Ok day, spent most of the day hanging out with friends.

Spent 2h writing some tampermonkey scripts to automate work for someone, once finished it'll save them a lot of time. Some useful js is

function $$(x) {
  return Array.from(document.querySelectorAll(x));
}
function $(x) {
  return document.querySelector(x);
}

Then you can do stuff like

$$("span")
  .find((s) => s.textContent === "Print")
  .click();

Which is nice. Also when automating stuff it's better to loop until elements are found (the way selenium does it) then to hook DOMContentLoaded, since DOMContentLoaded is meaningless for single page apps.

Spent an hour reading rudin, Managed to build some intuition for his inscrutable proof of the weierstrass approximation theorem.

Got home, spent 40m trying to fix stupid wifi card issue again (I hate computers). Spent an hour procrastinating by reading random blogs and looking at books on amazon.

Spent an hour writing cut the enemy. Should have been faster but I got sidetracked reading the book of five rings.

Spent another 30m doing nothing, spent an hour listening to a podcast about ai alignment while eating dinner.

Spent an hour on youtube and discord, was 9pm, went running because I diden't go earlier (sigma grindset amirite). It felt good to do something difficult after an unproductive day.

Spent 30m on discord, REALLY should have been journaling because it's late, then spent an hour journaling (mostly writing up the WAT theorem, agh I need to go to sleep earlier!)

WAT theorem

I should extract this to a post later, it can be followed by someone with basic calculus knowledge if I patiently explain each step

Rudin sets

With $c_n$ picked such that $\int_{-1}^{1} Q_n(x)dx = 1$. This seems wierd, but if you read the proof backwards you see the key step: Since $Q_n$ goes to zero when $x$ isn't close to the origin (see desmos) we can split the error integral into three parts, both $(-1,-\delta)$ and $(\delta,1)$ parts go to zero (because $Q_n$ goes to zero away from the origin) and the $(\delta, -\delta)$ part can be delt with using continuity.

Set the approximating polynomials $P_n$ as

(The $P_n$ can be shown to be polynomial by messing with the integral then differentiating, I think. not sure about this part)

Now because $\int_{-1}^1 Q_n(t)dt = 1$ and $f(x)$ is constant we have

Meaning we can combine integrals to write the difference as an integral, then split it up! (where $M = \sup |f(x)|$)

Here's where our choice of $Q_n$ pays off, for large enough $n$ both $2M$ integrals become less then $\epsilon/2$ meaning we have $|P_n(x)-f(x)| < \epsilon$ as desired.

How would you have came up with this? Well, if you notice that writing $P_n$ as an integral allows us to view the error in a nice form you might be able to guess $Q_n$ to make the proof work.

Writing the above took too long, it's 11pm now I should be in bed. I found some neat stuff to study tomorrow though

• https://nonagon.org/ExLibris/bernstein-proves-weierstrass
• https://en.m.wikipedia.org/wiki/Weierstrass_transform
• https://en.m.wikipedia.org/wiki/Bernstein_polynomial