so you want a VERY HARD math question?!

Try this one next: https://youtu.be/hztqL8104d0

Stop giving hard questions. You're making me feel like I'm bad at math.

Wow randomly scrolling through YouTube and seeing this guy used to be my math teacher a year ago glad to see you are successful on YouTube you are a great professor.

I absolutely love math when I'm not the one doing it.

This video gives us a very important lesson that many ML practioners overlook- Context is important in Math. When it comes to fields like Machine Learning, people sometimes blindly apply techniques without evaulating context. The part about 1 not being a valid solution encapsulates that perfectly.

I just started the integration chapter in calc 1 today, can't wait to understand most of the stuff on this channel!

As an Engineering student and having to deal with these problem solving, I am glad that I am already done with calculus with the help of this guy's math marathon. Thanks man.

Nothing more ominous then 36 boxes of expo markers sitting in the background.

I like functions like the Lambert W function. When there's an equation you can't solve, just invent a new function :D

I was just playing around with number guesses yesterday in a sleep deprived state, and happened to find that pi^(1/3) is a very close approximation to the value you get for x. Not sure what this means, but it's cool!

The different branches of Lambert W correspond to the complex branches of the logarithm. An easier way to identify the solution, and the fact that it's unique, is graphically: we are looking for where the function f(x) = x^2 - 1 - 3 ln x takes the value 0 in the range x>0. This function tends to infinity as x-> 0+ and as x-> infinity, and by elementary calculus has a unique local minimum at x=(3/2)^(1/2), where f is negative. Hence (by the intermediate value theorem) it must have a zero on either side of the minimum. On the left, there is the easy solution x=1, which can be ruled out because (as noted in the video) it doesn't solve the original problem, which requires ln x to be non-zero. So the required solution must be bigger than square root of 1.5, and also less than 2, since f(2)>0. To get a better approximation, set x=1+y and expand as a Taylor series about y=0: f = - y ÷ 5/2 y^2 - y^3 + O(y^4). (This is a convergent series for |y|<1.) We already ruled out the solution y=0 (x=1), but truncating the series at the 2nd term gives the approx y=2/5 (x=1.4), and at the 3rd term we solve a quadratic and pick the smaller root y=1/2 (x=1.5). A better value could be obtained by further truncation, interpolation between these two values, or doing a Newton-Raphson iteration.

Thank you for making these videos. I watch daily just to enjoy the beauty of integrals. Well explained and engaging!

Great question! Loved your introduction of Lambert W function here..

I'm starting to understand things from your videos, which (for me) is incredible because before i didn't understand at all. Keep up the good work sir 👍

What a fun little problem. It was nice to see the non-principle branch get utilized for a change!

If you plug y = (x^2 -1)/ln(x) and y = 3 into a graphing calculator and then use the intersect function you get 1.464. This video is still super cool because he got an exact answer using some math I didn't know existed, so keep up the great work!

Been learning cal one on my own watching this dudes vids is always an inspiration to keep on learning. Good luck with your maths chaps !!!

Fascinating video. Frankly, I did a naive geometrical interpretation and got 1.5. I pretty much went back to the definition of what an integral was as the "area under a curve" and realized if I just made the "curve" a straight line and "integrated" it's just the area of a box with a base of 2. Using X = (1.5)^(1/t) obviously just cancels out the t's and makes it a constant being integrated - thereby giving me 1.5t from 0 to 2 which gives 3. Apparently, that's a fairly close approximation, but I don't know if I just go lucky.

I was so close to solving it! I knew I had to use the product log function and I knew there would be some tricky branch stuff and I got it so close to the required form but I couldn't work out to raise both sides to the -2/3 will certainly keep it in mind for next time

Woow this was sick!! Missed these crazy videos a bit on this channel between all the Calc 1 uploads recently!! :) great vid!!