Integrating the impossible 

Yes

Glad this comment exists, I was about to say that

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 🙂

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 RU-vids, but I do engineering problems and try and check each step, via example computed numerically, or other means)

@@user-gs6lp9ko1c 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 RU-vid algorithm likes.

*Threethagorean

*Pythreegorean

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

Dixon elliptic functions?

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

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.

some people think in LaTeX

Damn, that works so well too.

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.

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

Euler substitution for dummies 🙂

You forgot to add the 50% safety margin.

Ehm what the sigma

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.

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

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

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.

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

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

@@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.

"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"

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

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

Thnx ​@@Keithfert490

L

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.