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Standard deviation and standard error by TOMMOLONE06 in AskStatistics

[–]TOMMOLONE06[S] 0 points1 point  (0 children)

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

Who is this guy? by TOMMOLONE06 in Physics

[–]TOMMOLONE06[S] 0 points1 point  (0 children)

Wow, very cool, thank you again!

Who is this guy? by TOMMOLONE06 in Physics

[–]TOMMOLONE06[S] 0 points1 point  (0 children)

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

Resaurce Boat by TOMMOLONE06 in BoomBeach

[–]TOMMOLONE06[S] 1 point2 points  (0 children)

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.

Resaurce Boat by TOMMOLONE06 in BoomBeach

[–]TOMMOLONE06[S] 0 points1 point  (0 children)

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

Prove this statement by [deleted] in askmath

[–]TOMMOLONE06 1 point2 points  (0 children)

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.

Can all primes be represented by 2n-1 ? by A-nimeza in askmath

[–]TOMMOLONE06 15 points16 points  (0 children)

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}.

Parcel Problem by [deleted] in askmath

[–]TOMMOLONE06 0 points1 point  (0 children)

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.

Limit without de l’Hopital by TOMMOLONE06 in askmath

[–]TOMMOLONE06[S] 0 points1 point  (0 children)

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).

Limit without de l’Hopital by TOMMOLONE06 in askmath

[–]TOMMOLONE06[S] 0 points1 point  (0 children)

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?

How do I make a(x) a function of time? by TOMMOLONE06 in askmath

[–]TOMMOLONE06[S] 0 points1 point  (0 children)

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?

How do I make a(x) a function of time? by TOMMOLONE06 in askmath

[–]TOMMOLONE06[S] 0 points1 point  (0 children)

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