution of a Hamiltonian system with H(x,y)= |x|^3+|y|^3 as the Hamiltonian. Since the hamiltonian is conserved the flow would paramatrize the curve, and the system is given by x'=3|y|y, y'=-3|x|x.

[@dukenukem9770]

This is badass! I’ll definitely integrate this into my son’s curriculum…

[@tomholroyd7519]

I love the generalization to sin_3, sin_n

[@fguizini]

Sorry but, deriving the function y=(atan (x)+x*(1+x^3)^(1/3))/2 I couldn't find the same curve as the original function y=(1+x^3)^(1/3)! What was wrong?

[@איתמרעינת]

Shouldn't the integral be changed to jus sec_3(t) instead of cubed?

[@Keithfert490]

No, the other two powers of sec_3(t) come from the change from dx to dt in the substitutio

[@איתמרעינת]

Thnx ​@@Keithfert490

[@lurkmoar3926]

Hi Michael, the region |x_1|^s + |x_2|^s +...+|x_n|^s 0, and it is convex for s>1. It is empty for s=0 (unless n=1). It is unbounded and nonconvex for s0, it has volume 2^n Gamma(1 + 1/s)^n /Gamma(1+n/s). I like your lectures and I am a subscriber. I particularly like today's and yesterday's lectures. Good job! 👍

[@MGSchmahl]

At 6:28, he writes x'=-y³, y'=x³ when it should be x'=-y², y'=x² instead. However, the solution is correct because the only time he uses this is at 12:38, where he uses the correct version. It's fairly common for him to make errors that have no effect on the final result. It's common enough that it's easy to wonder if it's intentional.

[@Alan-zf2tt]

It can be difficult doin math in exam conditions but in "live" recordings with no script no prompts and limited wish or opportunity to re-edit or do another take the simple answer is: heck it is a mature audience, a learned audience maybe they will just figure it out for themselves?? Which reminds me: ever heard of the constipated mathematician? Worked it out with a pencil 🙂

[@TomFarrell-p9z]

I don't think it's intentional, just because I usually make as many math mistakes myself. He has it right in his notes (usually), so it's a "typo" sort of error putting it on the board. (I don't do KZbins, but I do engineering problems and try and check each step, via example computed numerically, or other means)

[@MGSchmahl]

@@TomFarrell-p9z I don't actually think it's intentional either. You can see him referring to his notes quite frequently, which is why these "typos" never end up ruining the final result. However, there is little incentive to edit these out, because invariably someone comments on it, which drives engagement, which the KZbin algorithm likes.

[@allanjmcpherson]

I understand that calling it the 3-Pythagorean Identity generalizes easily, but it feels like a wasted opportunity not to call it the Trithagorean Identity.

[@DrYankeeDoodle]

*Threethagorean

[@davidmillikan42]

*Pythreegorean

[@Midori179]

shouldnt the differential equations be x'=-y^2 and y'=x^2?

[@nguyenquangkiet2103]

Dixon elliptic functions?

[@mohamedfarouk9654]

I believe this is the first indefinite integral I see on this channel in a very long time.

[@VideoFusco]

It is not true that |x|^3+|y|^3=1 cannot be parameterized using elementary functions| Just set x=|cos t|/CubeRoot[cos t] and x=|sin t|/CubeRoot[sin t] and it turns out that |x|^3+|y|^3=1 for every real t (these functions have discontinuities that can be eliminated in the zeros of sin t and cos t).

[@potaatobaked7013]

while it can be parameterized with elementary functions, it cannot be parameterized with elementary functions that have the derivative properties that allow us to solve the problem

[@The1RandomFool]

You can parameterize the curve with x = sgn(cos(t))*cos(t)^(2/3) and y = sgn(sin(t))*sin(t)^(2/3). If you substitute those into |x|^3 + |y|^3 = 1 you get cos(t)^2 + sin(t)^2 = 1 which is always true.

[@potaatobaked7013]

while it can be parameterized like this, these functions don't have derivative properties that allow us to solve the problem

[@Kapomafioso]

6:28 not sure where you got the cubes from, it should be squared. You're solving a different problem than originally stated.

[@PleegWat]

And he must have done wrong during his initial working out, or he would have found out next time he checked his notes. But there's more wrong than just writing sin₃ while using sin₄ definitions; I'm pretty sure the whole thing ends up being invalid.

[@Walczyk]

how do i evaluate arctan3?? i’m missing that math

[@HideyukiWatanabe]

6:28 "cube" is just a typo for "square." 9:00 The identity is derived from the "square" system; so the whole thing is valid. dtan₃t/dt = d/dt (sin₃t/cos₃t) = (sin₃'tcos₃t - sin₃t cos₃'t)/cos₃²t = (cos₃³t+sin₃³t)/cos₃²t = 1/cos₃²t = sec₃²t

[@mitcigamer4289]

Engineer approach: x^3 + 1 is basically x^3 so the integrand is x, so the integral is x^2 / 2 + c QED

[@dank.]

Damn, that works so well too.

[@iabervon]

Bad engineer! You have to say which regime x is in before making that approximation. For small x, the answer is x+C, and I think it's ((3x-1)^(4/3))/4+C around 1, although I may have made a mistake in calculating that one.

[@mhm6421]

x^2/2 is approximately x, but x is approximately 0, so it's 0 + c

[@jamesfortune243]

Euler substitution for dummies 🙂

[@_nemo171]

You forgot to add the 50% safety margin.

[@bingchilling8384]

Fanum tax yeah i just stole, They call me Drake cuz i got a pole, I integrated the impossible (whoah, whoah, whoah, whoah)

[@halchen1439]

Ehm what the sigma

[@major__kong]

I think I followed the last two videos. But I'm an engineer not a mathematician. So my head hurts :-)

[@charleyhoward4594]

same here - u got to admit he's amazing ...

[@ej3281]

at 6:28 you "move things around" and suddenly your x^2 turns into an x^3, and the y^2 turns into a y^3. You lost me after that.

[@bjornfeuerbacher5514]

If you correct the third powers to second powers and work onwards with these (correct) formulas, everything else is ok.

[@paulkohl9267]

Excellent video. Lovely generalization. Thank you.

[@charleyhoward4594]

this bothered me immensely ; this is most Definitely NOT 'Integrating the impossible' - just changing the domain of func.

[@VideoFusco]

Once defined cos3(t) e sin3(t) what is the range of variations of t to complete a period? Is it still [0,2pi]? I don't think so, but this must be cleared to understand the geometrical meaning of t for this new functions.

[@goodplacetostop2973]

21:53

[@kylesendgikoski4231]

"Next time on (Dragon Ball Z) Michael Penn does wierd math, watch Michael fix some errors he made last time and finish the problem by showing the point of the IVP with some actual description of what sin_x(t) actually is"

[@mauricepanero]

Could you prove those differential rules geographically ?

[@_skyslayer]

If you replace '1' with 'x' it's also possible

[@othila9902]

Assume 1=0 I hope this joke is clear enough lol

[@r75shell]

How can I numerically calculate those new functions? I can't assume that in cos_3(t) there t is angle nor arc length. I should prove it first.

[@bjornfeuerbacher5514]

There are lots of methods for solving a system of differential equations numerically, the simplest one being Euler's method.

[@r75shell]

@@bjornfeuerbacher5514Yep, but those mostly provide some approximation with hard estimation of error. You need to pick step-size, number of steps... For common trigonometry functions there are very efficient arbitrary precision numerical methods. Well after playing around with this, I was able to find efficient way of computing arctan, sin, cos version of this. Surprisingly, arctan was easiest. As mentioned in the video, arctan derivative is function representable by elementary functions. Thus, you can derive Taylor series at point x=0. Then, you can integrate this series using rules for polynomials to get series for our arctan. You can use it to calculate with arbitrary precision up to x=0.5. Then, you can derive Taylor series at point x=0.5 in similar fashion. There would be fractional powers but you may get rid of them if you factor out (1+0.5^3)*(2/3) -- by which you may multiply in the end. In this way you'll get a way to calculate any arctan within range [0,1]. What about >1 ? Well, arctan(x) is integral from 0 to x of (1+t^3)^(-2/3) dt, then substitute here t = 1/u, and you can easy show that arctan(x) = arctan(infinity) - arctan(1/x). Then, you may also show similarly that arctan(infinity) = 2arctan(1). This is all you need to calculate arctan. calculating sin, cos is also doable with taylor series, but much harder. At x=0 is straightforward, but for x=0.5 for example is horrible mess. Maybe it's possible to extend ideas how normal sin / cos calculated.

[@filippocamporeale3139]

How did we know that sin_3 and cos_3 are differentiable when we built the differential equation?

[@aniruddhvasishta8334]

It's basically part of their definition. He wants them to have nice properties like the original sin and cos so that the calculus works out and he can solve the integral.

[@wesleydeng71]

So these are fat sin() and cos.().

[@roeebaer]

There's a mistake, see other comments

[@CarloGinex]

How about the same thing but for the nth root of(1-x^n)

[@xinpingdonohoe3978]

It inspires us to define some cos_n and sin_n functions, sort of like the many polylogarithms.

[@erfanmohagheghian707]

I highly recommend you take some time and watch your videos carefully before you upload them. The differential equations you got should have squared, not cubed terms, though it doesn't cause any issues to the rest of the problem. You're smart and knowledgeable and admirable.

[@pierreabbat6157]

A Mordell curve is a cubic of the form y²=x³+a, where a is some number. What is the area between two Mordell curves y²=x³+a and y²=x³+b?

[@omivela94]

Hi Michael, is there another way to define sin_3 and cos_3 in an explicit form involving sin and cos of x? or any other expressions involving functions of x?

[@Calcprof]

Integral can also be solved with the Appell Hypergeometric function. (which then gives a double series) I suppose that you could get a series directly (binomial!).

[@topquark22]

Un tour de force! This is like the kind of carefully-constructed proofs that I did in university. It's like a Swiss watch when all the pieces come together!

[@ilyasb4792]

Could be interesting to see if these kind of functions have some kind of taylor expansion which maybe can give us an idea on how to approximate the area of a slice of the "fat circle".

[@vincentbutton5926]

Math went wrong at 6m30. What's he calling this? Squirtle math, squirrel math...

[@peterhall6656]

Do Michael's misprints follow a Poisson distribution?

[@Dionisi0]

your maths are wrong, your initial function is not bounded to anything

[@illumexhisoka6181]

Can you find the integral from 0 to 1 of the nth root of (1-x^n) ??

[@xviruzz_platinum151]

Since you have used the differential equations x'=-y^2 and y'=x^2 the functions you've found are not the sin_3x and cos_3x but the sqx and the csqx, the ones you found in the last video. Awesome video though :D.

[@nationalstudyacademykim5030]

Crazy stuff!!! touche Professor Penn!!!

[@tomholroyd7519]

very interesting connection to pythagoras

[@lloydgush]

Oh, squine and cosquine.

[@christopherravlin6051]

there should be NO cubes

[@ramonido]

Oh such a beautiful solution

[@stephenhamer8192]

Very interesting

[@MGoebel-c8e]

Going back to the beginning, i don’t quite see what the “fat circle” and its equation have in common. What is the =1 on the right hand side? There is no constant length like the radius in the case of a circle. One can define two functions yielding the same constant value when added together of course, but how exactly does that help to parameterize a “fat” circle?

[@toriknorth3324]

it depends on what distance metric (norm) you use for the metric space. most of the time we use euclidean distance (also called the L2 norm) with the usual d^2 = x^2 + y^2 formula to calculate the distance d between the points (0, 0) and (x, y). what that means is that a circle in the L2 metric space looks the same as a regular circle, but a circle in the L3 metric space (where distance is calculated with the equation d^3 = x^3 + y^3) looks like the squircle he drew. so, the distance between the origin and points on the squircle isn't constant in L2, but it is constant in L3.

[@MGoebel-c8e]

@@toriknorth3324 Still challenging to wrap my mind around but very much appreciate your answer!

[@hydropage2855]

You messed up the solution set to the initial values big time

[@TomFarrell-p9z]

Very interesting for this retired engineer; thank you Michael! Spent much of my career working with atmospheric turbulence, where there happen to be a number of integrals with integrands taken to n/3 powers. I may explore solving some of them with this technique. Of course, it would be nice to have canned special functions based on the sin_3 and cos_3 family of functions, so we could solve the definite integral version as we do with trig functions, the error function, etc. more efficiently than simply using canned numerical integration. Time consuming to work out, but perhaps worth it for someone who has to use an integral often.

[@nigellbutlerrr2638]

He needs to learn what Integration is 😮😮

[@nigellbutlerrr2638]

When a person says " anti derivative". Then. I ignore Everything which they state. Basically Using the Correct International terms should be respected. This person has zero to teach.😮😮

[@reeeeeplease1178]

L

[@robertveith6383]

Original poster, write sentences. The standard terms are "derivative" and "anti-derivative." You have much to learn. Your post is being reported for misinformation/disinformation, not to mention you are being uncivil.