Experimental Mathematics

Lo and behold, the next Problem of the Month has arrived! This one is much trickier than the last few! Have a gander:

Let \(\alpha\ \in \mathbf{R}^+\) and \(f \colon \mathbf{R} \rightarrow \mathbf{R}\) a function in differentiability class \(1\) defined on \([0, \alpha]\). Further, suppose: \(f'(x)>0\) for \(x \in [0, \alpha]\); \(f'(x)\) is strictly monotone; \(a \in [0, \alpha]\); \(b \in [0, \alpha]\); and \(f(0)=0\). If \[m=\min(f'(a), f'(f^{-1}(b)))\] and \[M=\max(f'(a), f'(f^{-1}(b)))\] then prove \[ \frac{m}{2}(a-f^{-1}(b))^2\leq\int_{0}^{a}f(x)\,\text{d}x + \int_{0}^{b}f^{-1}(b)\,\text{d}x - ab \leq\frac{M}{2}(a-f^{-1}(b))^2 \] with equality holding iff \(f(a)=b\).

What a doozy! But I'm confident I can solve it. Off to the races!

"But my dear boy, that means you've won!"

—Willy Wonka, Willy Wonka & the Chocolate Factory

Woohoo! My errant readings have borne fruit yet again. In mulling over this month's Problem of the Month, I kept getting this nagging feeling that I've seen a related result before, and I was correct! A little less than two months ago I read about the history of proof theory, a field the foundations of which L.E.J Brouwer greatly figured in (see [1, Introduction] for an excellent summary). One of his most outstanding results is his fixed-point theorem; I had come across it while learning about him and realized that it bore a marked resemblance to the problem at hand¹. Once I had made this connection, I found a proof of the relevant theorem, Banach's fixed-point theorem, and cleaned it up. And wouldn't you know it, I was the first person to submit a correct solution! You can see my submission here.

Another month, another problem! Below is this month's installment.

Let \((S, \text{d})\) be a complete metric space, \(\mu \in \mathbf{R}\) such that \(\mu \in [0, 1)\), and \(f \colon \mathbf{R} \rightarrow \mathbf{R}\) a function such that \[ \forall x, y \in s :\text{d}(f(x), f(y)) \leq \mu \, \text{d}(f(x), f(y)) \] Prove that there exists \(x \in S\) such that \(f(x) = x\).

Hmm... I think I have a key hint to this somewhere in my brain, but I'll have to go digging for it...

"The laws of nature are all conditional statements and they relate only to a very small part of our knowledge of the world."

—Eugene Wigner, The Unreasonable Effectiveness of Mathematics in the Natural Sciences

Spoilers for issue #8 of The Ultimates (2024), if you care about that. Anyways, I just finished reading it and, as expected, it was a wonderful issue! It was a fun detour from the main plot and gave Chavez a character beyond "amnesiac time traveller." I'm writing because an interesting new character made an appearance: Ultimate Nullifier.

He's a very small character in the grand scheme of things, but I'm always a sucker for superpowers based loosely on science. As his name suggests, he's a reimagining of the Ultimate Nullifier from Earth-616, a strange weapon from the planet Taa-II with immense power. In the (new) Ultimate universe, the Nullifier is a hyperintelligent future boy; with his N-Guns, he's able to "nullify anything [he] fully understand[s]." What a creative power! It reminds me of Sylar from Heroes (2006) and DC's Jesse Quick.

If one wanted to analyze his superpowers (and I do), there's a lot to be said about them! First off, what does it mean to "fully understand" something? In physics, as in most empirical sciences, you're ultimately dealing with mathematical models. But therein lies the treachery of science! At the end of the day, the models are simply that—models; they're not reality¹². And this is without making mention of the fact that our best theories of physics are incompatible and lack a complete axiomatization.

Even putting such a huge issue aside, there aren't concrete borders in the noosphere. Where does one subject end and another begin? For example, could you be considered as fully understanding cytomics if you don't also have a complete conception of proteomics, which in turn relies on quantum dynamics? It's an intractable rabbit hole!

Of course, this is all beating around the most important question: who, or what, is deciding this? Maybe the N-Guns have a psychic link with Nullifier and need just enough information to "lock on" to their target; maybe there's a divine entity in the future which arbitrates such decisions. I don't know! We're talking about make-believe science here.

Ultimately, there's not a real answer. But there doesn't need to be! When I speak of such inconsistencies, it's not out of a desire to tear down the material for not intricately detailing a one-off character's technobabble powers. Rather, I'm enjoying it so much that I want to contend with the ideas it's raising. These are fascinating concepts! This is simply my way of engaging with them.

Well, that's enough comic talk for 405 words. Email me! I'd love to hear any thoughts you have on this.

Here's a fun, little thinker: how many gifts are given out in the classic Christmas song, "The Twelve Days of Christmas"? My high school physics teacher once gave my class this problem for extra credit; we were only allotted 1 minute to solve it and, in my panic, I simply brute-forced it. It worked, of course, but it bothered me that I didn't know an elegant solution. Until now! Mull it over for a bit before I explain the result I arrived at¹.

This is simply the sum of successive triangular numbers! Recall that for positive integer \(n\), the \(n\)th triangular number, \(T_n\), is given by \(\frac{n(n+1)}{2}\). So we're trying to find a closed-form formula for \[\sum_{k = 1}^{n}\frac{k(k+1)}{2}\] Applying standard sum rules and the formula for the sum of the first \(n\) squares, \begin{align} \sum_{k = 1}^{n}\frac{k(k+1)}{2} &=\frac{1}{2}\left[\left(\sum_{k=1}^{n}k^2\right) + \left(\sum_{k=1}^{n}k\right)\right] \\ &=\frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right) \\ &= \frac{n(n+1)(n+2)}{6} \end{align} And voila! We have what we desired. Now we can just plug in \(n = 12\) and solve by hand, which yields \(364\). How ironic! A present for every day bar Christmas.

Thesis, antithesis, synthesis...

Ta-da! I was the first person to submit a correct solution, so I get a free gift card. Perhaps, however, the real prize was the math we did along the way¹. Here is a link to my submission. Note that you cannot simply use integration by substition and evaluate the limit of the resulting integral—the inner function does not have a continuous derivative on the interval.

My university's math department recently released its last Problem of the Month for the year¹; I have some time, so why not take a crack at it? In addition to getting to do some fun math, I'll also get a gift card if I submit the first (correct) solution. The problem is as follows:

For \(R \in \mathbf{R}^+\), let \(C_R\) be: \[C_{R}=\left\{(x, y) \in \mathbf{R}^2:x^2+y^2=R^2\right\}\] Prove the below. \[\lim\limits_{R \to \infty}\int_{C_R}f(x, y)\,\text{d}s = 0\]

I have some ideas, and will post my solution when the winner is announced. Don't go cheating, future reader, and just immediately look at my attempt—give it a shot!