Want to understand mathematics? This place is for you! Free videos + free ebooks = better education. Chris Tisdell is mathematician, educator, author and former DJ. He is a champion of free and open online learning resources.
It's an interesting question and something that I have thought about. One solution is to first trisect the acute angle (call it 3α) to form α. What you then want to do is trisect the larger angle (which is 180-3α). Note that the larger angle is 3*(60-α), so trisecting it will give 60-α. But, we already have α from our earlier work, so form an equilaterial triangle in the "right" place and there you have it! I will probably do a full video about this.
We can solve it by reducing of order using substitution y'=u Examples for series solution (1-x^2)y'' - 2xy'+n(n+1)y=0 , y(1) = 1 y'' - 2xy' + 2ny = 0 x^2y'' + xy' + (x^2 - n^2)y = 0
Wasn't the whole point to split a 2nd order equation into a system of two 1st order equations because 1st order equations are simpler to solve than 2nd order ones? Now you're kinda reversing this process, getting back to the 2nd order problem, making it harder. With 2nd order constant coefficients ODE it might seem to be "easy to solve", because all you need is solving quadratic equations. But for a 3rd order equation this might be harder to solve than three 1st order equations, and for 5th order you might get into trouble of solving a quintic :q And if the coefficients are not constant, you're pretty much dead in the waters for most of the interesting/practical cases.
There's a caveat in this general solution that is important to mention: When the parameter v is 0, the function term drops out, which means that there's a singular solution that is not captured by the general solution that you showed. And indeed, if you substitute v=0 in your general solution, you'll get: y = A·cos(0·arccos(x)) + B·sin(0·arccos(x)) = A·cos(0) + B·sin(0) = A·1 + B·0 = A which is just a one-dimensional space containing all constants. Being a one-dimensional space, it cannot contain all solutions to this second-order equation, of course. One example of such a solution is y=arcsin(x), which indeed is a valid solution for the v=0 case, but it cannot be expressed with your general solution. So how can we get this singular solution? Well, either by reduction of order, or by using the simplified version of the differential equation, y" + k²·y = 0, which in this case reduces to just y" = 0. Integrating it twice with respect to θ, we get: y = A + B·θ and from our substitution x = cosθ we know that θ=arccos(x) so the general solution for the singular case is: y = A + B·arccos(x) We can also use the well-known identity: arccos(x) = π/2 - arcsin(x) to get the other singular solution I mentioned earlier: y = A + B·(π/2 - arcsin(x)) = A + B·π/2 - B·arcsin(x) = (A + B·π/2) - B·arcsin(x) = C - B·arcsin(x)
I can't believe how many videos I am going through daily. I am sitting at the dining table, still listening, and sipping wine after dinner. oh, my eyes
Was sent to this video by a Pikuma course on 3D graphics. This video did help, but I felt like there were several places where a bunch of steps were skipped, and it took me a bit to figure out the missing pieces. Also, the low framerate made things hard to follow at times. Not incredibly intuitive overall. Also, the thing about assuming that lambda is 1 at the end to "get rid of it" was a bit confusing. I suppose the point is that lambda can be whatever we want, and when we assume it's the scalar 1 (it is a scalar and not a vector, right?), then it makes the math easy. I suppose I'm on board with that (assuming I'm understanding it correctly), but confusing.
I would love to see a series from this guy in which he explores the general solvability of equations in terms of, encoding the (generally unexplored) concepts of Liouville, Hadamard, Niels Hendrik Abel, Dirichlet, Du Hammel, Galois, Existence & theory of Algebraically Closed/Solvable Differential Equations - IN EXTREME DEPTH
Make videos about constructible numbers and other mathematical objects also, like, the Euclidean geometry is always awesome, but go tangent, parallel & orthogonal in all directions, including theory of Liouville Hadamard Duhamel Wronskian general Solvability theory and expressible/Algebraic /Galois Theory & the Likes I would love to see a proper series in the direction one day Overall - learned a great deal from you my friend
Chris you are one BAMF! such a groovy set. I was lucky enough to have you for MATH1131 a couple of years ago, thank you for being by far the most engaging lecturer I've had and for these epic tunes.
Nicely done. However, please finish the job. You found the angle that is 1/3 that of the original angle, but for completeness you should go ahead and apply that to finish trisecting the original arc.
