Integrating the impossible

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Michael Penn

Michael Penn

Күн бұрын

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Пікірлер: 89
@Keithfert490
@Keithfert490 6 ай бұрын
I believe it should be x'=y^2 and y'=-x^2, instead of cubed
@krisbrandenberger544
@krisbrandenberger544 6 ай бұрын
Yes
@aniruddhvasishta8334
@aniruddhvasishta8334 6 ай бұрын
Glad this comment exists, I was about to say that
@lurkmoar3926
@lurkmoar3926 6 ай бұрын
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
@MGSchmahl 6 ай бұрын
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
@Alan-zf2tt 6 ай бұрын
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
@TomFarrell-p9z 6 ай бұрын
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
@MGSchmahl 6 ай бұрын
@@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
@allanjmcpherson 6 ай бұрын
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
@DrYankeeDoodle 6 ай бұрын
*Threethagorean
@davidmillikan42
@davidmillikan42 6 ай бұрын
*Pythreegorean
@Midori179
@Midori179 6 ай бұрын
shouldnt the differential equations be x'=-y^2 and y'=x^2?
@nguyenquangkiet2103
@nguyenquangkiet2103 4 ай бұрын
Dixon elliptic functions?
@The1RandomFool
@The1RandomFool 6 ай бұрын
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
@potaatobaked7013 6 ай бұрын
while it can be parameterized like this, these functions don't have derivative properties that allow us to solve the problem
@blqbfish2643
@blqbfish2643 6 ай бұрын
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
@Kapomafioso
@Kapomafioso 6 ай бұрын
6:28 not sure where you got the cubes from, it should be squared. You're solving a different problem than originally stated.
@PleegWat
@PleegWat 6 ай бұрын
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.
@HideyukiWatanabe
@HideyukiWatanabe 6 ай бұрын
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
@VideoFusco
@VideoFusco 6 ай бұрын
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
@potaatobaked7013 6 ай бұрын
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
@mohamedfarouk9654
@mohamedfarouk9654 6 ай бұрын
I believe this is the first indefinite integral I see on this channel in a very long time.
@whatelseison8970
@whatelseison8970 Ай бұрын
The Dixon elliptic functions cm() and sm() satisfy cm^3+sm^3=1. They form an equilateral triangular lattice in the complex plane.
@mitcigamer4289
@mitcigamer4289 6 ай бұрын
Engineer approach: x^3 + 1 is basically x^3 so the integrand is x, so the integral is x^2 / 2 + c QED
@dank.
@dank. 6 ай бұрын
Damn, that works so well too.
@iabervon
@iabervon 6 ай бұрын
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
@mhm6421 6 ай бұрын
x^2/2 is approximately x, but x is approximately 0, so it's 0 + c
@jamesfortune243
@jamesfortune243 6 ай бұрын
Euler substitution for dummies 🙂
@_nemo171
@_nemo171 6 ай бұрын
You forgot to add the 50% safety margin.
@Demo-critus
@Demo-critus 6 ай бұрын
"Underscore 3" should be "subscript 3" 🙂
@tomholroyd7519
@tomholroyd7519 6 ай бұрын
some people think in LaTeX
@hiddebalk98
@hiddebalk98 6 ай бұрын
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.
@disgruntledtoons
@disgruntledtoons 5 ай бұрын
I'd do the engineer approach: Work out the Taylor series until the margin of error is good enough, and call it a day.
@paulkohl9267
@paulkohl9267 6 ай бұрын
Excellent video. Lovely generalization. Thank you.
@Walczyk
@Walczyk 6 ай бұрын
how do i evaluate arctan3?? i’m missing that math
@peterhall6656
@peterhall6656 6 ай бұрын
Do Michael's misprints follow a Poisson distribution?
@ilyasb4792
@ilyasb4792 6 ай бұрын
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".
@VideoFusco
@VideoFusco 6 ай бұрын
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.
@bingchilling8384
@bingchilling8384 6 ай бұрын
Fanum tax yeah i just stole, They call me Drake cuz i got a pole, I integrated the impossible (whoah, whoah, whoah, whoah)
@halchen1439
@halchen1439 5 ай бұрын
Ehm what the sigma
@Calcprof
@Calcprof 6 ай бұрын
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!).
@dukenukem9770
@dukenukem9770 6 ай бұрын
This is badass! I’ll definitely integrate this into my son’s curriculum…
@pierreabbat6157
@pierreabbat6157 6 ай бұрын
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?
@topquark22
@topquark22 6 ай бұрын
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!
@nationalstudyacademykim5030
@nationalstudyacademykim5030 6 ай бұрын
Crazy stuff!!! touche Professor Penn!!!
@major__kong
@major__kong 6 ай бұрын
I think I followed the last two videos. But I'm an engineer not a mathematician. So my head hurts :-)
@charleyhoward4594
@charleyhoward4594 5 ай бұрын
same here - u got to admit he's amazing ...
@tomholroyd7519
@tomholroyd7519 6 ай бұрын
I love the generalization to sin_3, sin_n
@_skyslayer
@_skyslayer 6 ай бұрын
If you replace '1' with 'x' it's also possible
@MGoebel-c8e
@MGoebel-c8e 6 ай бұрын
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
@toriknorth3324 6 ай бұрын
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
@MGoebel-c8e 6 ай бұрын
@@toriknorth3324 Still challenging to wrap my mind around but very much appreciate your answer!
@mauricepanero
@mauricepanero 6 ай бұрын
Could you prove those differential rules geographically ?
@illumexhisoka6181
@illumexhisoka6181 6 ай бұрын
Can you find the integral from 0 to 1 of the nth root of (1-x^n) ??
@fguizini
@fguizini 6 ай бұрын
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?
@filippocamporeale3139
@filippocamporeale3139 6 ай бұрын
How did we know that sin_3 and cos_3 are differentiable when we built the differential equation?
@aniruddhvasishta8334
@aniruddhvasishta8334 6 ай бұрын
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.
@r75shell
@r75shell 6 ай бұрын
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
@bjornfeuerbacher5514 6 ай бұрын
There are lots of methods for solving a system of differential equations numerically, the simplest one being Euler's method.
@r75shell
@r75shell 5 ай бұрын
@@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.
@ej3281
@ej3281 6 ай бұрын
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
@bjornfeuerbacher5514 6 ай бұрын
If you correct the third powers to second powers and work onwards with these (correct) formulas, everything else is ok.
@CarloGinex
@CarloGinex 6 ай бұрын
How about the same thing but for the nth root of(1-x^n)
@xinpingdonohoe3978
@xinpingdonohoe3978 6 ай бұрын
It inspires us to define some cos_n and sin_n functions, sort of like the many polylogarithms.
@christopherravlin6051
@christopherravlin6051 6 ай бұрын
there should be NO cubes
@roeebaer
@roeebaer 5 ай бұрын
There's a mistake, see other comments
@SnoopyMath
@SnoopyMath 6 ай бұрын
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?
@xviruzz_platinum151
@xviruzz_platinum151 6 ай бұрын
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.
@tomholroyd7519
@tomholroyd7519 6 ай бұрын
very interesting connection to pythagoras
@ramonido
@ramonido 6 ай бұрын
Oh such a beautiful solution
@איתמרעינת
@איתמרעינת 6 ай бұрын
Shouldn't the integral be changed to jus sec_3(t) instead of cubed?
@Keithfert490
@Keithfert490 6 ай бұрын
No, the other two powers of sec_3(t) come from the change from dx to dt in the substitutio
@איתמרעינת
@איתמרעינת 6 ай бұрын
Thnx ​@@Keithfert490
@erfanmohagheghian707
@erfanmohagheghian707 6 ай бұрын
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.
@charleyhoward4594
@charleyhoward4594 5 ай бұрын
this bothered me immensely ; this is most Definitely NOT 'Integrating the impossible' - just changing the domain of func.
@lloydgush
@lloydgush 6 ай бұрын
Oh, squine and cosquine.
@goodplacetostop2973
@goodplacetostop2973 6 ай бұрын
21:53
@kylesendgikoski4231
@kylesendgikoski4231 6 ай бұрын
"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"
@hydropage2855
@hydropage2855 6 ай бұрын
You messed up the solution set to the initial values big time
@vincentbutton5926
@vincentbutton5926 6 ай бұрын
Math went wrong at 6m30. What's he calling this? Squirtle math, squirrel math...
@TomFarrell-p9z
@TomFarrell-p9z 6 ай бұрын
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.
@wesleydeng71
@wesleydeng71 6 ай бұрын
So these are fat sin() and cos.().
@stephenhamer8192
@stephenhamer8192 6 ай бұрын
Very interesting
@othila9902
@othila9902 6 ай бұрын
Assume 1=0 I hope this joke is clear enough lol
@Dionisi0
@Dionisi0 6 ай бұрын
your maths are wrong, your initial function is not bounded to anything
@nigellbutlerrr2638
@nigellbutlerrr2638 6 ай бұрын
He needs to learn what Integration is 😮😮
@nigellbutlerrr2638
@nigellbutlerrr2638 6 ай бұрын
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
@reeeeeplease1178 6 ай бұрын
L
@robertveith6383
@robertveith6383 6 ай бұрын
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.
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