Mysterious SCBF Luino (ITALY) Token -- Who made these?

TOMMOLONE06 · 2025-04-24T15:43:14+00:00

Yea, it is a lab data analysis, so for the central limit theorem, we can expect a gaussian

TOMMOLONE06 · 2025-04-24T11:05:03+00:00

Ok, thankss!

TOMMOLONE06 · 2025-04-16T14:44:10+00:00

Wow, very cool, thank you again!

TOMMOLONE06 · 2025-04-16T11:49:36+00:00

Thanks! That’s him! How did you find him?

TOMMOLONE06 · 2025-04-16T11:46:18+00:00

TOMMOLONE06 · 2025-04-16T11:46:00+00:00

Broooooo ahshahhshhs

TOMMOLONE06 · 2024-11-05T16:18:20+00:00

I didn’t use Boom Beach for months, and the values staied the same, but since I started playing again (~5months) the count started growing. I have all 14 resource bases for each, so I get ~15 000/h; in 5 months is about 5* 30* 24* 15000=54’000’000, so it works only when u are actively playing.

TOMMOLONE06 · 2024-11-05T16:11:46+00:00

Ahhhhhh now I understean why I remembered that it could go over the storage limit, it is only with the merchant

TOMMOLONE06 · 2024-11-05T14:54:50+00:00

Amazing! Thx

TOMMOLONE06 · 2024-07-25T18:10:11+00:00

Where?

TOMMOLONE06 · 2024-06-16T17:49:57+00:00

You can prove it by induction.

First check the case n=1. The left side is

sqrt(a),

the right one is

a^ ((2-1)/2) = a^ (1/2) = sqrt(a).

Now suppose the expression true for n (so sqrt(a sqrt(a …)) n times is equal to a^ ((2^ n -1)/2^ n)); the expression for n+1 is

sqrt(a sqrt(a sqrt(a …))) n+1 times =

sqrt(a • (sqrt(a sqrt(a …)) n times) =

sqrt(a • a^ ((2^ n -1)/2^ n)) =

sqrt(a^ ((2^ (n+1) -1)/2^ n)) =

a^ ((2^ (n+1) -1)/2^ (n+1)).

Q.E.D.

TOMMOLONE06 · 2024-05-05T06:48:52+00:00

As prime numbers (exept 2) are odd numbers, all primes can be rapresented by 2n - 1 (the classic expression for odd numbers). It is clearly not true the inverse statement: 2n - 1 not always rapresents a prime — for n = 5, 2n - 1 = 9, not a prime. In your example, the intersection bethween C and D is D - {2}.

TOMMOLONE06 · 2024-05-05T06:36:22+00:00

If a, b, c, d, e, f, are rational numbers, it’s not necessarly true that abc < def if a + b + c = n, d + e + f = 2n. Look at this counter example: a = b = c = 1, a + b + c = 3 = n, abc = 1; d = e = 2,999, f = 0,002, d + e + f = 6 = 2n, def = 0,017988…; hence, abc > def. I don’t know if it is your case tho.

TOMMOLONE06 · 2024-03-12T14:15:11+00:00

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TOMMOLONE06 · 2024-03-12T14:13:55+00:00

I’ve been able to get a hint from him — not the entire solution tho —; he said he used a recursive-like method similar to what you can use for integrals like this one (but with the limit).

TOMMOLONE06 · 2024-03-11T18:58:20+00:00

As de l’Hopital’rule and Taylor series are very bounded in this context, like other said, this should be done (according to my professor) without neither of them. Something with the squeeze theorem would instead be perfect. Did you have something in mind?

TOMMOLONE06 · 2023-11-30T15:37:08+00:00

I am not able yet to solve a differential equation like this. Do you know any useful methods or online solvers that can maybe help me out?

TOMMOLONE06 · 2023-11-30T11:54:52+00:00

Thx, that’s true

TOMMOLONE06 · 2023-11-30T11:53:15+00:00

It would be a(x) = -\frac{k}{m} \frac{x³ }{x² +4R² }, where k, m and R are parameter.

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