The Universe of Discourse

Wed, 19 Jan 2022

Pranking the Italian Senate

The news today contains the story “Italian Senate Accidentally Plays 30 Seconds Of NSFW Tifa Lockhart Video” although I have not been able to find any source I would consider reliable. TheGamer reports:

The conference was hosted Monday by Nobel Prize winner Giorgio Parisi and featured several Italian senators. At some point during the Zoom call, a user … broke into the call and started broadcasting hentai videos.

Assuming this is accurate, it is disappointing on so many levels. Most obviously because if this was going to happen at all one would hope that it was an embarrassing mistake on the part of someone who was invited to the call, perhaps even the Nobel laureate, and not just some juvenile vandal who ran into the room with a sock on his dick.

If someone was going to go to the trouble of pulling this prank at all, why some run-of-the mill computer-generated video? Why not something really offensive? Or thematically appropriate, such as a scene from one of Cicciolina's films?

I think the guy who did this should feel ashamed of his squandered opportunity, and try a little harder next time. The world is watching!

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The squares are kinda Fibonacci-like

I got a cute little surprise today. I was thinking: suppose someone gives you a large square integer and asks you to find the next larger square. You can't really do any better than to extract the square root, add 1, and square the result. But if someone gives you two consecutive square numbers, you can find the next one with much less work. Say the two squares are !!b = n^2!! and !!a = n^2+2n+1!!, where !!n!! is unknown. Then you want to find !!n^2+4n+4!!, which is simply !!2a-b+2!!. No square rooting is required.

So the squares can be defined by the recurrence $$\begin{align} s_0 & = 0 \\ s_1 & = 1 \\ s_{n+1} & = 2s_n - s_{n-1} + 2\tag{$\ast$} \end{align} $$

This looks a great deal like the Fibonacci recurrence:

$$\begin{align} f_0 & = 0 \\ f_1 & = 1 \\ f_{n+1} & = f_n + f_{n-1} \end{align} $$

and I was a bit surprised because I thought all those Fibonacci-ish recurrences turned out to be approximately exponential. For example, !!f_n = O(\phi^n)!! where !!\phi=\frac12(1 + \sqrt 5)!!. And actually the !!f_0!! and !!f_1!! values don't matter, whatever you start with you get !!f_n = O(\phi^n)!!; the differences are small and are hidden in the Landau sign.

Similarly, if the recurrence is !!g_{n+1} = 2g_n + g_{n-1}!! you get !!g_n = O((1+\sqrt2)^n)!!, exponential again. So I was surprised that !!(\ast)!! produced squares instead of something exponential.

But as it turns out, it is producing something exponential. Sort of. Kind of. Not really.


There are a number of ways to explain the appearance of the !!\phi!! constant in the Fibonacci sequence. Feel free to replace this one with whatever you prefer: The Fibonacci recurrence can be written as $$\left[\matrix{1&1\\1&0}\right] \left[\matrix{f_n\\f_{n-1}}\right] = \left[\matrix{f_{n+1}\\f_n}\right] $$ so that $$\left[\matrix{1&1\\1&0}\right]^n \left[\matrix{1\\0}\right] = \left[\matrix{f_{n+1}\\f_n}\right] $$

and !!\phi!! appears because it is the positive eigenvalue of the square matrix !!\sm1,1,1,0!!. Similarly, !!1+\sqrt2!! is the positive eigenvalue of the matrix !!\sm 2,1,1,0!! that arises in connection with the !!g_n!! sequences that obey !!g_{n+1} = 2g_n + g_{n-1}!!.

For !!s_n!! the recurrence !!(\ast)!! is !!s_{n+1} = 2s_n - s_{n-1} + 2!!, Briefly disregarding the 2, we get the matrix form

$$\left[\matrix{2&-1\\1&0}\right]^n \left[\matrix{s_1\\s_0}\right] = \left[\matrix{s_{n+1}\\s_n}\right] $$

and the eigenvalues of !!\sm2,-1,1,0!! are exactly !!1!!. Where the Fibonacci sequence had !!f_n \approx k\cdot\phi^n!! we get instead !!s_n \approx k\cdot1^n!!, and instead of exploding, the exponential part remains well-behaved and the lower-order contributions remain significant.

If the two initial terms are !!t_0!! and !!t_1!!, then !!n!!th term of the sequence is simply !!t_0 + n(t_1-t_0)!!. That extra !!+2!! I temporarily disregarded in the previous paragraph is making all the interesting contributions: $$0, 0, 2, 6, 12, 20, \ldots, n(n-1) \ldots$$ and when you add the !!t_0 + n(t_1-t_0)!! and put !!t_0=0, t_1=1!! you get the squares.

So the squares can be considered a sort of Fibonacci-ish approximately exponential sequence, except that the exponential part doesn't matter because the base of the exponent is !!1!!.

How about that.

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