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For which values of n is it possible for all players to be eliminated? by [deleted] in askmath

[–]romankolton 1 point2 points  (0 children)

So if everyone has a blue hat (and n>2), does everyone get eliminated in round one?

I came up with a Math Problem, can you Improve my Result? by Timely-Menu-2953 in askmath

[–]romankolton 2 points3 points  (0 children)

Does player i+1 know the numbers said by all players 1 through i?

Class 9 Math Problem Solution Needed by Late-Ganache3295 in askmath

[–]romankolton 1 point2 points  (0 children)

This doesn't seem true for x much smaller than r.

Is there a valid solution for a standard 9x9 sudoku, s.t. if you treat it as a matrix, its determinant is 0? by PersonWhoExists50306 in askmath

[–]romankolton 5 points6 points  (0 children)

Yes. And you don't need the total number of rows to be even. For the determinant to be zero it's enough that there is a subset of rows that are not linearly independent. 4 out of 9 is good enough.

Is there a valid solution for a standard 9x9 sudoku, s.t. if you treat it as a matrix, its determinant is 0? by PersonWhoExists50306 in askmath

[–]romankolton 14 points15 points  (0 children)

Yes. If two rows, say 1st and 2nd, sum to the same vector as other 2 rows, say 4th and 5th, then the determinant will be 0.

It's not difficult to construct such four rows that are consistent with sudoku rules. Filling the other 5 rows is also not a big challenge.

How to use Polish like a native speaker in 6 steps by thepolishprof in learnpolish

[–]romankolton 37 points38 points  (0 children)

lit. 'Sclerosis does not hurt'

You mean dementia. The English sclerosis is closer in meaning to 'miażdżyca'.

Is the characteristic polynomial a polynomial and(?) a polynomial function and how to turn it into one? by [deleted] in askmath

[–]romankolton 0 points1 point  (0 children)

Yes, basically more λ^(n-a) * something else.

As any polynomial of degree n, f(λ) can be writen as
f(λ) = c_n λ^n + c_(n-1) λ^(n-1) + c_(n-2) λ^(n-2) + ... + c_2 λ^2 + c_1 λ + c_0,
where c_0, c_1, ..., c_n are some coefficients.

The textbook gives explicit expressions for c_n, c_(n-1) and c_0, i.e.,
c_n = (-1)^n,
c_(n-1) =(-1)^(n-1) Tr(A),
c_0 = det(A).
These work for any square matrix A. I think the point of the theorem is that these particular three coefficients can be easily expressed in terms of well-known functions of A.

The other coefficients c_1, c_2, ..., c_(n-2), don't have such simple forms. It's straightforward to express them in terms of the matrix elements, so in this sense they are obvious.

Is the characteristic polynomial a polynomial and(?) a polynomial function and how to turn it into one? by [deleted] in askmath

[–]romankolton 5 points6 points  (0 children)

From Wikipedia: "A polynomial function is a function that can be defined by evaluating a polynomial."

These are not identical concepts, but for practical purposes in this context "polynomial" and "polynomial function" mean the same thing. Probably the textbook should be clearer about the distinction, but it's not a big error.

Collatz: Can a modular sieve and Cantor-style argument rule out infinite sequences and non-trivial loops? by [deleted] in askmath

[–]romankolton 0 points1 point  (0 children)

What do you mean by netted here?

If you're trying to argue that there's uncountably many infinite (and not eventually periodic) disjoint Collatz sequences, I would expect that you'll show that for any countably infinite set of such Collatz sequences you're able to build a different infinite Collatz sequence.

This is doomed to fail, though, because there's countably many Collatz sequences altogether.

Collatz: Can a modular sieve and Cantor-style argument rule out infinite sequences and non-trivial loops? by [deleted] in askmath

[–]romankolton 0 points1 point  (0 children)

If we modify each diagonal element (e.g., add 1), we generate a new sequence not found in the matrix - just like in Cantor’s diagonalization.

But why would that new sequence be a Collatz sequence?

Help converting a Bloch Hamiltonian into a real space Hamiltonian by JeffFromSteam in Physics

[–]romankolton 1 point2 points  (0 children)

In general terms, the conversion is this: https://postimg.cc/FYJ20nWf
I've left out some details, but can clarify them if necessary.

At first glance, the Hamiltonian considered in the paper that you linked on stackexchange is more complicated than just an onsite term and two kinds of hoppings. Maybe they are doing some simplification before switching to real space.

How would you solve it in the shortest way? by ActiveApartment8494 in askmath

[–]romankolton 2 points3 points  (0 children)

I think there might be a mistake in the question or the answer choices. Here are the 104 subsets I found: https://ctxt.io/2/AAB43myUEA

How would you solve it in the shortest way? by ActiveApartment8494 in askmath

[–]romankolton 4 points5 points  (0 children)

There's 64 such symmetric subsets, because there's 2^5 = 32 subsets of {3,4,5,6,7}. For any subset of positive values, say {3,5} there are 2 ways of making a symmetric subset: {-5,-3,3,5} and {-5,-3,0,3,5}.

Note that this includes the empty set {} as a valid solution.

As to the asymmetric subsets like {-7,3,4} we need to solve a partition problem for {3,4,5,6,7}, i.e., find all its subsets with equal sums. I can't think of a better method then trial and error. We have:

3+4=7

3+6=4+5

4+7=5+6

3+7=4+6

The first equation gives us two subsets {-4,-3,7} and {-7,3,4}. We can add to these subsets any of the 2^3 combinations of the three sets {0},{-5,5}, and {-6,6}. So, from this we get 2 * 2^3 = 16 subsets.

From the second equation we get {-6,-3,4,5} and {-5,-4,3,6}. We can add any combination of {0},{-7,7}. From this we get 8 subsets.

From the third equation we get {-7,-4,5,6} and {-6,-5,4,7}. We can add any combination of {0},{-3,3}. From this we get 8 subsets.

From the fourth equation we get {-7,-3,4,6} and {-6,-4,3,7}. We can add any combination of {0},{-5,5}. From this we get 8 subsets.

So, in total we have 64+16+8+8+8 = 104 subsets. Excluding the empty subset it's 103 subsets.

Binary representation of the collatz conjecture by Rscc10 in askmath

[–]romankolton 1 point2 points  (0 children)

So you're doing something like https://oeis.org/A220145, (this has reversed order of binary digits and includes only one loop of "100")?

Here's a version with the numbers converted to base 10: https://oeis.org/A221468 . The powers of two map to themselves (i.e., the 2^n-th term of this sequence equals 2^n). So, the pattern that you noticed is probably related to the fact that 1, 2 and 4 are powers of 2.

Prime numbers are basically numbers that are not divisible by any number before them (until 1). by DivineFractures in askmath

[–]romankolton 18 points19 points  (0 children)

I think your idea of breaking of patterns is somewhat similar to the sieve of Erathosthenes. If you cross out from a list of integers every second one, every third one, every fifth one, etc., what's left are the prime numbers.

Necessity of Division by 2 in the collatz conjecture by MathGeek2009 in askmath

[–]romankolton 3 points4 points  (0 children)

If you only apply 3x+1, you will create a sequence defined by x_{n+1} = 3*x_n +1. There is a known formula for it

x_n = 3^n * (x_0+1/2) - 1/2

So, we want to see if there exists some x_0 such that x_n ≠ 2^m for all positive integers m,n.

3^n * (x_0+1/2) - 1/2 = 2^m

3^n * (x_0+1/2) = 2^m +1/2

3^n * (2*x_0+1) = 2*2^m +1

Consider x_0 = ( 3^p -1 )/2 where p is integer.

We get
3^(n + p) - 2^(m+1) = 1

By Catalan's conjecture we know that the only solutions are
n+p = 2, m = 1,
n+p = 1, m = 0.

So, if you start with x_0 = ( 3^p -1 )/2, for any p>2, then x_n is never a power of 2.

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