Learning habits

15 Sep 2025

You don’t learn math: you get used to it

This was something my calculus teacher used to say and I really agree on that: it’s more about a slow burn.

I spent a lot of time trying to redefine a studying routine. I never had one: I used to attend the lesson, listen to the teacher, do some exercise and, boom!, information acquired. Now, after something like a quarter of a century without any exam and without following any live lesson, finding a way to process and digest all the information is a complex job.

If depression is out of the equation (and more often than not it isn’t), making a tiny amount of exercise every day, as I always suggested to every single student I had, goes a long way.

That’s something where AI can get useful. I’m testing various AI asking them to teach me how combat works in various game systems I know: it’s a valid test, because it’s a niche knowledge that I’m able to judge.

And AI still fail here. Hard. Even attaching the rules of the game. And I can recognize the mistakes because I know the systems - if I was trying to learn something a bit advanced (and any subject in a math degree course is a bit advanced) I could not be able to find the problems and I’d end up learning crap.

But asking AI to craft like 50 easy exercises on a subject kinda works. I get the tiny work I need to do daily and any hallucination AI can produce is limited and easily detectable - it’s a second exercise, in a way, being on my toes to catch any weirdness.

For example, these were my exercises from yesterday

Transport equation

Equation:

$u_t + x u_x = t, \quad x \in \mathbb{R}, t > 0$

Data:

$u(x,0) = x^2$

Solution

Characteristics:

\[\frac{dx}{dt} = x \implies x(t) = x_0 e^{t}, \quad x_0 = x e^{-t}\] \[\Rightarrow u^0(x,t)=\mathcal F(x e^{-t})\]

Then

\[\frac{du}{dt} = t \implies u = x_0^2 + \frac{t^2}{2}\]

implying

\[u(x,t) = x^2 e^{-2t} + \frac{t^2}{2}\]

Wave equation

Equation:

$u_{tt} - 4 u_{xx} = 0, \quad x \in [0,1], t>0$

Data and border:

$u(x,0) = \sin(\pi x), \quad u_t(x,0) = x(1-x), \quad u(0,t) = u(1,t) = 0$

Solution

Formula:

\[u(x,t) = \sum_{n=1}^{\infty} \big[A_n \cos(\omega_n \pi ct) + B_n \sin(\omega_n \pi ct)\big] \sin(\omega_n \pi x)\]

leading to two steps, the first one is easy, $u(x,0)=0$ meaning that $\sin(\omega_n\pi c0)=0$, only then $n=1$ survives, leading to

\[A_1 = 1\]

while the second one needs

\[B_n=\frac1{\omega_nc}\int\limits_0^Lu_t(x,0)\sin(\omega_nx)\,dx= \frac1{n\pi c}\int\limits_0^Lx(1-x)\sin(\omega_nx)\,dx\]

using this we get the result

\[u(x,t) = \cos(c\pi t)\sin(\pi x) + \sum_{n\in\mathbb N} \left[\frac{(-1)^n-1}{n^3 \pi^3 c} \sin(n\pi ct)\sin(n\pi x)\right]\]

Slowly, this is fixing the various processes in my mind. It seems to be working.