uxor

ciredutempsEsme, 20 days ago to random French

Envie de travailler de le lowtech 😭

ciredutempsEsme, 22 days ago to random French

Look at this bug qui a fait sa déclaration à 9h du matin (bon a cause du chat et parce que j'ai procrastiné but still)
C'est très satisfaisant de déclarer ma cotisation syndicale cette année.
Et j'ai ete espantee par la case 7EB. C'est quoi ce truc qui n'est pas coché par défaut et genre pour les riches ?

uxor, 22 days ago to random French

Qui passe tranquillou devant la véranda ?🦌 Petite visite du jardin et du potager tant qu'à faire 😂

johncarlosbaez, 30 days ago (edited 30 days ago) to random

Hardcore math puzzle:

Suppose raindrops are falling on your head, randomly and independently, at an average rate of one per minute. What's the average of the 𝑐𝑢𝑏𝑒 of the number of raindrops that fall on your head in one minute?

The probability that (k) raindrops fall on your head in a minute is given by the Poisson distribution of mean 1, so it's
[ \frac{1}{ek!} ]
I could explain this but let's move on. The puzzle asks us to compute the expected value of (k^3) for this probability distribution, which is
[ \sum_{k=0}^\infty \frac{k^3}{ek!} ]
The heart of the puzzle is to figure out this sum. It turns out that
[ \sum_{k = 0}^\infty \frac{k^n}{k!} = B_n e ]
where (B_n) is the (n)th 'Bell number': the number of partitions of an (n)-element set into nonempty subsets. This is called 'Dobiński's formula'. I'll prove it in my next post. Now let's just use it!

We're interested in the case (n = 3). There are 5 partitions of a 3-element set
[ {{1,2,3}}, ]
[ {{1,2}, {3}}, ; {{2,3}, {1}}, ; {{3,1}, {2}}, ]
[ {{1}, {2}, {3}} ]
so (B_3 = 5).

So, the average of the cube of the number of raindrops that fall on your head in one minute is 𝟓.

Wild, huh? From probability theory to combinatorics.

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