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February 24Edit

Material Drop ChanceEdit

Hi there,

I am currently attemping to "farm"(mass-collect) resources in in a video game. I am currently attemping to collect a material that has a 2.5% of dropping when an enemy is killed. The enemy in question takes around 3-5 minutes to kill. On average, how long would it take before I get the material that I need?

Thanks for your help, 31.127.153.3 (talk) 13:00, 24 February 2021 (UTC)

If an event has a probability of 2.5% to succeed, it means that it succeeds on the average 25 times out of 1000 tries, or once on every 40 tries. If the events are independent (the outcomes of prior tries have no influence on those of a new try), and the time spent on vanquishing a hostile combatant also does not impact the drop chance, this will then require 40 times the average time of a single fight (not counting time spent on other things between fights). Estimating that at 4 minutes as the halfway between the estimate extremes, we arrive at 40 times 4 = 160 minutes.  --Lambiam 14:05, 24 February 2021 (UTC)

Thank you for your help! 31.127.153.3 (talk) 14:35, 24 February 2021 (UTC)

February 25Edit

Nelson rulesEdit

Nelson rules present 8 rules descriptively. Included are illustrations which in general are fake. For example in rule 2, it is obvious that the mean is in the wrong place, and as soon as that is corrected then for the shown data, there won't be "nine or more" points on the same side of the mean.

I would like to redo the graphs with real data, to which end, in each case, I would like to find 20 integer values that meet the relevant criterion. I have no idea how to go about generating such sets of data other than running a program using rand() and waiting until I get a set that happens to meet the need.

Is there a better way of generating set of values? -- 09:30, 25 February 2021 (UTC)

The figures show sections from much longer time series. The mean (and the standard deviations) are those of the long time series and therefore are mostly based on data points that are not shown. I don't think the figures need or should be altered. --Wrongfilter (talk) 10:18, 25 February 2021 (UTC)
I agree that there is no need to alter the chart examples. Any example should eventually appear in a sufficiently long purely Gaussian random sequence, for which one should not use rand (). (You get a decent approximation of the standard normal distribution using rand()-rand()+rand()-rand()+rand()-rand()+rand()-rand()+rand()-rand()+rand()-rand(), that is, twelve calls, half of which count as negative.) Among the innumerable examples, you probably do not want one that is statistically very unlikely (such as having excursions beyond five sigma). But those that are statistically the most likely, will also be very atypical. For example, the most likely occurrences of nine points in a row on the same side have them infinitesimally displaced from the mean, and three points in a row more than two sigma from the mean will be exactly two sigma away. In the time it will take you to come up with devising a smart criterion for what you want, the dumbest program you can imagine will have produced zillions of examples.  --Lambiam 13:41, 25 February 2021 (UTC)
I'd add that the test is meant to be applied to data with a normal distribution, which is a continuous distribution so integer data would not work. Similar tests might be developed for non-normal distributions, but they would have different criteria. --RDBury (talk) 14:23, 25 February 2021 (UTC)
The formula I gave above assumed that 0 ≤ rand() < 1, like random.random() in Python, but I see that in C++ rand() returns a number in the range 0 to RAND_MAX, so to get variance 1 the results need to be divided by RAND_MAX+1.  --Lambiam 14:31, 25 February 2021 (UTC)
Another way to generate normal deviates is the Box–Muller transform. The rand() function does generate numbers between 0 and 1 in most languages; I guess C++, being built a bit closer to the hardware, lets you code that in yourself if that's what you want. --RDBury (talk) 14:44, 25 February 2021 (UTC)

February 26Edit

Prove that any positive integer, can be decomposed to at most norm-equal 2 elements in gauss integerEdit

Hello, I'm trying to prove that any positive integer, can be decomposed to at most norm-equal 2 elements in gauss integer. My thought, write down the number as integer product, and then decompose each prime. My problem that I'm unable to handle gaussian prime numbers such 3. --Exx8 (talk) 22:48, 26 February 2021 (UTC)

I'm confused. First, I assume you mean decompose as a product, in other words factor; you can also decompose things as a sum, so you need to specify. Also, every integer is also a Gaussian integer, so if X is an integer then X taken as the product of 1 factor is a decomposition of X. One is less than two so that would be a decomposition of X into at most two Gaussian integer factors. So did you mean exactly two factors? If so then it can't be done, there is no such factorization for 3. This is a consequence of unique factorization which holds in the Gaussian integers. --RDBury (talk) 02:34, 27 February 2021 (UTC)
Yes as a product. Decomposition to 1 number is not decomposition to 2 numbers.

I meant that if X=YZ, and YZ are equal in their norm, then there is no A,B such X=AB, |A|=|B|,A≠Y, B≠Z up to multiplication by unity.--Exx8 (talk) 06:52, 27 February 2021 (UTC)

I take the claim to be that if  , where the two factors of either product have equal norms, then the factorizations on either side of the equality can be transformed into each other by multiplying these factors by units. Now   fits the antecedent of the implication, yet there is no unit   such that  . The claim holds in the complex numbers:   has norm  . But it is not a Gaussian integer, so it is not a unit in the ring of Gaussian integers.  --Lambiam 08:31, 27 February 2021 (UTC)

February 27Edit

A proof that there is no q such: q²=k²+1 by gaussian integersEdit

I'm trying to prove that there is no a real integer q such: q²=k²+1 via gaussian integers. I know there is a simpler proof with just decomposing it to equation of 1. But does anyone see a gaussian integer related-proof? Thanks --Exx8 (talk) 22:50, 27 February 2021 (UTC)

The question is unclear. The equation   has a solution in   when   or   Do you mean, prove that the equation   has no general solution in   for all values of  ? Then it is sufficient to give just a single counterexample, like the unsolvability of   Or does the use of "real integer" in the question mean that we should understand the variables   and   to range over  , and is your question whether the (simple) proof that this has no non-trivial solutions in   can be made complicated by involving the Gaussian integers? I have no idea what "equation of 1" means, and what the "it" is that can be decomposed to this equation. If you want us to consider your questions, you should really put more effort in formulating them clearly.  --Lambiam 07:46, 28 February 2021 (UTC)

let  . prove that :   with gauss integer properties. meaning use gauss-integer properties and theorems.--Exx8 (talk) 08:10, 28 February 2021 (UTC)

The k=0 case is still an exception so you'd have to work around that. If you add k≠0 as an assumption then it's hard to see how Gaussian integers wouldn't be making a mountain out of a molehill. If q2-k2=1 then (q-k)(q+k)=1, q-k=q+k=±1, whence q=±1, k=0. --RDBury (talk) 10:15, 28 February 2021 (UTC)

Yes of course, it is for k>=1. I know the proof for with equations. Can you see how to solve with gaussian integers?--Exx8 (talk) 12:38, 28 February 2021 (UTC)

Expounding on RDB's proof, solutions in the integers are also solution in the Gaussian integers, so the absence of solutions in the latter ring implies absence of solutions in the integers. Starting from   we see that   and   are both units and each other's conjugates, which, next to the solution   given already, only leaves   The excursion through the complex plane did not lead over a formidable mountain, but is an unhelpful detour nevertheless.  --Lambiam 12:47, 28 February 2021 (UTC)
Is there any connection to that in the complex number, we have a decomposition into to linear polynomials? I wonder if it can be somehow related to the gaussian integers...--Exx8 (talk) 13:02, 28 February 2021 (UTC)
I don't see a connection. In the ring   the polynomial   can be factored as   while the same polynomial is irreducible in    --Lambiam 15:07, 28 February 2021 (UTC)
Well is there something unique with   is there a reason which says that we cannot decompose   somehow else?--Exx8 (talk) 16:29, 28 February 2021 (UTC)
Suppose   By equating the coefficients, we have that   and   so   and therefore   This means that   This is nothing special for this polynomial: in general, the factorization of a polynomial in   into a multiset of linear factors is unique up to multiplication by a scalar (polynomial of degree 0).  --Lambiam 18:23, 28 February 2021 (UTC)

March 2Edit

Notation questionEdit

All the following notation is gleaned from WP. Given a differentiable manifold M with tangent bundle TM and cotangent bundle TM, the set of sections of TM (also called vector fields) are denoted Γ(TM), and similarly of TM are Γ(TM). The exterior algebra on Γ(TM) is denoted Ω(M) (the differential forms on M). Is there a suitable equivalent notation for the exterior algebra of Γ(TM), i.e., the dual of Ω(M)? —Quondum 13:45, 2 March 2021 (UTC)

March 3Edit